''' 【课程2.14】 数值计算和统计基础 常用数学、统计方法 '''
# 基本参数:axis、skipna import numpy as np import pandas as pd df = pd.DataFrame({'key1':[4,5,3,np.nan,2], 'key2':[1,2,np.nan,4,5], 'key3':[1,2,3,'j','k']}, index = ['a','b','c','d','e']) print(df) print(df['key1'].dtype,df['key2'].dtype,df['key3'].dtype) print('-----') m1 = df.mean() print(m1,type(m1)) print('单独统计一列:',df['key2'].mean()) print('-----') # np.nan :空值 # .mean()计算均值 # 只统计数字列 # 可以通过索引单独统计一列 m2 = df.mean(axis=1) print(m2) print('-----') # axis参数:默认为0,以列来计算,axis=1,以行来计算,这里就按照行来汇总了 m3 = df.mean(skipna=False) print(m3) print('-----') # skipna参数:是否忽略NaN,默认True,如False,有NaN的列统计结果仍未NaN
输出:
key1 key2 key3 a 4.0 1.0 1 b 5.0 2.0 2 c 3.0 NaN 3 d NaN 4.0 j e 2.0 5.0 k float64 float64 object ----- key1 3.5 key2 3.0 dtype: float64 <class 'pandas.core.series.Series'> 单独统计一列: 3.0 ----- a 2.5 b 3.5 c 3.0 d 4.0 e 3.5 dtype: float64 ----- key1 NaN key2 NaN dtype: float64 -----
# 主要数学计算方法,可用于Series和DataFrame(1) df = pd.DataFrame({'key1':np.arange(10), 'key2':np.random.rand(10)*10}) print(df) print('-----') print(df.count(),'→ count统计非Na值的数量 ') print(df.min(),'→ min统计最小值 ',df['key2'].max(),'→ max统计最大值 ') print(df.quantile(q=0.75),'→ quantile统计分位数,参数q确定位置 ') print(df.sum(),'→ sum求和 ') print(df.mean(),'→ mean求平均值 ') print(df.median(),'→ median求算数中位数,50%分位数 ') print(df.std(),' ',df.var(),'→ std,var分别求标准差,方差 ') print(df.skew(),'→ skew样本的偏度 ') print(df.kurt(),'→ kurt样本的峰度 ')
输出:
key1 key2 0 0 4.667989 1 1 4.336625 2 2 0.746852 3 3 9.670919 4 4 8.732045 5 5 0.013751 6 6 8.963752 7 7 0.279303 8 8 8.586821 9 9 8.899657 ----- key1 10 key2 10 dtype: int64 → count统计非Na值的数量 key1 0.000000 key2 0.013751 dtype: float64 → min统计最小值 9.67091932107 → max统计最大值 key1 6.750000 key2 8.857754 dtype: float64 → quantile统计分位数,参数q确定位置 key1 45.000000 key2 54.897714 dtype: float64 → sum求和 key1 4.500000 key2 5.489771 dtype: float64 → mean求平均值 key1 4.500000 key2 6.627405 dtype: float64 → median求算数中位数,50%分位数 key1 3.027650 key2 3.984945 dtype: float64 key1 9.166667 key2 15.879783 dtype: float64 → std,var分别求标准差,方差 key1 0.000000 key2 -0.430166 dtype: float64 → skew样本的偏度 key1 -1.200000 key2 -1.800296 dtype: float64 → kurt样本的峰度
# 主要数学计算方法,可用于Series和DataFrame(2) df['key1_s'] = df['key1'].cumsum() df['key2_s'] = df['key2'].cumsum() print(df,'→ cumsum样本的累计和 ') df['key1_p'] = df['key1'].cumprod() df['key2_p'] = df['key2'].cumprod() print(df,'→ cumprod样本的累计积 ') print(df.cummax(),' ',df.cummin(),'→ cummax,cummin分别求累计最大值,累计最小值 ') # 会填充key1,和key2的值
输出:
key1 key2 key1_s key2_s 0 0 4.667989 0 4.667989 1 1 4.336625 1 9.004614 2 2 0.746852 3 9.751466 3 3 9.670919 6 19.422386 4 4 8.732045 10 28.154431 5 5 0.013751 15 28.168182 6 6 8.963752 21 37.131934 7 7 0.279303 28 37.411236 8 8 8.586821 36 45.998057 9 9 8.899657 45 54.897714 → cumsum样本的累计和 key1 key2 key1_s key2_s key1_p key2_p 0 0 4.667989 0 4.667989 0 4.667989 1 1 4.336625 1 9.004614 0 20.243318 2 2 0.746852 3 9.751466 0 15.118767 3 3 9.670919 6 19.422386 0 146.212377 4 4 8.732045 10 28.154431 0 1276.733069 5 5 0.013751 15 28.168182 0 17.556729 6 6 8.963752 21 37.131934 0 157.374157 7 7 0.279303 28 37.411236 0 43.955024 8 8 8.586821 36 45.998057 0 377.433921 9 9 8.899657 45 54.897714 0 3359.032396 → cumprod样本的累计积 key1 key2 key1_s key2_s key1_p key2_p 0 0.0 4.667989 0.0 4.667989 0.0 4.667989 1 1.0 4.667989 1.0 9.004614 0.0 20.243318 2 2.0 4.667989 3.0 9.751466 0.0 20.243318 3 3.0 9.670919 6.0 19.422386 0.0 146.212377 4 4.0 9.670919 10.0 28.154431 0.0 1276.733069 5 5.0 9.670919 15.0 28.168182 0.0 1276.733069 6 6.0 9.670919 21.0 37.131934 0.0 1276.733069 7 7.0 9.670919 28.0 37.411236 0.0 1276.733069 8 8.0 9.670919 36.0 45.998057 0.0 1276.733069 9 9.0 9.670919 45.0 54.897714 0.0 3359.032396 key1 key2 key1_s key2_s key1_p key2_p 0 0.0 4.667989 0.0 4.667989 0.0 4.667989 1 0.0 4.336625 0.0 4.667989 0.0 4.667989 2 0.0 0.746852 0.0 4.667989 0.0 4.667989 3 0.0 0.746852 0.0 4.667989 0.0 4.667989 4 0.0 0.746852 0.0 4.667989 0.0 4.667989 5 0.0 0.013751 0.0 4.667989 0.0 4.667989 6 0.0 0.013751 0.0 4.667989 0.0 4.667989 7 0.0 0.013751 0.0 4.667989 0.0 4.667989 8 0.0 0.013751 0.0 4.667989 0.0 4.667989 9 0.0 0.013751 0.0 4.667989 0.0 4.667989 → cummax,cummin分别求累计最大值,累计最小值
# 唯一值:.unique() s = pd.Series(list('asdvasdcfgg')) sq = s.unique() print(s) print(sq,type(sq)) print(pd.Series(sq)) # 得到一个唯一值数组 # 通过pd.Series重新变成新的Series sq.sort() print(sq) # 重新排序
输出:
0 a 1 s 2 d 3 v 4 a 5 s 6 d 7 c 8 f 9 g 10 g dtype: object ['a' 's' 'd' 'v' 'c' 'f' 'g'] <class 'numpy.ndarray'> 0 a 1 s 2 d 3 v 4 c 5 f 6 g dtype: object ['a' 'c' 'd' 'f' 'g' 's' 'v']
# 值计数:.value_counts() sc = s.value_counts(sort = False) # 也可以这样写:pd.value_counts(sc, sort = False) print(sc) # 得到一个新的Series,计算出不同值出现的频率 # sort参数:排序,默认为True
输出:
s 2 d 2 v 1 c 1 a 2 g 2 f 1 dtype: int64
# 成员资格:.isin() s = pd.Series(np.arange(10,15)) df = pd.DataFrame({'key1':list('asdcbvasd'), 'key2':np.arange(4,13)}) print(s) print(df) print('-----') print(s.isin([5,14])) print(df.isin(['a','bc','10',8])) # 用[]表示 # 得到一个布尔值的Series或者Dataframe
输出:
0 10 1 11 2 12 3 13 4 14 dtype: int32 key1 key2 0 a 4 1 s 5 2 d 6 3 c 7 4 b 8 5 v 9 6 a 10 7 s 11 8 d 12 ----- 0 False 1 False 2 False 3 False 4 True dtype: bool key1 key2 0 True False 1 False False 2 False False 3 False False 4 False True 5 False False 6 True False 7 False False 8 False False
''' 【课程2.15】 文本数据 Pandas针对字符串配备的一套方法,使其易于对数组的每个元素进行操作 '''
# 通过str访问,且自动排除丢失/ NA值 s = pd.Series(['A','b','C','bbhello','123',np.nan,'hj']) df = pd.DataFrame({'key1':list('abcdef'), 'key2':['hee','fv','w','hija','123',np.nan]}) print(s) print(df) print('-----') print(s.str.count('b')) print(df['key2'].str.upper()) print('-----') # 直接通过.str调用字符串方法 # 可以对Series、Dataframe使用 # 自动过滤NaN值 df.columns = df.columns.str.upper() print(df) # df.columns是一个Index对象,也可使用.str
输出:
0 A 1 b 2 C 3 bbhello 4 123 5 NaN 6 hj dtype: object key1 key2 0 a hee 1 b fv 2 c w 3 d hija 4 e 123 5 f NaN ----- 0 0.0 1 1.0 2 0.0 3 2.0 4 0.0 5 NaN 6 0.0 dtype: float64 0 HEE 1 FV 2 W 3 HIJA 4 123 5 NaN Name: key2, dtype: object ----- KEY1 KEY2 0 a hee 1 b fv 2 c w 3 d hija 4 e 123 5 f NaN
# 字符串常用方法(1) - lower,upper,len,startswith,endswith s = pd.Series(['A','b','bbhello','123',np.nan]) print(s.str.lower(),'→ lower小写 ') print(s.str.upper(),'→ upper大写 ') print(s.str.len(),'→ len字符长度 ') print(s.str.startswith('b'),'→ 判断起始是否为a ') print(s.str.endswith('3'),'→ 判断结束是否为3 ')
输出:
0 a 1 b 2 bbhello 3 123 4 NaN dtype: object → lower小写 0 A 1 B 2 BBHELLO 3 123 4 NaN dtype: object → upper大写 0 1.0 1 1.0 2 7.0 3 3.0 4 NaN dtype: float64 → len字符长度 0 False 1 True 2 True 3 False 4 NaN dtype: object → 判断起始是否为a 0 False 1 False 2 False 3 True 4 NaN dtype: object → 判断结束是否为3
# 字符串常用方法(2) - strip s = pd.Series([' jack', 'jill ', ' jesse ', 'frank']) df = pd.DataFrame(np.random.randn(3, 2), columns=[' Column A ', ' Column B '], index=range(3)) print(s) print(df) print('-----') print(s.str.strip()) # 去除字符串中的空格 print(s.str.lstrip()) # 去除字符串中的左空格 print(s.str.rstrip()) # 去除字符串中的右空格 df.columns = df.columns.str.strip() print(df) # 这里去掉了columns的前后空格,但没有去掉中间空格
输出:
0 jack 1 jill 2 jesse 3 frank dtype: object Column A Column B 0 0.647766 0.094747 1 0.342940 -0.660643 2 1.183315 -0.143729 ----- 0 jack 1 jill 2 jesse 3 frank dtype: object 0 jack 1 jill 2 jesse 3 frank dtype: object 0 jack 1 jill 2 jesse 3 frank dtype: object Column A Column B 0 0.647766 0.094747 1 0.342940 -0.660643 2 1.183315 -0.143729
# 字符串常用方法(3) - replace df = pd.DataFrame(np.random.randn(3, 2), columns=[' Column A ', ' Column B '], index=range(3)) df.columns = df.columns.str.replace(' ','-') print(df) # 替换 df.columns = df.columns.str.replace('-','hehe',n=1) print(df) # n:替换个数
输出:
-Column-A- -Column-B- 0 1.855227 -0.519479 1 -0.400376 -0.421383 2 -0.293797 -0.432481 heheColumn-A- heheColumn-B- 0 1.855227 -0.519479 1 -0.400376 -0.421383 2 -0.293797 -0.432481
# 字符串常用方法(4) - split、rsplit s = pd.Series(['a,b,c','1,2,3',['a,,,c'],np.nan]) print(s.str.split(',')) print('-----') # 类似字符串的split print(s.str.split(',')[0]) print('-----') # 直接索引得到一个list print(s.str.split(',').str[0]) print(s.str.split(',').str.get(1)) print('-----') # 可以使用get或[]符号访问拆分列表中的元素 print(s.str.split(',', expand=True)) print(s.str.split(',', expand=True, n = 1)) print(s.str.rsplit(',', expand=True, n = 1)) print('-----') # 可以使用expand可以轻松扩展此操作以返回DataFrame # n参数限制分割数 # rsplit类似于split,反向工作,即从字符串的末尾到字符串的开头 df = pd.DataFrame({'key1':['a,b,c','1,2,3',[':,., ']], 'key2':['a-b-c','1-2-3',[':-.- ']]}) print(df['key2'].str.split('-')) # Dataframe使用split
输出:
0 [a, b, c] 1 [1, 2, 3] 2 NaN 3 NaN dtype: object ----- ['a', 'b', 'c'] ----- 0 a 1 1 2 NaN 3 NaN dtype: object 0 b 1 2 2 NaN 3 NaN dtype: object ----- 0 1 2 0 a b c 1 1 2 3 2 NaN None None 3 NaN None None 0 1 0 a b,c 1 1 2,3 2 NaN None 3 NaN None 0 1 0 a,b c 1 1,2 3 2 NaN None 3 NaN None ----- 0 [a, b, c] 1 [1, 2, 3] 2 NaN Name: key2, dtype: object
# 字符串索引 s = pd.Series(['A','b','C','bbhello','123',np.nan,'hj']) df = pd.DataFrame({'key1':list('abcdef'), 'key2':['hee','fv','w','hija','123',np.nan]}) print(s.str[0]) # 取第一个字符串 print(s.str[:2]) # 取前两个字符串 print(df['key2'].str[0]) # str之后和字符串本身索引方式相同
输出:
0 A 1 b 2 C 3 b 4 1 5 NaN 6 h dtype: object 0 A 1 b 2 C 3 bb 4 12 5 NaN 6 hj dtype: object 0 h 1 f 2 w 3 h 4 1 5 NaN Name: key2, dtype: object
''' 【课程2.16】 合并 merge、join Pandas具有全功能的,高性能内存中连接操作,与SQL等关系数据库非常相似 pd.merge(left, right, how='inner', on=None, left_on=None, right_on=None, left_index=False, right_index=False, sort=True, suffixes=('_x', '_y'), copy=True, indicator=False) '''
# merge合并 → 类似excel的vlookup df1 = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'], 'A': ['A0', 'A1', 'A2', 'A3'], 'B': ['B0', 'B1', 'B2', 'B3']}) df2 = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'], 'C': ['C0', 'C1', 'C2', 'C3'], 'D': ['D0', 'D1', 'D2', 'D3']}) df3 = pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'], 'key2': ['K0', 'K1', 'K0', 'K1'], 'A': ['A0', 'A1', 'A2', 'A3'], 'B': ['B0', 'B1', 'B2', 'B3']}) df4 = pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'], 'key2': ['K0', 'K0', 'K0', 'K0'], 'C': ['C0', 'C1', 'C2', 'C3'], 'D': ['D0', 'D1', 'D2', 'D3']}) print(pd.merge(df1, df2, on='key')) print('------') # left:第一个df # right:第二个df # on:参考键 print(pd.merge(df3, df4, on=['key1','key2'])) # 多个链接键
输出:
A B key C D 0 A0 B0 K0 C0 D0 1 A1 B1 K1 C1 D1 2 A2 B2 K2 C2 D2 3 A3 B3 K3 C3 D3 ------ A B key1 key2 C D 0 A0 B0 K0 K0 C0 D0 1 A2 B2 K1 K0 C1 D1 2 A2 B2 K1 K0 C2 D2
# 参数how → 合并方式 print(pd.merge(df3, df4,on=['key1','key2'], how = 'inner')) print('------') # inner:默认,取交集 print(pd.merge(df3, df4, on=['key1','key2'], how = 'outer')) print('------') # outer:取并集,数据缺失范围NaN print(pd.merge(df3, df4, on=['key1','key2'], how = 'left')) print('------') # left:按照df3为参考合并,数据缺失范围NaN print(pd.merge(df3, df4, on=['key1','key2'], how = 'right')) # right:按照df4为参考合并,数据缺失范围NaN
输出:
A B key1 key2 C D 0 A0 B0 K0 K0 C0 D0 1 A2 B2 K1 K0 C1 D1 2 A2 B2 K1 K0 C2 D2 ------ A B key1 key2 C D 0 A0 B0 K0 K0 C0 D0 1 A1 B1 K0 K1 NaN NaN 2 A2 B2 K1 K0 C1 D1 3 A2 B2 K1 K0 C2 D2 4 A3 B3 K2 K1 NaN NaN 5 NaN NaN K2 K0 C3 D3 ------ A B key1 key2 C D 0 A0 B0 K0 K0 C0 D0 1 A1 B1 K0 K1 NaN NaN 2 A2 B2 K1 K0 C1 D1 3 A2 B2 K1 K0 C2 D2 4 A3 B3 K2 K1 NaN NaN ------ A B key1 key2 C D 0 A0 B0 K0 K0 C0 D0 1 A2 B2 K1 K0 C1 D1 2 A2 B2 K1 K0 C2 D2 3 NaN NaN K2 K0 C3 D3
# 参数 left_on, right_on, left_index, right_index → 当键不为一个列时,可以单独设置左键与右键 df1 = pd.DataFrame({'lkey':list('bbacaab'), 'data1':range(7)}) df2 = pd.DataFrame({'rkey':list('abd'), 'date2':range(3)}) print(pd.merge(df1, df2, left_on='lkey', right_on='rkey')) print('------') # df1以‘lkey’为键,df2以‘rkey’为键 df1 = pd.DataFrame({'key':list('abcdfeg'), 'data1':range(7)}) df2 = pd.DataFrame({'date2':range(100,105)}, index = list('abcde')) print(pd.merge(df1, df2, left_on='key', right_index=True)) # df1以‘key’为键,df2以index为键 # left_index:为True时,第一个df以index为键,默认False # right_index:为True时,第二个df以index为键,默认False # 所以left_on, right_on, left_index, right_index可以相互组合: # left_on + right_on, left_on + right_index, left_index + right_on, left_index + right_index
输出:
data1 lkey date2 rkey 0 0 b 1 b 1 1 b 1 b 2 6 b 1 b 3 2 a 0 a 4 4 a 0 a 5 5 a 0 a ------ data1 key date2 0 0 a 100 1 1 b 101 2 2 c 102 3 3 d 103 5 5 e 104
# 参数 sort df1 = pd.DataFrame({'key':list('bbacaab'), 'data1':[1,3,2,4,5,9,7]}) df2 = pd.DataFrame({'key':list('abd'), 'date2':[11,2,33]}) x1 = pd.merge(df1,df2, on = 'key', how = 'outer') x2 = pd.merge(df1,df2, on = 'key', sort=True, how = 'outer') print(x1) print(x2) print('------') # sort:按照字典顺序通过 连接键 对结果DataFrame进行排序。默认为False,设置为False会大幅提高性能 print(x2.sort_values('data1')) # 也可直接用Dataframe的排序方法:sort_values,sort_index
输出:
data1 key date2 0 1.0 b 2.0 1 3.0 b 2.0 2 7.0 b 2.0 3 2.0 a 11.0 4 5.0 a 11.0 5 9.0 a 11.0 6 4.0 c NaN 7 NaN d 33.0 data1 key date2 0 2.0 a 11.0 1 5.0 a 11.0 2 9.0 a 11.0 3 1.0 b 2.0 4 3.0 b 2.0 5 7.0 b 2.0 6 4.0 c NaN 7 NaN d 33.0 ------ data1 key date2 3 1.0 b 2.0 0 2.0 a 11.0 4 3.0 b 2.0 6 4.0 c NaN 1 5.0 a 11.0 5 7.0 b 2.0 2 9.0 a 11.0 7 NaN d 33.0
# pd.join() → 直接通过索引链接 left = pd.DataFrame({'A': ['A0', 'A1', 'A2'], 'B': ['B0', 'B1', 'B2']}, index=['K0', 'K1', 'K2']) right = pd.DataFrame({'C': ['C0', 'C2', 'C3'], 'D': ['D0', 'D2', 'D3']}, index=['K0', 'K2', 'K3']) print(left) print(right) print(left.join(right)) print(left.join(right, how='outer')) print('-----') # 等价于:pd.merge(left, right, left_index=True, right_index=True, how='outer') df1 = pd.DataFrame({'key':list('bbacaab'), 'data1':[1,3,2,4,5,9,7]}) df2 = pd.DataFrame({'key':list('abd'), 'date2':[11,2,33]}) print(df1) print(df2) print(pd.merge(df1, df2, left_index=True, right_index=True, suffixes=('_1', '_2'))) print(df1.join(df2['date2'])) print('-----') # suffixes=('_x', '_y')默认 left = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'], 'B': ['B0', 'B1', 'B2', 'B3'], 'key': ['K0', 'K1', 'K0', 'K1']}) right = pd.DataFrame({'C': ['C0', 'C1'], 'D': ['D0', 'D1']}, index=['K0', 'K1']) print(left) print(right) print(left.join(right, on = 'key')) # 等价于pd.merge(left, right, left_on='key', right_index=True, how='left', sort=False); # left的‘key’和right的index
输出:
A B K0 A0 B0 K1 A1 B1 K2 A2 B2 C D K0 C0 D0 K2 C2 D2 K3 C3 D3 A B C D K0 A0 B0 C0 D0 K1 A1 B1 NaN NaN K2 A2 B2 C2 D2 A B C D K0 A0 B0 C0 D0 K1 A1 B1 NaN NaN K2 A2 B2 C2 D2 K3 NaN NaN C3 D3 ----- data1 key 0 1 b 1 3 b 2 2 a 3 4 c 4 5 a 5 9 a 6 7 b date2 key 0 11 a 1 2 b 2 33 d data1 key_1 date2 key_2 0 1 b 11 a 1 3 b 2 b 2 2 a 33 d data1 key date2 0 1 b 11.0 1 3 b 2.0 2 2 a 33.0 3 4 c NaN 4 5 a NaN 5 9 a NaN 6 7 b NaN ----- A B key 0 A0 B0 K0 1 A1 B1 K1 2 A2 B2 K0 3 A3 B3 K1 C D K0 C0 D0 K1 C1 D1 A B key C D 0 A0 B0 K0 C0 D0 1 A1 B1 K1 C1 D1 2 A2 B2 K0 C0 D0 3 A3 B3 K1 C1 D1
''' 【课程2.17】 连接与修补 concat、combine_first 连接 - 沿轴执行连接操作 pd.concat(objs, axis=0, join='outer', join_axes=None, ignore_index=False, keys=None, levels=None, names=None, verify_integrity=False, copy=True) '''
# 连接:concat s1 = pd.Series([1,2,3]) s2 = pd.Series([2,3,4]) s3 = pd.Series([1,2,3],index = ['a','c','h']) s4 = pd.Series([2,3,4],index = ['b','e','d']) print(pd.concat([s1,s2])) print(pd.concat([s3,s4]).sort_index()) print('-----') # 默认axis=0,行+行 print(pd.concat([s3,s4], axis=1)) print('-----') # axis=1,列+列,成为一个Dataframe
输出:
0 1 1 2 2 3 0 2 1 3 2 4 dtype: int64 a 1 b 2 c 2 d 4 e 3 h 3 dtype: int64 ----- 0 1 a 1.0 NaN b NaN 2.0 c 2.0 NaN d NaN 4.0 e NaN 3.0 h 3.0 NaN -----
# 连接方式:join,join_axes s5 = pd.Series([1,2,3],index = ['a','b','c']) s6 = pd.Series([2,3,4],index = ['b','c','d']) print(pd.concat([s5,s6], axis= 1)) print(pd.concat([s5,s6], axis= 1, join='inner')) print(pd.concat([s5,s6], axis= 1, join_axes=[['a','b','d']])) # join:{'inner','outer'},默认为“outer”。如何处理其他轴上的索引。outer为联合和inner为交集。 # join_axes:指定联合的index
输出:
0 1 a 1.0 NaN b 2.0 2.0 c 3.0 3.0 d NaN 4.0 0 1 b 2 2 c 3 3 0 1 a 1.0 NaN b 2.0 2.0 d NaN 4.0
# 覆盖列名 sre = pd.concat([s5,s6], keys = ['one','two']) print(sre,type(sre)) print(sre.index) print('-----') # keys:序列,默认值无。使用传递的键作为最外层构建层次索引 sre = pd.concat([s5,s6], axis=1, keys = ['one','two']) print(sre,type(sre)) # axis = 1, 覆盖列名
输出:
one a 1 b 2 c 3 two b 2 c 3 d 4 dtype: int64 <class 'pandas.core.series.Series'> MultiIndex(levels=[['one', 'two'], ['a', 'b', 'c', 'd']], labels=[[0, 0, 0, 1, 1, 1], [0, 1, 2, 1, 2, 3]]) ----- one two a 1.0 NaN b 2.0 2.0 c 3.0 3.0 d NaN 4.0 <class 'pandas.core.frame.DataFrame'>
# 修补 pd.combine_first() df1 = pd.DataFrame([[np.nan, 3., 5.], [-4.6, np.nan, np.nan],[np.nan, 7., np.nan]]) df2 = pd.DataFrame([[-42.6, np.nan, -8.2], [-5., 1.6, 4]],index=[1, 2]) print(df1) print(df2) print(df1.combine_first(df2)) print('-----') # 根据index,df1的空值被df2替代 # 如果df2的index多于df1,则更新到df1上,比如index=['a',1] df1.update(df2) print(df1) # update,直接df2覆盖df1,相同index位置
输出:
0 1 2 0 NaN 3.0 5.0 1 -4.6 NaN NaN 2 NaN 7.0 NaN 0 1 2 1 -42.6 NaN -8.2 2 -5.0 1.6 4.0 0 1 2 0 NaN 3.0 5.0 1 -4.6 NaN -8.2 2 -5.0 7.0 4.0 ----- 0 1 2 0 NaN 3.0 5.0 1 -42.6 NaN -8.2 2 -5.0 1.6 4.0
''' 【课程2.18】 去重及替换 .duplicated / .replace '''
# 去重 .duplicated s = pd.Series([1,1,1,1,2,2,2,3,4,5,5,5,5]) print(s.duplicated()) print(s[s.duplicated() == False]) print('-----') # 判断是否重复 # 通过布尔判断,得到不重复的值 s_re = s.drop_duplicates() print(s_re) print('-----') # drop.duplicates移除重复 # inplace参数:是否替换原值,默认False df = pd.DataFrame({'key1':['a','a',3,4,5], 'key2':['a','a','b','b','c']}) print(df.duplicated()) print(df['key2'].duplicated()) # Dataframe中使用duplicated
输出:
0 False 1 True 2 True 3 True 4 False 5 True 6 True 7 False 8 False 9 False 10 True 11 True 12 True dtype: bool 0 1 4 2 7 3 8 4 9 5 dtype: int64 ----- 0 1 4 2 7 3 8 4 9 5 dtype: int64 ----- 0 False 1 True 2 False 3 False 4 False dtype: bool 0 False 1 True 2 False 3 True 4 False Name: key2, dtype: bool
# 替换 .replace s = pd.Series(list('ascaazsd')) print(s.replace('a', np.nan)) print(s.replace(['a','s'] ,np.nan)) print(s.replace({'a':'hello world!','s':123})) # 可一次性替换一个值或多个值 # 可传入列表或字典
输出:
0 NaN 1 s 2 c 3 NaN 4 NaN 5 z 6 s 7 d dtype: object 0 NaN 1 NaN 2 c 3 NaN 4 NaN 5 z 6 NaN 7 d dtype: object 0 hello world! 1 123 2 c 3 hello world! 4 hello world! 5 z 6 123 7 d dtype: object
''' 【课程2.19】 数据分组 分组统计 - groupby功能 ① 根据某些条件将数据拆分成组 ② 对每个组独立应用函数 ③ 将结果合并到一个数据结构中 Dataframe在行(axis=0)或列(axis=1)上进行分组,将一个函数应用到各个分组并产生一个新值,然后函数执行结果被合并到最终的结果对象中。 df.groupby(by=None, axis=0, level=None, as_index=True, sort=True, group_keys=True, squeeze=False, **kwargs) '''
# 分组 df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar','foo', 'bar', 'foo', 'foo'], 'B' : ['one', 'one', 'two', 'three', 'two', 'two', 'one', 'three'], 'C' : np.random.randn(8), 'D' : np.random.randn(8)}) print(df) print('------') print(df.groupby('A'), type(df.groupby('A'))) print('------') # 直接分组得到一个groupby对象,是一个中间数据,没有进行计算 a = df.groupby('A').mean() b = df.groupby(['A','B']).mean() c = df.groupby(['A'])['D'].mean() # 以A分组,算D的平均值 print(a,type(a),' ',a.columns) print(b,type(b),' ',b.columns) print(c,type(c)) # 通过分组后的计算,得到一个新的dataframe # 默认axis = 0,以行来分组 # 可单个或多个([])列分组
输出:
A B C D 0 foo one 0.891765 -1.121156 1 bar one -1.272769 1.188977 2 foo two 0.198131 0.543673 3 bar three -0.827655 -1.608699 4 foo two -1.114089 -0.696145 5 bar two -0.345336 0.718507 6 foo one -0.207091 -0.922269 7 foo three -0.431760 -0.123696 ------ <pandas.core.groupby.DataFrameGroupBy object at 0x0000000004B65E10> <class 'pandas.core.groupby.DataFrameGroupBy'> ------ C D A bar -0.815253 0.099595 foo -0.132609 -0.463918 <class 'pandas.core.frame.DataFrame'> Index(['C', 'D'], dtype='object') C D A B bar one -1.272769 1.188977 three -0.827655 -1.608699 two -0.345336 0.718507 foo one 0.342337 -1.021713 three -0.431760 -0.123696 two -0.457979 -0.076236 <class 'pandas.core.frame.DataFrame'> Index(['C', 'D'], dtype='object') A bar 0.099595 foo -0.463918 Name: D, dtype: float64 <class 'pandas.core.series.Series'>
# 分组 - 可迭代对象 df = pd.DataFrame({'X' : ['A', 'B', 'A', 'B'], 'Y' : [1, 4, 3, 2]}) print(df) print(df.groupby('X'), type(df.groupby('X'))) print('-----') print(list(df.groupby('X')), '→ 可迭代对象,直接生成list ') print(list(df.groupby('X'))[0], '→ 以元祖形式显示 ') for n,g in df.groupby('X'): print(n) print(g) print('###') print('-----') # n是组名,g是分组后的Dataframe print(df.groupby(['X']).get_group('A'),' ') print(df.groupby(['X']).get_group('B'),' ') print('-----') # .get_group()提取分组后的组 grouped = df.groupby(['X']) print(grouped.groups) print(grouped.groups['A']) # 也可写:df.groupby('X').groups['A'] print('-----') # .groups:将分组后的groups转为dict # 可以字典索引方法来查看groups里的元素 sz = grouped.size() print(sz,type(sz)) print('-----') # .size():查看分组后的长度 df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar','foo', 'bar', 'foo', 'foo'], 'B' : ['one', 'one', 'two', 'three', 'two', 'two', 'one', 'three'], 'C' : np.random.randn(8), 'D' : np.random.randn(8)}) grouped = df.groupby(['A','B']).groups print(df) print(grouped) print(grouped[('foo', 'three')]) # 按照两个列进行分组
输出:
X Y 0 A 1 1 B 4 2 A 3 3 B 2 <pandas.core.groupby.DataFrameGroupBy object at 0x00000000091B6F28> <class 'pandas.core.groupby.DataFrameGroupBy'> ----- [('A', X Y 0 A 1 2 A 3), ('B', X Y 1 B 4 3 B 2)] → 可迭代对象,直接生成list ('A', X Y 0 A 1 2 A 3) → 以元祖形式显示 A X Y 0 A 1 2 A 3 ### B X Y 1 B 4 3 B 2 ### ----- X Y 0 A 1 2 A 3 X Y 1 B 4 3 B 2 ----- {'B': [1, 3], 'A': [0, 2]} [0, 2] ----- X A 2 B 2 dtype: int64 <class 'pandas.core.series.Series'> ----- A B C D 0 foo one -0.668695 0.247781 1 bar one -0.125374 2.259134 2 foo two -0.112052 1.618603 3 bar three -0.098986 0.150488 4 foo two 0.912286 -1.260029 5 bar two 1.096757 -0.571223 6 foo one -0.090907 -1.671482 7 foo three 0.088176 -0.292702 {('bar', 'two'): [5], ('foo', 'two'): [2, 4], ('bar', 'one'): [1], ('foo', 'three'): [7], ('bar', 'three'): [3], ('foo', 'one'): [0, 6]} [7]
# 其他轴上的分组 df = pd.DataFrame({'data1':np.random.rand(2), 'data2':np.random.rand(2), 'key1':['a','b'], 'key2':['one','two']}) print(df) print(df.dtypes) print('-----') for n,p in df.groupby(df.dtypes, axis=1): print(n) print(p) print('##') # 按照值类型分列
输出:
data1 data2 key1 key2 0 0.454580 0.692637 a one 1 0.496928 0.214309 b two data1 float64 data2 float64 key1 object key2 object dtype: object ----- float64 data1 data2 0 0.454580 0.692637 1 0.496928 0.214309 ## object key1 key2 0 a one 1 b two ##
# 通过字典或者Series分组 df = pd.DataFrame(np.arange(16).reshape(4,4), columns = ['a','b','c','d']) print(df) print('-----') mapping = {'a':'one','b':'one','c':'two','d':'two','e':'three'} by_column = df.groupby(mapping, axis = 1) print(by_column.sum()) print('-----') # mapping中,a、b列对应的为one,c、d列对应的为two,以字典来分组 s = pd.Series(mapping) print(s,' ') print(s.groupby(s).count()) # s中,index中a、b对应的为one,c、d对应的为two,以Series来分组
输出:
a b c d 0 0 1 2 3 1 4 5 6 7 2 8 9 10 11 3 12 13 14 15 ----- one two 0 1 5 1 9 13 2 17 21 3 25 29 ----- a one b one c two d two e three dtype: object one 2 three 1 two 2 dtype: int64
# 通过函数分组 df = pd.DataFrame(np.arange(16).reshape(4,4), columns = ['a','b','c','d'], index = ['abc','bcd','aa','b']) print(df,' ') print(df.groupby(len).sum()) # 按照字母长度分组
输出:
a b c d abc 0 1 2 3 bcd 4 5 6 7 aa 8 9 10 11 b 12 13 14 15 a b c d 1 12 13 14 15 2 8 9 10 11 3 4 6 8 10
# 分组计算函数方法 s = pd.Series([1, 2, 3, 10, 20, 30], index = [1, 2, 3, 1, 2, 3]) grouped = s.groupby(level=0) # 唯一索引用.groupby(level=0),将同一个index的分为一组 print(grouped) print(grouped.first(),'→ first:非NaN的第一个值 ') print(grouped.last(),'→ last:非NaN的最后一个值 ') print(grouped.sum(),'→ sum:非NaN的和 ') print(grouped.mean(),'→ mean:非NaN的平均值 ') print(grouped.median(),'→ median:非NaN的算术中位数 ') print(grouped.count(),'→ count:非NaN的值 ') print(grouped.min(),'→ min、max:非NaN的最小值、最大值 ') print(grouped.std(),'→ std,var:非NaN的标准差和方差 ') print(grouped.prod(),'→ prod:非NaN的积 ')
输出:
<pandas.core.groupby.SeriesGroupBy object at 0x00000000091992B0> 1 1 2 2 3 3 dtype: int64 → first:非NaN的第一个值 1 10 2 20 3 30 dtype: int64 → last:非NaN的最后一个值 1 11 2 22 3 33 dtype: int64 → sum:非NaN的和 1 5.5 2 11.0 3 16.5 dtype: float64 → mean:非NaN的平均值 1 5.5 2 11.0 3 16.5 dtype: float64 → median:非NaN的算术中位数 1 2 2 2 3 2 dtype: int64 → count:非NaN的值 1 1 2 2 3 3 dtype: int64 → min、max:非NaN的最小值、最大值 1 6.363961 2 12.727922 3 19.091883 dtype: float64 → std,var:非NaN的标准差和方差 1 10 2 40 3 90 dtype: int64 → prod:非NaN的积
# 多函数计算:agg() df = pd.DataFrame({'a':[1,1,2,2], 'b':np.random.rand(4), 'c':np.random.rand(4), 'd':np.random.rand(4),}) print(df) print(df.groupby('a').agg(['mean',np.sum])) print(df.groupby('a')['b'].agg({'result1':np.mean, 'result2':np.sum})) # 函数写法可以用str,或者np.方法 # 可以通过list,dict传入,当用dict时,key名为columns
输出:
a b c d 0 1 0.357911 0.318324 0.627797 1 1 0.964829 0.500017 0.570063 2 2 0.116608 0.194164 0.049509 3 2 0.933123 0.542615 0.718640 b c d mean sum mean sum mean sum a 1 0.661370 1.322739 0.409171 0.818341 0.598930 1.19786 2 0.524865 1.049730 0.368390 0.736780 0.384075 0.76815 result2 result1 a 1 1.322739 0.661370 2 1.049730 0.524865
''' 【课程2.20】 分组转换及一般性“拆分-应用-合并” transform / apply '''
# 数据分组转换,transform df = pd.DataFrame({'data1':np.random.rand(5), 'data2':np.random.rand(5), 'key1':list('aabba'), 'key2':['one','two','one','two','one']}) k_mean = df.groupby('key1').mean() print(df) print(k_mean) print(pd.merge(df,k_mean,left_on='key1',right_index=True).add_prefix('mean_')) # .add_prefix('mean_'):添加前缀 print('-----') # 通过分组、合并,得到一个包含均值的Dataframe print(df.groupby('key2').mean()) # 按照key2分组求均值 print(df.groupby('key2').transform(np.mean)) # data1、data2每个位置元素取对应分组列的均值 # 字符串不能进行计算
输出:
data1 data2 key1 key2 0 0.003727 0.390301 a one 1 0.744777 0.130300 a two 2 0.887207 0.679309 b one 3 0.448585 0.169208 b two 4 0.448045 0.993775 a one data1 data2 key1 a 0.398850 0.504792 b 0.667896 0.424258 mean_data1_x mean_data2_x mean_key1 mean_key2 mean_data1_y mean_data2_y 0 0.003727 0.390301 a one 0.398850 0.504792 1 0.744777 0.130300 a two 0.398850 0.504792 4 0.448045 0.993775 a one 0.398850 0.504792 2 0.887207 0.679309 b one 0.667896 0.424258 3 0.448585 0.169208 b two 0.667896 0.424258 ----- data1 data2 key2 one 0.446326 0.687795 two 0.596681 0.149754 data1 data2 0 0.446326 0.687795 1 0.596681 0.149754 2 0.446326 0.687795 3 0.596681 0.149754 4 0.446326 0.687795
# 一般化Groupby方法:apply df = pd.DataFrame({'data1':np.random.rand(5), 'data2':np.random.rand(5), 'key1':list('aabba'), 'key2':['one','two','one','two','one']}) print(df.groupby('key1').apply(lambda x: x.describe())) # apply直接运行其中的函数 # 这里为匿名函数,直接描述分组后的统计量 def f_df1(d,n): return(d.sort_index()[:n]) def f_df2(d,k1): return(d[k1]) print(df.groupby('key1').apply(f_df1,2),' ') print(df.groupby('key1').apply(f_df2,'data2')) print(type(df.groupby('key1').apply(f_df2,'data2'))) # f_df1函数:返回排序后的前n行数据 # f_df2函数:返回分组后表的k1列,结果为Series,层次化索引 # 直接运行f_df函数 # 参数直接写在后面,也可以为.apply(f_df,n = 2))
输出:
data1 data2 key1 a count 3.000000 3.000000 mean 0.561754 0.233470 std 0.313439 0.337209 min 0.325604 0.026906 25% 0.383953 0.038906 50% 0.442303 0.050906 75% 0.679829 0.336753 max 0.917355 0.622599 b count 2.000000 2.000000 mean 0.881906 0.547206 std 0.079357 0.254051 min 0.825791 0.367564 25% 0.853849 0.457385 50% 0.881906 0.547206 75% 0.909963 0.637026 max 0.938020 0.726847 data1 data2 key1 key2 key1 a 0 0.325604 0.050906 a one 1 0.917355 0.622599 a two b 2 0.825791 0.726847 b one 3 0.938020 0.367564 b two key1 a 0 0.050906 1 0.622599 4 0.026906 b 2 0.726847 3 0.367564 Name: data2, dtype: float64 <class 'pandas.core.series.Series'>
''' 【课程2.21】 透视表及交叉表 类似excel数据透视 - pivot table / crosstab '''
# 透视表:pivot_table # pd.pivot_table(data, values=None, index=None, columns=None, aggfunc='mean', fill_value=None, margins=False, dropna=True, margins_name='All') date = ['2017-5-1','2017-5-2','2017-5-3']*3 rng = pd.to_datetime(date) df = pd.DataFrame({'date':rng, 'key':list('abcdabcda'), 'values':np.random.rand(9)*10}) print(df) print('-----') print(pd.pivot_table(df, values = 'values', index = 'date', columns = 'key', aggfunc=np.sum)) # 也可以写 aggfunc='sum' print('-----') # data:DataFrame对象 # values:要聚合的列或列的列表 # index:数据透视表的index,从原数据的列中筛选 # columns:数据透视表的columns,从原数据的列中筛选 # aggfunc:用于聚合的函数,默认为numpy.mean,支持numpy计算方法 print(pd.pivot_table(df, values = 'values', index = ['date','key'], aggfunc=len)) print('-----') # 这里就分别以date、key共同做数据透视,值为values:统计不同(date,key)情况下values的平均值 # aggfunc=len(或者count):计数
输出:
date key values 0 2017-05-01 a 5.886424 1 2017-05-02 b 9.906472 2 2017-05-03 c 8.617297 3 2017-05-01 d 8.972318 4 2017-05-02 a 7.990905 5 2017-05-03 b 8.131856 6 2017-05-01 c 2.823731 7 2017-05-02 d 2.394605 8 2017-05-03 a 0.667917 ----- key a b c d date 2017-05-01 5.886424 NaN 2.823731 8.972318 2017-05-02 7.990905 9.906472 NaN 2.394605 2017-05-03 0.667917 8.131856 8.617297 NaN ----- date key 2017-05-01 a 1.0 c 1.0 d 1.0 2017-05-02 a 1.0 b 1.0 d 1.0 2017-05-03 a 1.0 b 1.0 c 1.0 Name: values, dtype: float64 -----
# 交叉表:crosstab # 默认情况下,crosstab计算因子的频率表,比如用于str的数据透视分析 # pd.crosstab(index, columns, values=None, rownames=None, colnames=None, aggfunc=None, margins=False, dropna=True, normalize=False) df = pd.DataFrame({'A': [1, 2, 2, 2, 2], 'B': [3, 3, 4, 4, 4], 'C': [1, 1, np.nan, 1, 1]}) print(df) print('-----') print(pd.crosstab(df['A'],df['B'])) print('-----') # 如果crosstab只接收两个Series,它将提供一个频率表。 # 用A的唯一值,统计B唯一值的出现次数 print(pd.crosstab(df['A'],df['B'],normalize=True)) print('-----') # normalize:默认False,将所有值除以值的总和进行归一化 → 为True时候显示百分比 print(pd.crosstab(df['A'],df['B'],values=df['C'],aggfunc=np.sum)) print('-----') # values:可选,根据因子聚合的值数组 # aggfunc:可选,如果未传递values数组,则计算频率表,如果传递数组,则按照指定计算 # 这里相当于以A和B界定分组,计算出每组中第三个系列C的值 print(pd.crosstab(df['A'],df['B'],values=df['C'],aggfunc=np.sum, margins=True)) print('-----') # margins:布尔值,默认值False,添加行/列边距(小计)
输出:
A B C 0 1 3 1.0 1 2 3 1.0 2 2 4 NaN 3 2 4 1.0 4 2 4 1.0 ----- B 3 4 A 1 1 0 2 1 3 ----- B 3 4 A 1 0.2 0.0 2 0.2 0.6 ----- B 3 4 A 1 1.0 NaN 2 1.0 2.0 ----- B 3 4 All A 1 1.0 NaN 1.0 2 1.0 2.0 3.0 All 2.0 2.0 4.0 -----
''' 【课程2.22】 数据读取 核心:read_table, read_csv, read_excel '''
# 读取普通分隔数据:read_table # 可以读取txt,csv import os os.chdir('C:/Users/Hjx/Desktop/') data1 = pd.read_table('data1.txt', delimiter=',',header = 0, index_col=1) print(data1) # delimiter:用于拆分的字符,也可以用sep:sep = ',' # header:用做列名的序号,默认为0(第一行) # index_col:指定某列为行索引,否则自动索引0, 1, ..... # read_table主要用于读取简单的数据,txt/csv
输出:
va1 va3 va4 va2 2 1 3 4 3 2 4 5 4 3 5 6 5 4 6 7
# 读取csv数据:read_csv # 先熟悉一下excel怎么导出csv data2 = pd.read_csv('data2.csv',engine = 'python') print(data2.head()) # engine:使用的分析引擎。可以选择C或者是python。C引擎快但是Python引擎功能更加完备。 # encoding:指定字符集类型,即编码,通常指定为'utf-8' # 大多数情况先将excel导出csv,再读取
输出:
省级政区代码 省级政区名称 地市级政区代码 地市级政区名称 年份 党委书记姓名 出生年份 出生月份 籍贯省份代码 籍贯省份名称 0 130000 河北省 130100 石家庄市 2000 陈来立 NaN NaN NaN NaN 1 130000 河北省 130100 石家庄市 2001 吴振华 NaN NaN NaN NaN 2 130000 河北省 130100 石家庄市 2002 吴振华 NaN NaN NaN NaN 3 130000 河北省 130100 石家庄市 2003 吴振华 NaN NaN NaN NaN 4 130000 河北省 130100 石家庄市 2004 吴振华 NaN NaN NaN NaN ... 民族 教育 是否是党校教育(是=1,否=0) 专业:人文 专业:社科 专业:理工 专业:农科 专业:医科 入党年份 工作年份 0 ... NaN 硕士 1.0 NaN NaN NaN NaN NaN NaN NaN 1 ... NaN 本科 0.0 0.0 0.0 1.0 0.0 0.0 NaN NaN 2 ... NaN 本科 0.0 0.0 0.0 1.0 0.0 0.0 NaN NaN 3 ... NaN 本科 0.0 0.0 0.0 1.0 0.0 0.0 NaN NaN 4 ... NaN 本科 0.0 0.0 0.0 1.0 0.0 0.0 NaN NaN [5 rows x 23 columns]
# 读取excel数据:read_excel data3 = pd.read_excel('地市级党委书记数据库(2000-10).xlsx',sheetname='中国人民共和国地市级党委书记数据库(2000-10)',header=0) print(data3) # io :文件路径。 # sheetname:返回多表使用sheetname=[0,1],若sheetname=None是返回全表 → ① int/string 返回的是dataframe ②而none和list返回的是dict # header:指定列名行,默认0,即取第一行 # index_col:指定列为索引列,也可以使用u”strings”
输出:
省级政区代码 省级政区名称 地市级政区代码 地市级政区名称 年份 党委书记姓名 出生年份 出生月份 籍贯省份代码 0 130000 河北省 130100 石家庄市 2000 陈来立 NaN NaN NaN 1 130000 河北省 130100 石家庄市 2001 吴振华 NaN NaN NaN 2 130000 河北省 130100 石家庄市 2002 吴振华 NaN NaN NaN 3 130000 河北省 130100 石家庄市 2003 吴振华 NaN NaN NaN 4 130000 河北省 130100 石家庄市 2004 吴振华 NaN NaN NaN 5 130000 河北省 130100 石家庄市 2005 吴振华 NaN NaN NaN 6 130000 河北省 130100 石家庄市 2006 吴振华 NaN NaN NaN 7 130000 河北省 130100 石家庄市 2007 吴显国 NaN NaN NaN 8 130000 河北省 130100 石家庄市 2008 吴显国 NaN NaN NaN 9 130000 河北省 130100 石家庄市 2009 车俊 NaN NaN NaN 10 130000 河北省 130100 石家庄市 2010 孙瑞彬 NaN NaN NaN 11 130000 河北省 130200 唐山市 2000 白润璋 NaN NaN NaN