pandas的拼接操作
#重点
pandas的拼接分为两种：
- 级联：pd.concat, pd.append
- 合并：pd.merge, pd.join
0. 回顾numpy的级联
import numpy as np
import pandas as pd
from pandas import Series,DataFrame
============================================
练习12：
- 生成2个3*3的矩阵，对其分别进行两个维度上的级联
============================================
nd1 =np.array([1,2,3])
nd2 =np.array([-1,-2,-3,-4])
np.concatenate([nd1,nd2])
array([ 1, 2, 3, -1, -2, -3, -4])
nd3 = np.array([[-1,-2,-3],[0,2,4]])
nd1 + nd3
array([[0, 0, 0],
[1, 4, 7]])
nd1.shape
(3,)
nd3.shape
(2, 3)
nd1 + nd2
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-10-cffcceec071c> in <module>()
----> 1 nd1 + nd2
ValueError: operands could not be broadcast together with shapes (3,) (4,)
为方便讲解，我们首先定义一个生成DataFrame的函数：
def make_df(cols,inds):
data = {c:[c+str(i) for i in inds] for c in cols}
return DataFrame(data,index = inds)
#当c = a c:a1 a2 a3
#当c =b c: b1 b2 b3
df1 = make_df(list("abc"),[1,2,3])
df1
#
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
df2 = make_df(list('abc'),[4,5,6])
df2
| a | b | c | |
|---|---|---|---|
| 4 | a4 | b4 | c4 |
| 5 | a5 | b5 | c5 |
| 6 | a6 | b6 | c6 |
1. 使用pd.concat()级联
pandas使用pd.concat函数，与np.concatenate函数类似，只是多了一些参数：
pd.concat(objs, axis=0, join='outer', join_axes=None, ignore_index=False,
keys=None, levels=None, names=None, verify_integrity=False,
copy=True)
1) 简单级联
和np.concatenate一样，优先增加行数（默认axis=0）
pd.concat([df1,df2])
#在级联的时候，一定要注意他的轴！！！
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
| 4 | a4 | b4 | c4 |
| 5 | a5 | b5 | c5 |
| 6 | a6 | b6 | c6 |
df1
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
df3 =make_df(list("def"),[1,2,3])
df3
| d | e | f | |
|---|---|---|---|
| 1 | d1 | e1 | f1 |
| 2 | d2 | e2 | f2 |
| 3 | d3 | e3 | f3 |
df1 + df3
| a | b | c | d | e | f | |
|---|---|---|---|---|---|---|
| 1 | NaN | NaN | NaN | NaN | NaN | NaN |
| 2 | NaN | NaN | NaN | NaN | NaN | NaN |
| 3 | NaN | NaN | NaN | NaN | NaN | NaN |
pd.concat([df1, df3], axis = 1)
| a | b | c | d | e | f | |
|---|---|---|---|---|---|---|
| 1 | a1 | b1 | c1 | d1 | e1 | f1 |
| 2 | a2 | b2 | c2 | d2 | e2 | f2 |
| 3 | a3 | b3 | c3 | d3 | e3 | f3 |
pd.concat([df1,df2],axis = 1)
| a | b | c | a | b | c | |
|---|---|---|---|---|---|---|
| 1 | a1 | b1 | c1 | NaN | NaN | NaN |
| 2 | a2 | b2 | c2 | NaN | NaN | NaN |
| 3 | a3 | b3 | c3 | NaN | NaN | NaN |
| 4 | NaN | NaN | NaN | a4 | b4 | c4 |
| 5 | NaN | NaN | NaN | a5 | b5 | c5 |
| 6 | NaN | NaN | NaN | a6 | b6 | c6 |
可以通过设置axis来改变级联方向
注意index在级联时可以重复
df1
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
df4 = make_df(list('abc'),[2,3,4])
df4
| a | b | c | |
|---|---|---|---|
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
| 4 | a4 | b4 | c4 |
pd.concat([df1,df4])
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
| 4 | a4 | b4 | c4 |
也可以选择忽略ignore_index，重新索引
pd.concat([df1,df4],ignore_index=True)
| a | b | c | |
|---|---|---|---|
| 0 | a1 | b1 | c1 |
| 1 | a2 | b2 | c2 |
| 2 | a3 | b3 | c3 |
| 3 | a2 | b2 | c2 |
| 4 | a3 | b3 | c3 |
| 5 | a4 | b4 | c4 |
或者使用多层索引 keys
concat([x,y],keys=['x','y'])
pd.concat([df1,df4],keys = ["三班","四班"])
| a | b | c | ||
|---|---|---|---|---|
| 三班 | 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 | |
| 3 | a3 | b3 | c3 | |
| 四班 | 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 | |
| 4 | a4 | b4 | c4 |
============================================
练习13：
-
想一想级联的应用场景？
-
使用昨天的知识，建立一个期中考试张三、李四的成绩表ddd
-
假设新增考试学科"计算机"，如何实现？
-
新增王老五同学的成绩，如何实现？
============================================
2) 不匹配级联
不匹配指的是级联的维度的索引不一致。例如纵向级联时列索引不一致，横向级联时行索引不一致
df1
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
df5 = make_df(list("abcd"),[3,4,5,6])
df5
| a | b | c | d | |
|---|---|---|---|---|
| 3 | a3 | b3 | c3 | d3 |
| 4 | a4 | b4 | c4 | d4 |
| 5 | a5 | b5 | c5 | d5 |
| 6 | a6 | b6 | c6 | d6 |
pd.concat([df1,df5])
| a | b | c | d | |
|---|---|---|---|---|
| 1 | a1 | b1 | c1 | NaN |
| 2 | a2 | b2 | c2 | NaN |
| 3 | a3 | b3 | c3 | NaN |
| 3 | a3 | b3 | c3 | d3 |
| 4 | a4 | b4 | c4 | d4 |
| 5 | a5 | b5 | c5 | d5 |
| 6 | a6 | b6 | c6 | d6 |
有3种连接方式：
- 外连接：补NaN（默认模式）
#上面的这种情况 默认的这种情况！！！！
#join='outer'
- 内连接：只连接匹配的项
pd.concat([df1,df5],join = "inner")
#只匹配你能够匹配上去的项
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
| 3 | a3 | b3 | c3 |
| 4 | a4 | b4 | c4 |
| 5 | a5 | b5 | c5 |
| 6 | a6 | b6 | c6 |
- 连接指定轴 join_axes
df6 = make_df(list("abcz"), [3,4,7,8])
df6
| a | b | c | z | |
|---|---|---|---|---|
| 3 | a3 | b3 | c3 | z3 |
| 4 | a4 | b4 | c4 | z4 |
| 7 | a7 | b7 | c7 | z7 |
| 8 | a8 | b8 | c8 | z8 |
type(df6.columns)
pandas.core.indexes.base.Index
df6.columns
Index(['a', 'b', 'c', 'z'], dtype='object')
pd.concat([df6,df5,df2,df1], join_axes=[df6.columns])
#axis 轴 axes 轴面
#join_axes list of Index objects
| a | b | c | z | |
|---|---|---|---|---|
| 3 | a3 | b3 | c3 | z3 |
| 4 | a4 | b4 | c4 | z4 |
| 7 | a7 | b7 | c7 | z7 |
| 8 | a8 | b8 | c8 | z8 |
| 3 | a3 | b3 | c3 | NaN |
| 4 | a4 | b4 | c4 | NaN |
| 5 | a5 | b5 | c5 | NaN |
| 6 | a6 | b6 | c6 | NaN |
| 4 | a4 | b4 | c4 | NaN |
| 5 | a5 | b5 | c5 | NaN |
| 6 | a6 | b6 | c6 | NaN |
| 1 | a1 | b1 | c1 | NaN |
| 2 | a2 | b2 | c2 | NaN |
| 3 | a3 | b3 | c3 | NaN |
============================================
练习14：
假设【期末】考试ddd2的成绩没有张三的，只有李四、王老五、赵小六的，使用多种方法级联
============================================
3) 使用append()函数添加
由于在后面级联的使用非常普遍，因此有一个函数append专门用于在后面添加
s1 = ["123"]
s1.append('456')
s1
['123', '456']
#append和concat非常类似
df1.append(df2)
| a | b | c | |
|---|---|---|---|
| 1 | a1 | b1 | c1 |
| 2 | a2 | b2 | c2 |
| 3 | a3 | b3 | c3 |
| 4 | a4 | b4 | c4 |
| 5 | a5 | b5 | c5 |
| 6 | a6 | b6 | c6 |
df5
| a | b | c | d | |
|---|---|---|---|---|
| 3 | a3 | b3 | c3 | d3 |
| 4 | a4 | b4 | c4 | d4 |
| 5 | a5 | b5 | c5 | d5 |
| 6 | a6 | b6 | c6 | d6 |
df5.append(df1)
| a | b | c | d | |
|---|---|---|---|---|
| 3 | a3 | b3 | c3 | d3 |
| 4 | a4 | b4 | c4 | d4 |
| 5 | a5 | b5 | c5 | d5 |
| 6 | a6 | b6 | c6 | d6 |
| 1 | a1 | b1 | c1 | NaN |
| 2 | a2 | b2 | c2 | NaN |
| 3 | a3 | b3 | c3 | NaN |
============================================
练习15：
新建一个只有张三李四王老五的期末考试成绩单ddd3，使用append()与期中考试成绩表ddd级联
============================================
2. 使用pd.merge()合并
#重点
#必须是两个DataFrame有相同属性的时候才能进行merge
merge与concat的区别在于，merge需要依据某一共同的行或列来进行合并
使用pd.merge()合并时，会自动根据两者相同column名称的那一列，作为key来进行合并。
注意每一列元素的顺序不要求一致
1) 一对一合并
df1 = DataFrame({"age":[30,22,36],"work":['tech',"accounting","sell"],"sex":["男","女","女"]}, index = list("abc"))
df1
| age | sex | work | |
|---|---|---|---|
| a | 30 | 男 | tech |
| b | 22 | 女 | accounting |
| c | 36 | 女 | sell |
df2 = DataFrame({"home":["上海","安徽","山东"],"work":['tech',"accounting","sell"],"weight":[60,50,45]},
index = list("abc"))
df2
| home | weight | work | |
|---|---|---|---|
| a | 上海 | 60 | tech |
| b | 安徽 | 50 | accounting |
| c | 山东 | 45 | sell |
pd.concat([df1,df2],axis = 1)
| age | sex | work | home | weight | work | |
|---|---|---|---|---|---|---|
| a | 30 | 男 | tech | 上海 | 60 | tech |
| b | 22 | 女 | accounting | 安徽 | 50 | accounting |
| c | 36 | 女 | sell | 山东 | 45 | sell |
df1.merge(df2)
| age | sex | work | home | weight | |
|---|---|---|---|---|---|
| 0 | 30 | 男 | tech | 上海 | 60 |
| 1 | 22 | 女 | accounting | 安徽 | 50 |
| 2 | 36 | 女 | sell | 山东 | 45 |
2) 多对一合并
df1
| age | sex | work | |
|---|---|---|---|
| a | 30 | 男 | tech |
| b | 22 | 女 | accounting |
| c | 36 | 女 | sell |
df3 = DataFrame({"home":["深圳","北京","上海","安徽","山东"],
"work":["tech","tech","tech","accounting","sell"],
"weight":[60,75,80,54,63]},index = list("abcde"))
df3
| home | weight | work | |
|---|---|---|---|
| a | 深圳 | 60 | tech |
| b | 北京 | 75 | tech |
| c | 上海 | 80 | tech |
| d | 安徽 | 54 | accounting |
| e | 山东 | 63 | sell |
df1.merge(df3)
| age | sex | work | home | weight | |
|---|---|---|---|---|---|
| 0 | 30 | 男 | tech | 深圳 | 60 |
| 1 | 30 | 男 | tech | 北京 | 75 |
| 2 | 30 | 男 | tech | 上海 | 80 |
| 3 | 22 | 女 | accounting | 安徽 | 54 |
| 4 | 36 | 女 | sell | 山东 | 63 |
3) 多对多合并
df5 = DataFrame({"age":[28,30,22,36], "work":['tech',"tech","accounting","sell"],"sex":["女","男","女","女"]}, index = list("abce"))
df5
| age | sex | work | |
|---|---|---|---|
| a | 28 | 女 | tech |
| b | 30 | 男 | tech |
| c | 22 | 女 | accounting |
| e | 36 | 女 | sell |
df3
| home | weight | work | |
|---|---|---|---|
| a | 深圳 | 60 | tech |
| b | 北京 | 75 | tech |
| c | 上海 | 80 | tech |
| d | 安徽 | 54 | accounting |
| e | 山东 | 63 | sell |
df3.merge(df5)
| home | weight | work | age | sex | |
|---|---|---|---|---|---|
| 0 | 深圳 | 60 | tech | 28 | 女 |
| 1 | 深圳 | 60 | tech | 30 | 男 |
| 2 | 北京 | 75 | tech | 28 | 女 |
| 3 | 北京 | 75 | tech | 30 | 男 |
| 4 | 上海 | 80 | tech | 28 | 女 |
| 5 | 上海 | 80 | tech | 30 | 男 |
| 6 | 安徽 | 54 | accounting | 22 | 女 |
| 7 | 山东 | 63 | sell | 36 | 女 |
4) key的规范化
- 使用on=显式指定哪一列为key,当有多个key相同时使用
df5
| age | sex | work | |
|---|---|---|---|
| a | 28 | 女 | tech |
| b | 30 | 男 | tech |
| c | 22 | 女 | accounting |
| e | 36 | 女 | sell |
df6 = DataFrame({"age":[30,27,36],"work":["tech","leader","sell"],"hoppy":["sixdog","diaofish","playcat"]}, index = list("abc"))
df6
| age | hoppy | work | |
|---|---|---|---|
| a | 30 | sixdog | tech |
| b | 27 | diaofish | leader |
| c | 36 | playcat | sell |
df5.merge(df6, on = "age", suffixes=["_总部","_分部"])
| age | sex | work_总部 | hoppy | work_分部 | |
|---|---|---|---|---|---|
| 0 | 30 | 男 | tech | sixdog | tech |
| 1 | 36 | 女 | sell | playcat | sell |
df5.merge(df6,on = "work")
| age_x | sex | work | age_y | hoppy | |
|---|---|---|---|---|---|
| 0 | 28 | 女 | tech | 30 | sixdog |
| 1 | 30 | 男 | tech | 30 | sixdog |
| 2 | 36 | 女 | sell | 36 | playcat |
- 使用left_on和right_on指定左右两边的列作为key，当左右两边的key都不想等时使用
df5
| age | sex | work | |
|---|---|---|---|
| a | 28 | 女 | tech |
| b | 30 | 男 | tech |
| c | 22 | 女 | accounting |
| e | 36 | 女 | sell |
df7 = DataFrame({"年龄":[30,22,36],"工作":["tech","accounting","sell"],"性别":["男","女","女"]},index = list("abc"))
df7
| 工作 | 年龄 | 性别 | |
|---|---|---|---|
| a | tech | 30 | 男 |
| b | accounting | 22 | 女 |
| c | sell | 36 | 女 |
df5.merge(df7,left_on = "work", right_on = "工作")
| age | sex | work | 工作 | 年龄 | 性别 | |
|---|---|---|---|---|---|---|
| 0 | 28 | 女 | tech | tech | 30 | 男 |
| 1 | 30 | 男 | tech | tech | 30 | 男 |
| 2 | 22 | 女 | accounting | accounting | 22 | 女 |
| 3 | 36 | 女 | sell | sell | 36 | 女 |
df5
| age | sex | work | |
|---|---|---|---|
| a | 28 | 女 | tech |
| b | 30 | 男 | tech |
| c | 22 | 女 | accounting |
| e | 36 | 女 | sell |
s = df5[["age"]]*1000
s.columns = ["salary"]
s
#可以对列的名字进行修改
| salary | |
|---|---|
| a | 28000 |
| b | 30000 |
| c | 22000 |
| e | 36000 |
df5.merge(s, left_index = True,right_index=True)
| age | sex | work | salary | |
|---|---|---|---|---|
| a | 28 | 女 | tech | 28000 |
| b | 30 | 男 | tech | 30000 |
| c | 22 | 女 | accounting | 22000 |
| e | 36 | 女 | sell | 36000 |
pd.concat([df5,s],axis = 1)
| age | sex | work | salary | |
|---|---|---|---|---|
| a | 28 | 女 | tech | 28000 |
| b | 30 | 男 | tech | 30000 |
| c | 22 | 女 | accounting | 22000 |
| e | 36 | 女 | sell | 36000 |
============================================
练习16：
-
假设有两份成绩单，除了ddd是张三李四王老五之外，还有ddd4是张三和赵小六的成绩单，如何合并？
-
如果ddd4中张三的名字被打错了，成为了张十三，怎么办？
-
自行练习多对一，多对多的情况
============================================
5) 内合并与外合并
- 内合并：只保留两者都有的key（默认模式）
df3
| home | weight | work | |
|---|---|---|---|
| a | 深圳 | 60 | tech |
| b | 北京 | 75 | tech |
| c | 上海 | 80 | tech |
| d | 安徽 | 54 | accounting |
| e | 山东 | 63 | sell |
df5
| age | sex | work | |
|---|---|---|---|
| a | 28 | 女 | tech |
| b | 30 | 男 | tech |
| c | 22 | 女 | accounting |
| e | 36 | 女 | sell |
df6
| age | hoppy | work | |
|---|---|---|---|
| a | 30 | sixdog | tech |
| b | 27 | diaofish | leader |
| c | 36 | playcat | sell |
df3
| home | weight | work | |
|---|---|---|---|
| a | 深圳 | 60 | tech |
| b | 北京 | 75 | tech |
| c | 上海 | 80 | tech |
| d | 安徽 | 54 | accounting |
| e | 山东 | 63 | sell |
df3.merge(df6)
| home | weight | work | age | hoppy | |
|---|---|---|---|---|---|
| 0 | 深圳 | 60 | tech | 30 | sixdog |
| 1 | 北京 | 75 | tech | 30 | sixdog |
| 2 | 上海 | 80 | tech | 30 | sixdog |
| 3 | 山东 | 63 | sell | 36 | playcat |
- 外合并 how='outer'：补NaN
df3.merge(df6,how = "outer")
| home | weight | work | age | hoppy | |
|---|---|---|---|---|---|
| 0 | 深圳 | 60.0 | tech | 30.0 | sixdog |
| 1 | 北京 | 75.0 | tech | 30.0 | sixdog |
| 2 | 上海 | 80.0 | tech | 30.0 | sixdog |
| 3 | 安徽 | 54.0 | accounting | NaN | NaN |
| 4 | 山东 | 63.0 | sell | 36.0 | playcat |
| 5 | NaN | NaN | leader | 27.0 | diaofish |
- 左合并、右合并：how='left'，how='right'，
df3
| home | weight | work | |
|---|---|---|---|
| a | 深圳 | 60 | tech |
| b | 北京 | 75 | tech |
| c | 上海 | 80 | tech |
| d | 安徽 | 54 | accounting |
| e | 山东 | 63 | sell |
df6
| age | hoppy | work | |
|---|---|---|---|
| a | 30 | sixdog | tech |
| b | 27 | diaofish | leader |
| c | 36 | playcat | sell |
df3.merge(df6, how = "left")
| home | weight | work | age | hoppy | |
|---|---|---|---|---|---|
| 0 | 深圳 | 60 | tech | 30.0 | sixdog |
| 1 | 北京 | 75 | tech | 30.0 | sixdog |
| 2 | 上海 | 80 | tech | 30.0 | sixdog |
| 3 | 安徽 | 54 | accounting | NaN | NaN |
| 4 | 山东 | 63 | sell | 36.0 | playcat |
df3.merge(df6, how = "right")
| home | weight | work | age | hoppy | |
|---|---|---|---|---|---|
| 0 | 深圳 | 60.0 | tech | 30 | sixdog |
| 1 | 北京 | 75.0 | tech | 30 | sixdog |
| 2 | 上海 | 80.0 | tech | 30 | sixdog |
| 3 | 山东 | 63.0 | sell | 36 | playcat |
| 4 | NaN | NaN | leader | 27 | diaofish |
============================================
练习17：
-
如果只有张三赵小六语数英三个科目的成绩，如何合并？
-
考虑应用情景，使用多种方式合并ddd与ddd4
============================================
6) 列冲突的解决
当列冲突时，即有多个列名称相同时，需要使用on=来指定哪一个列作为key，配合suffixes指定冲突列名
可以使用suffixes=自己指定后缀
============================================
练习18：
假设有两个同学都叫李四，ddd5、ddd6都是张三和李四的成绩表，如何合并？
============================================
作业
3. 案例分析：美国各州人口数据分析
首先导入文件，并查看数据样本
pop = pd.read_csv("./state-population.csv")
pop.head(20)
| state/region | ages | year | population | |
|---|---|---|---|---|
| 0 | AL | under18 | 2012 | 1117489.0 |
| 1 | AL | total | 2012 | 4817528.0 |
| 2 | AL | under18 | 2010 | 1130966.0 |
| 3 | AL | total | 2010 | 4785570.0 |
| 4 | AL | under18 | 2011 | 1125763.0 |
| 5 | AL | total | 2011 | 4801627.0 |
| 6 | AL | total | 2009 | 4757938.0 |
| 7 | AL | under18 | 2009 | 1134192.0 |
| 8 | AL | under18 | 2013 | 1111481.0 |
| 9 | AL | total | 2013 | 4833722.0 |
| 10 | AL | total | 2007 | 4672840.0 |
| 11 | AL | under18 | 2007 | 1132296.0 |
| 12 | AL | total | 2008 | 4718206.0 |
| 13 | AL | under18 | 2008 | 1134927.0 |
| 14 | AL | total | 2005 | 4569805.0 |
| 15 | AL | under18 | 2005 | 1117229.0 |
| 16 | AL | total | 2006 | 4628981.0 |
| 17 | AL | under18 | 2006 | 1126798.0 |
| 18 | AL | total | 2004 | 4530729.0 |
| 19 | AL | under18 | 2004 | 1113662.0 |
pop.shape
(2544, 4)
areas = pd.read_csv("./state-areas.csv")
areas
| state | area (sq. mi) | |
|---|---|---|
| 0 | Alabama | 52423 |
| 1 | Alaska | 656425 |
| 2 | Arizona | 114006 |
| 3 | Arkansas | 53182 |
| 4 | California | 163707 |
| 5 | Colorado | 104100 |
| 6 | Connecticut | 5544 |
| 7 | Delaware | 1954 |
| 8 | Florida | 65758 |
| 9 | Georgia | 59441 |
| 10 | Hawaii | 10932 |
| 11 | Idaho | 83574 |
| 12 | Illinois | 57918 |
| 13 | Indiana | 36420 |
| 14 | Iowa | 56276 |
| 15 | Kansas | 82282 |
| 16 | Kentucky | 40411 |
| 17 | Louisiana | 51843 |
| 18 | Maine | 35387 |
| 19 | Maryland | 12407 |
| 20 | Massachusetts | 10555 |
| 21 | Michigan | 96810 |
| 22 | Minnesota | 86943 |
| 23 | Mississippi | 48434 |
| 24 | Missouri | 69709 |
| 25 | Montana | 147046 |
| 26 | Nebraska | 77358 |
| 27 | Nevada | 110567 |
| 28 | New Hampshire | 9351 |
| 29 | New Jersey | 8722 |
| 30 | New Mexico | 121593 |
| 31 | New York | 54475 |
| 32 | North Carolina | 53821 |
| 33 | North Dakota | 70704 |
| 34 | Ohio | 44828 |
| 35 | Oklahoma | 69903 |
| 36 | Oregon | 98386 |
| 37 | Pennsylvania | 46058 |
| 38 | Rhode Island | 1545 |
| 39 | South Carolina | 32007 |
| 40 | South Dakota | 77121 |
| 41 | Tennessee | 42146 |
| 42 | Texas | 268601 |
| 43 | Utah | 84904 |
| 44 | Vermont | 9615 |
| 45 | Virginia | 42769 |
| 46 | Washington | 71303 |
| 47 | West Virginia | 24231 |
| 48 | Wisconsin | 65503 |
| 49 | Wyoming | 97818 |
| 50 | District of Columbia | 68 |
| 51 | Puerto Rico | 3515 |
areas.shape
(52, 2)
abbr = pd.read_csv("./state-abbrevs.csv")
abbr.head()
| state | abbreviation | |
|---|---|---|
| 0 | Alabama | AL |
| 1 | Alaska | AK |
| 2 | Arizona | AZ |
| 3 | Arkansas | AR |
| 4 | California | CA |
abbr.shape
(51, 2)
合并pop与abbrevs两个DataFrame，分别依据state/region列和abbreviation列来合并。
为了保留所有信息，使用外合并。
#pop :2544行的数据 abbr 51的条数据
pop2 = pop.merge(abbr,left_on = "state/region", right_on = "abbreviation", how = "left")
pop2.head()
| state/region | ages | year | population | state | abbreviation | |
|---|---|---|---|---|---|---|
| 0 | AL | under18 | 2012 | 1117489.0 | Alabama | AL |
| 1 | AL | total | 2012 | 4817528.0 | Alabama | AL |
| 2 | AL | under18 | 2010 | 1130966.0 | Alabama | AL |
| 3 | AL | total | 2010 | 4785570.0 | Alabama | AL |
| 4 | AL | under18 | 2011 | 1125763.0 | Alabama | AL |
去除abbreviation的那一列（axis=1）
pop2.drop("abbreviation", axis = 1,inplace=True)
pop2
| state/region | ages | year | population | state | |
|---|---|---|---|---|---|
| 0 | AL | under18 | 2012 | 1117489.0 | Alabama |
| 1 | AL | total | 2012 | 4817528.0 | Alabama |
| 2 | AL | under18 | 2010 | 1130966.0 | Alabama |
| 3 | AL | total | 2010 | 4785570.0 | Alabama |
| 4 | AL | under18 | 2011 | 1125763.0 | Alabama |
| 5 | AL | total | 2011 | 4801627.0 | Alabama |
| 6 | AL | total | 2009 | 4757938.0 | Alabama |
| 7 | AL | under18 | 2009 | 1134192.0 | Alabama |
| 8 | AL | under18 | 2013 | 1111481.0 | Alabama |
| 9 | AL | total | 2013 | 4833722.0 | Alabama |
| 10 | AL | total | 2007 | 4672840.0 | Alabama |
| 11 | AL | under18 | 2007 | 1132296.0 | Alabama |
| 12 | AL | total | 2008 | 4718206.0 | Alabama |
| 13 | AL | under18 | 2008 | 1134927.0 | Alabama |
| 14 | AL | total | 2005 | 4569805.0 | Alabama |
| 15 | AL | under18 | 2005 | 1117229.0 | Alabama |
| 16 | AL | total | 2006 | 4628981.0 | Alabama |
| 17 | AL | under18 | 2006 | 1126798.0 | Alabama |
| 18 | AL | total | 2004 | 4530729.0 | Alabama |
| 19 | AL | under18 | 2004 | 1113662.0 | Alabama |
| 20 | AL | total | 2003 | 4503491.0 | Alabama |
| 21 | AL | under18 | 2003 | 1113083.0 | Alabama |
| 22 | AL | total | 2001 | 4467634.0 | Alabama |
| 23 | AL | under18 | 2001 | 1120409.0 | Alabama |
| 24 | AL | total | 2002 | 4480089.0 | Alabama |
| 25 | AL | under18 | 2002 | 1116590.0 | Alabama |
| 26 | AL | under18 | 1999 | 1121287.0 | Alabama |
| 27 | AL | total | 1999 | 4430141.0 | Alabama |
| 28 | AL | total | 2000 | 4452173.0 | Alabama |
| 29 | AL | under18 | 2000 | 1122273.0 | Alabama |
| ... | ... | ... | ... | ... | ... |
| 2514 | USA | under18 | 1999 | 71946051.0 | NaN |
| 2515 | USA | total | 2000 | 282162411.0 | NaN |
| 2516 | USA | under18 | 2000 | 72376189.0 | NaN |
| 2517 | USA | total | 1999 | 279040181.0 | NaN |
| 2518 | USA | total | 2001 | 284968955.0 | NaN |
| 2519 | USA | under18 | 2001 | 72671175.0 | NaN |
| 2520 | USA | total | 2002 | 287625193.0 | NaN |
| 2521 | USA | under18 | 2002 | 72936457.0 | NaN |
| 2522 | USA | total | 2003 | 290107933.0 | NaN |
| 2523 | USA | under18 | 2003 | 73100758.0 | NaN |
| 2524 | USA | total | 2004 | 292805298.0 | NaN |
| 2525 | USA | under18 | 2004 | 73297735.0 | NaN |
| 2526 | USA | total | 2005 | 295516599.0 | NaN |
| 2527 | USA | under18 | 2005 | 73523669.0 | NaN |
| 2528 | USA | total | 2006 | 298379912.0 | NaN |
| 2529 | USA | under18 | 2006 | 73757714.0 | NaN |
| 2530 | USA | total | 2007 | 301231207.0 | NaN |
| 2531 | USA | under18 | 2007 | 74019405.0 | NaN |
| 2532 | USA | total | 2008 | 304093966.0 | NaN |
| 2533 | USA | under18 | 2008 | 74104602.0 | NaN |
| 2534 | USA | under18 | 2013 | 73585872.0 | NaN |
| 2535 | USA | total | 2013 | 316128839.0 | NaN |
| 2536 | USA | total | 2009 | 306771529.0 | NaN |
| 2537 | USA | under18 | 2009 | 74134167.0 | NaN |
| 2538 | USA | under18 | 2010 | 74119556.0 | NaN |
| 2539 | USA | total | 2010 | 309326295.0 | NaN |
| 2540 | USA | under18 | 2011 | 73902222.0 | NaN |
| 2541 | USA | total | 2011 | 311582564.0 | NaN |
| 2542 | USA | under18 | 2012 | 73708179.0 | NaN |
| 2543 | USA | total | 2012 | 313873685.0 | NaN |
2544 rows × 5 columns
查看存在缺失数据的列。
使用.isnull().any()，只有某一列存在一个缺失数据，就会显示True。
cond = pop2.isnull().any(axis = 1)
pop2[cond]
| state/region | ages | year | population | state | |
|---|---|---|---|---|---|
| 2448 | PR | under18 | 1990 | NaN | NaN |
| 2449 | PR | total | 1990 | NaN | NaN |
| 2450 | PR | total | 1991 | NaN | NaN |
| 2451 | PR | under18 | 1991 | NaN | NaN |
| 2452 | PR | total | 1993 | NaN | NaN |
| 2453 | PR | under18 | 1993 | NaN | NaN |
| 2454 | PR | under18 | 1992 | NaN | NaN |
| 2455 | PR | total | 1992 | NaN | NaN |
| 2456 | PR | under18 | 1994 | NaN | NaN |
| 2457 | PR | total | 1994 | NaN | NaN |
| 2458 | PR | total | 1995 | NaN | NaN |
| 2459 | PR | under18 | 1995 | NaN | NaN |
| 2460 | PR | under18 | 1996 | NaN | NaN |
| 2461 | PR | total | 1996 | NaN | NaN |
| 2462 | PR | under18 | 1998 | NaN | NaN |
| 2463 | PR | total | 1998 | NaN | NaN |
| 2464 | PR | total | 1997 | NaN | NaN |
| 2465 | PR | under18 | 1997 | NaN | NaN |
| 2466 | PR | total | 1999 | NaN | NaN |
| 2467 | PR | under18 | 1999 | NaN | NaN |
| 2468 | PR | total | 2000 | 3810605.0 | NaN |
| 2469 | PR | under18 | 2000 | 1089063.0 | NaN |
| 2470 | PR | total | 2001 | 3818774.0 | NaN |
| 2471 | PR | under18 | 2001 | 1077566.0 | NaN |
| 2472 | PR | total | 2002 | 3823701.0 | NaN |
| 2473 | PR | under18 | 2002 | 1065051.0 | NaN |
| 2474 | PR | total | 2004 | 3826878.0 | NaN |
| 2475 | PR | under18 | 2004 | 1035919.0 | NaN |
| 2476 | PR | total | 2003 | 3826095.0 | NaN |
| 2477 | PR | under18 | 2003 | 1050615.0 | NaN |
| ... | ... | ... | ... | ... | ... |
| 2514 | USA | under18 | 1999 | 71946051.0 | NaN |
| 2515 | USA | total | 2000 | 282162411.0 | NaN |
| 2516 | USA | under18 | 2000 | 72376189.0 | NaN |
| 2517 | USA | total | 1999 | 279040181.0 | NaN |
| 2518 | USA | total | 2001 | 284968955.0 | NaN |
| 2519 | USA | under18 | 2001 | 72671175.0 | NaN |
| 2520 | USA | total | 2002 | 287625193.0 | NaN |
| 2521 | USA | under18 | 2002 | 72936457.0 | NaN |
| 2522 | USA | total | 2003 | 290107933.0 | NaN |
| 2523 | USA | under18 | 2003 | 73100758.0 | NaN |
| 2524 | USA | total | 2004 | 292805298.0 | NaN |
| 2525 | USA | under18 | 2004 | 73297735.0 | NaN |
| 2526 | USA | total | 2005 | 295516599.0 | NaN |
| 2527 | USA | under18 | 2005 | 73523669.0 | NaN |
| 2528 | USA | total | 2006 | 298379912.0 | NaN |
| 2529 | USA | under18 | 2006 | 73757714.0 | NaN |
| 2530 | USA | total | 2007 | 301231207.0 | NaN |
| 2531 | USA | under18 | 2007 | 74019405.0 | NaN |
| 2532 | USA | total | 2008 | 304093966.0 | NaN |
| 2533 | USA | under18 | 2008 | 74104602.0 | NaN |
| 2534 | USA | under18 | 2013 | 73585872.0 | NaN |
| 2535 | USA | total | 2013 | 316128839.0 | NaN |
| 2536 | USA | total | 2009 | 306771529.0 | NaN |
| 2537 | USA | under18 | 2009 | 74134167.0 | NaN |
| 2538 | USA | under18 | 2010 | 74119556.0 | NaN |
| 2539 | USA | total | 2010 | 309326295.0 | NaN |
| 2540 | USA | under18 | 2011 | 73902222.0 | NaN |
| 2541 | USA | total | 2011 | 311582564.0 | NaN |
| 2542 | USA | under18 | 2012 | 73708179.0 | NaN |
| 2543 | USA | total | 2012 | 313873685.0 | NaN |
96 rows × 5 columns
查看缺失数据
根据数据是否缺失情况显示数据，如果缺失为True，那么显示
找到有哪些state/region使得state的值为NaN，使用unique()查看非重复值
pop2.head()
| state/region | ages | year | population | state | |
|---|---|---|---|---|---|
| 0 | AL | under18 | 2012 | 1117489.0 | Alabama |
| 1 | AL | total | 2012 | 4817528.0 | Alabama |
| 2 | AL | under18 | 2010 | 1130966.0 | Alabama |
| 3 | AL | total | 2010 | 4785570.0 | Alabama |
| 4 | AL | under18 | 2011 | 1125763.0 | Alabama |
#让你查看哪一个州的有空值的 州的缩写
cond_state = pop2["state"].isnull()
cond_state
0 False
1 False
2 False
3 False
4 False
5 False
6 False
7 False
8 False
9 False
10 False
11 False
12 False
13 False
14 False
15 False
16 False
17 False
18 False
19 False
20 False
21 False
22 False
23 False
24 False
25 False
26 False
27 False
28 False
29 False
...
2514 True
2515 True
2516 True
2517 True
2518 True
2519 True
2520 True
2521 True
2522 True
2523 True
2524 True
2525 True
2526 True
2527 True
2528 True
2529 True
2530 True
2531 True
2532 True
2533 True
2534 True
2535 True
2536 True
2537 True
2538 True
2539 True
2540 True
2541 True
2542 True
2543 True
Name: state, Length: 2544, dtype: bool
pop2[cond_state]["state/region"].unique()
array(['PR', 'USA'], dtype=object)
为找到的这些state/region的state项补上正确的值，从而去除掉state这一列的所有NaN！
记住这样清除缺失数据NaN的方法！
合并各州面积数据areas，使用左合并。
思考一下为什么使用外合并？
继续寻找存在缺失数据的列
我们会发现area(sq.mi)这一列有缺失数据，为了找出是哪一行，我们需要找出是哪个state没有数据
去除含有缺失数据的行
查看数据是否缺失
找出2010年的全民人口数据,df.query(查询语句)
对查询结果进行处理，以state列作为新的行索引:set_index
计算人口密度。注意是Series/Series，其结果还是一个Series。
排序，并找出人口密度最高的五个州sort_values()
找出人口密度最低的五个州
要点总结：
- 统一用loc()索引
- 善于使用.isnull().any()找到存在NaN的列
- 善于使用.unique()确定该列中哪些key是我们需要的
- 一般使用外合并、左合并，目的只有一个：宁愿该列是NaN也不要丢弃其他列的信息
回顾：Series/DataFrame运算与ndarray运算的区别
- Series与DataFrame没有广播，如果对应index没有值，则记为NaN；或者使用add的fill_value来补缺失值
- ndarray有广播，通过重复已有值来计算