• 列表解析+判断语句求偶数+有序字典


    #列表解析+判断语句求偶数
    # print([i for i in range(1,101) if i %2==0])
    import random
    math_report=[random.randint(0,100) for i in range(20)]
    chinese_report=[random.randint(0,100) for i in range(20)]
    # print(math_report,chinese_report)
    print([ v for v in math_report if v > 59]) #数学成绩及格
    print([ j for j in chinese_report if j > 59]) #语文成绩及格
    print([var for var in zip(math_report,chinese_report) if sum(var) >=150]) #数学 语文总分大于150的
    #统计每个单词的长度
    s='this is test python java golang'
    print([len(word) for word in s.split()])

    #有序字典
    from collections import OrderedDict
    s='acdef'
    orderd=OrderedDict()
    for k in s:
    orderd[k] = 0
    for k,v in orderd.items():
    print(k,v)

    #数量统计
    # d={} #定义字典
    d=OrderedDict()
    s='this is test'
    for k in s: #遍历s 取k索引
    if k in d: #判断k索引是否在字典里面
    d[k] += 1 #每个索引+1 统计次数
    else:
    d[k]=1 #索引=1
    print(d)
    for k,v in d.items():
    print(k,v)

    #统计字符串里面的字符出现的次数
    s='asthisdjma'
    key=[chr(i)for i in range(ord('a'),ord('z')+1)] #生成a-z的值
    print(key)
    d=dict.fromkeys(key,0)
    print(d)
    for k in s:
    d[k]+=1
    print(d)

    b='qwerertsa'
    from collections import Counter
    c=Counter()
    for k in s:
    c[k] +=1
    print(c)


    #深入拷贝 修改列表里面的值 不会影响同一块地址的变化
    import copy
    #copy.deepcopy(x,memoryview=None,_nil=[])
    listx = [1,2,[3,4]]
    listy = copy.deepcopy(listx)
    print(listx,listy)
    listx[2][1] = 100
    print(listx,listy)
    print(id(listx[2]),id(listy[2]))
     
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  • 原文地址:https://www.cnblogs.com/zhangcaiwang1/p/16299630.html
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