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2018-10-24:统计字符串中各个字母/单词的个数

2018-10-24:统计字符串中各个字母/单词的个数

作者: 种树在此时 | 来源:发表于2018-10-31 00:13 被阅读0次

    b.txt文件中内容:When you are starting the selenium server, open a console in the directory with chromedriver and selenium server and execute the above command

    终稿如下:

    file = 'b.txt'

    with open(file) as f:
    line = f.read() # 读取文件,为一个字符串
    str = []
    outstr = {}
    for x in line:
    if x.isalpha():# 判断是否为字母,
    if x in outstr:# 判断这个字母是否添加到字典
    outstr[x] += 1# 添加到字典了,则值加一
    else:
    outstr[x] = 1 #若没有在字典,则追加一个键值对到字典
    print(outstr)

    ############################################################

    初稿如下:

    file = 'b.txt'

    with open(file) as f:
    line = f.read() # 读取文件,为一个字符串
    str = []
    outstr = {}
    for x in line:
    if x.isalpha():# 判断是否为字母,是字母再加入到str列表中去
    str.append(x)

    for j in range(len(str)-1,0,-1):

     #   for i in range(0,j):
    for j in range(0,len(str)-1):
        if str[j] in outstr:
                #if str[i] == str[j]:
            outstr[str[j]] += 1
        else:
            outstr[str[j]] = 1
    print(str)
    print(outstr)
    

    ###################################################################################################################################

    大神稿如下:

    letters = {}

    with open('b.txt') as f:
    for letter in f.read():
    '''判断是否是字母'''
    if letter.islower():
    '''判断是否存在元组中'''
    if letters.get(letter):
    '''是将数量+1'''
    letters[letter] += 1
    else:
    '''第一次加入,赋值为1'''
    letters[letter] = 1
    print(letters)

    单词版如下:

    file = 'b.txt'

    with open(file) as f:
    line = f.read().split(' ')

    outstr = {}
    str = []
    for x in line:
        str.append(x)
    for one in line:
        a = str.count(one)
        outstr[one] = a
    

    print(outstr)

    ###################################################################################################

    简单粗暴版如下:

    file = 'b.txt'

    with open(file) as f:
    line = f.read().replace(' ','')
    str = []
    for x in line:
    str.append(x)

    outstr = {}
    
    
    for one in str:
        a = str.count(one)
        outstr[one] = a
    

    print(str)
    print(outstr)

    ##########################################

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