Question

我已经编写了一个适用于长度为1的字符串的函数，但我不知道如何让它适用于较长的字符串。

def function(text, n):
    dict={}
    char_count=0

    for c in text:
        keys=dict.keys()
        if c.isalpha()==True:
            char_count+=1
            if c in keys:
                dict[c] +=1
            else:
                dict[c]=1
    for key in dict:
        dict[key]=dict[key]/char_count

    return dict

不太欢迎使用import：/

Answer 1

您可以创建一个生成器而不是遍历长度为n的每个子字符串。然后将它们用作跟踪计数的字典的键。

def substring_percentage(text, n):
    out = {}
    n_substrings = len(text)-n+1
    subs = (text[i:i+n] for i in range(n_substrings))
    for s in subs:
        if s in out:
            out[s] += 100 / n_substrings
        else:
            out[s] = 100 / n_substrings
    return out

试验：

s = 'I have an assignment to write a function that will receive a sentence and a number ' \
  +'and will return the percentage of the occurrences of strings of length of the given ' \
  +'number in the given string.'

pcts = substring_percentage(s, 4)
sorted(pcts.items(), key=lambda x: x[::-1], reverse=True)
# returns:
[('the ', 2.094240837696335),
 (' the', 2.094240837696335),
 (' of ', 2.094240837696335),
 ('n th', 1.5706806282722514),
 ...
 (' an ', 0.5235602094240838),
 (' a s', 0.5235602094240838),
 (' a n', 0.5235602094240838),
 (' a f', 0.5235602094240838)]

Answer 2

三个步骤：

将输入分成单个单词; Python的split函数将为您返回一个很好的列表。
制作相应的字长列表;在每个元素上使用len。
使用count函数计算每个长度的出现次数;将这些结果放在字典中。

例如，如果你从：

开始

sentence = "Now I will a rhyme construct "       + \
           "By chosen words the young instruct " + \
           "Cunningly ensured endeavour "        + \
           "Con it and remember ever "           + \
           "Widths of circle here you see "      + \
           "Stretchd out in strange obscurity "

将其分成单个单词。列出每个单词的长度;它看起来像这样：

[3, 1, 4, 1, 5, 9, 2, 6, 
 5, 3, 5, 8, 9, 7, 9, 3, 
 2, 3, 8, 4, 6, 2, 6, 4, 
 3, 3, 8, 3, 2, 7, 9]

然后计算此列表中每个数字的数量。这会让你感动吗？

计算string，python中子字符串出现的百分比

2 个答案: