附加一个计算单词长度的列表

Question

我正在从一个文本文件中提取单词，删除每个单词的 \n 并从这些单词中创建一个新列表。

现在我需要系统地逐字查找单词的长度，然后在该单词长度的计数中加 1，即我将从一个空计数开始：

length_of_words = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

那么如果剥离的单词列表包含 5x 7 个字母的单词和 3x 2 个字母的单词，我最终会得到：

length_of_words = [0,3,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

这归结为：

• 计算单词的长度，例如n
• 在 length_of_words[n-1] 的 length_of_words 中加一（因为它仍然以 1 个字母的单词开头，即第 0 项）

我真的被困在如何从本质上将列表中 1 项的值增加 1，而不仅仅是将 1 附加到列表的末尾。

我现在拥有的是这样的：

lines = open ('E:\Python\Assessment\dracula.txt', 'r'). readlines ()

stripped_list = [item.strip() for item in lines]

tally = [] #empty set of lengths
for lengths in range(1,20):
    tally.append(0)

print tally #original tally

for i in stripped_list:
    length_word = int(len(i))
    tally[length_word] = tally[length_word] + 1
print tally

Answer 1

collections.Counter 类对这类事情很有帮助：

>>> from collections import Counter
>>> words = 'the quick brown fox jumped over the lazy dog'.split()
>>> Counter(map(len, words))
Counter({3: 4, 4: 2, 5: 2, 6: 1})

您在问题中发布的代码可以按原样正常工作，所以我不确定您遇到了什么问题。

FWIW，这里有一些小的代码改进（更 Pythonic 风格）：

stripped_list = 'the quick brown fox jumped over the lazy dog'.split()

tally = [0] * 20
print tally #original tally

for i in stripped_list:
    length_word = len(i)
    tally[length_word] += 1
print tally

Answer 2

我认为您代码中的错误行是tally[length_word]，您忘记添加- 1

我还对您的代码进行了一些更改，使其更加 pythonic

#lines = open ('E:\Python\Assessment\dracula.txt', 'r'). readlines ()

#stripped_list = [item.strip() for item in lines]

with open('/home/facundo/tmp/words.txt') as i:
    stripped_list = [x.strip() for x in i.readlines()]

#tally = [] #empty set of lengths
#for lengths in range(1,20):
#    tally.append(0)

tally = [0] * 20

print tally #original tally

for i in stripped_list:
    #length_word = int(len(i))
    word_length = len(i)
    #tally[length_word] = tally[length_word] + 1
    if word_length > 0:
        tally[word_length - 1] += 1

print tally