string.punctuation 删除一个值

Question

所以我正在编写一个代码来计算#ed 的单词数量，如果一个单词没有#ed，它会忽略它。

当我运行代码：

import string
all = []
count = {}
word = []
line = input("Tweet: ").lower().strip(string.punctuation)
while line != '':
  word.extend(line.split())
  line = input("Tweet: ").lower().strip(string.punctuation)
for w in word:
  if w.startswith('#'):
    count[w] = count.get(w, 0) + 1
for word in sorted(count):
  print(word, count[word])

我输入#Python是#AWESOME！

它输出#awesome 1 但没有输出#python 1

我需要它来输出所有#ed 单词并计算使用了多少。我认为问题是由 string.punctuation.

引起的

Answer 1

你说对了一部分！

.strip(string.punctuation)

是罪魁祸首。

根据 the Python docs，# 是 string.punctuation 集的一部分。

同样来自 the Python docs，在 strip-函数上：

string.strip(s[, chars])

Return a copy of the string with leading and trailing characters removed. If chars is omitted or None, whitespace characters are removed. If given and not None, chars must be a string; the characters in the string will be stripped from the both ends of the string this method is called on.

因此，您要删除前导（例如第一个）#，并且保存在变量 line 中的字符串是 "python is #awesome"。您的 while 循环也永远不会退出，如 ".".strip(string.punctuation) == ""。看来您真的根本不想要 .strip-方法。如果只在标点符号时删除最后一个字符，请使用 "your string".rstrip(string.punctuation) 代替

Answer 2

这可能对你有用

import string
all = []
count = {}
word = []
line = input("Tweet: ").lower().rstrip(string.punctuation)
while line != '.':
  word.extend(line.split())
  line = input("Tweet: ").lower().rstrip(string.punctuation)
for w in word:
  if w.startswith('#'):
    count[w] = count.get(w, 0) + 1
for word in sorted(count):
  print(word, count[word])

或

import string
all = []
count = {}
word = []
line = input("Tweet: ").lower()
while line != '.':
  word.extend(line.split())
  line = input("Tweet: ").lower()
for w in word:
  if w.startswith('#'):
    w = w.rstrip(string.punctuation)
    count[w] = count.get(w, 0) + 1
for word in sorted(count):
  print(word, count[word])

string.punctuation 删除一个值

string.punctuation deleting a value

python

string

counter