在 Python 中僅從字符串中提取字符

Question

在 Python 中，我只想從字符串中提取字符。

考慮我有以下字符串，

input = "{('players',): 24, ('year',): 28, ('money',): 19, ('ipod',): 36, ('case',): 23, ('mini',): 46}"

我想要的結果是，

output =  "players year money ipod case mini"

我試圖只考慮字母來拆分，

word1 = st.split("[a-zA-Z]+")

但分裂並沒有發生。

Answer 1

你可以用 re 來做，但字符串拆分方法不需要正則表達式，它需要一個字符串。

這是使用 re 的一種方法：

import re
word1 = " ".join(re.findall("[a-zA-Z]+", st))

Answer 2

string.split() 不采用正則表達式。 你想要這樣的東西：

re.split("[^a-zA-Z]*", "your string")

並得到一個字符串：

" ".join(re.split("[^a-zA-Z]*", "your string"))

Answer 3

我認為你想要所有的單詞，而不是字符。

result = re.findall(r"(?i)\b[a-z]+\b", subject)

解釋：

"
\b       # Assert position at a word boundary
[a-z]    # Match a single character in the range between “a” and “z”
   +        # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\b       # Assert position at a word boundary
"

Answer 4

您可以采用遍歷字符串的方法，並使用isalpha函數來確定它是否是字母字符。 如果是，您可以將其附加到輸出字符串。

a = "Some57 996S/tr::--!!ing"
q = ""
for i in a:
    if i.isalpha():
        q = "".join([q,i])

Answer 5

這樣做呢？

>>> import ast
>>> " ".join([k[0] for k in ast.literal_eval("{('players',): 24, ('year',): 28, ('money',): 19, ('ipod',): 36, ('case',): 23, ('mini',): 46}").keys()])
'case mini year money ipod players'

Answer 6

或者，如果您想要所有字符而不考慮單詞或空格

    a = "Some57 996S/tr::--!!ing"
    q = ""
    for i in a:
        if i.isalpha():
            q = "".join([q,i])

打印 q 'SomeString'

Answer 7

import re
string = ''.join([i for i in re.findall('[\w +/.]', string) if i.isalpha()])

#'[\w +/.]' -> it will give characters numbers and punctuation, then 'if i.isalpha()' this condition will only get alphabets out of it and then join list to get expected result.
# It will remove spaces also.