[英]Regex to remove characters after multiple white spaces
Hi all I've got and input strings:大家好,我有并输入字符串:
list_of_strings = ["apple", "orange ca", "pear sa", "banana sth"]
And I want to remove everything after multiple white spaces (more than 1), so end result is:我想在多个空格(超过 1 个)之后删除所有内容,所以最终结果是:
final_list_of_strings = ["apple", "orange ca", "pear", "banana"]
I've tried regex:我试过正则表达式:
import re
regex_expression = r"(.*\s?)(\s{2,}.*)"
for name in list_of_strings:
regex_matching_groups = re.findall(regex_expression, name)
if regex_matching_groups:
name = regex_matching_groups[0][0]
but fails on multiple spaces ... Thank you for help!但在多个空间失败......谢谢你的帮助!
You can use re.sub
in a list comprehension:您可以在列表理解中使用re.sub
:
import re
list_of_strings = ["apple", "orange ca", "pear sa", "banana sth"]
list_of_strings = [re.sub(r'\s{2}.*', '', x, flags=re.S) for x in list_of_strings]
print(list_of_strings)
# -> ['apple', 'orange ca', 'pear', 'banana']
See the Python demo .请参阅Python 演示。
The \\s{2}.*
regex matches two whitespace chars and then the rest of the string (even if there are line break chars due to re.S
flag). \\s{2}.*
正则表达式匹配两个空白字符,然后匹配字符串的其余部分(即使由于re.S
标志存在re.S
)。
using a regular expression find the first word and optionally a second word with only one space between使用正则表达式查找第一个单词和可选的第二个单词,它们之间只有一个空格
list_of_strings = ["apple", "orange ca", "pear sa", "banana sth"]
my_list=[]
def find_phrase(list_of_strings):
for string in list_of_strings:
matches=re.findall(r"(\w+)( \w+)*", string)
if len(matches)>0:
my_list.append("".join([matches[0][0],matches[0][1]]))
return my_list
print(find_phrase(list_of_strings))
output:输出:
['apple', 'orange ca', 'pear', 'banana']
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