如何从列表中删除换行符?

Question

我有一个python列表，我需要使它成为一个没有换行符的字符串。我有这个['1n2n2n,n6n2n9n,n7n1n3']，我希望它是122,629,713在一个字符串，而不是列表。这是我的列表代码:

follower = [driver.find_elements_by_xpath('//*[@id="__next"]/div/div/div[3]/div[2]/div[2]/div/div')[0].text]
follow = ''
for item in follower:
item = item.strip('n')
item = item.strip('[')
item = item.strip(']')
item = item.strip("'")
follow = follow + item

Answer 1

for item in follower:
follow += item
print(follow.replace("n",""))

Answer 2

如果必须替换多个字符，最简单的方法是使用正则表达式。您可以使用|分隔字符串中的字符，并记住转义特殊字符。

import re
chars_to_remove = "n|'|[|]"
text = "1n2n2n,]n6n2n9n,[n7n1n3'"
re.sub(chars_to_remove, "", text)

输出:

'122,629,713'

Answer 3

我不确定你是否有一个字符串列表:

list_of_strings = ['1n2n2n,n6n2n9n,n7n1n3']

或者只是一个字符串:

string = '1n2n2n,n6n2n9n,n7n1n3'

如果您有一个字符串列表，只需使用for循环来获取每个字符串，然后使用下面的代码来执行您的要求，其他只需遍历字符串

# CASE 1: you have a list of strings
for string in list_of_strings:
fixed_string = ""
for char in string:
if char != 'n'
fixed_string += char
print(fixed string)

# CASE 2: you just have the string
fixed_string = ""
for char in string:
if char != 'n'
fixed_string += char
print(fixed_string)

两种情况下都应该遍历字符串，如果它们不是换行字符(n)，则将它们添加到新字符串中。

Answer 4

给定:

>>> l=['1n2n2n,n6n2n9n,n7n1n3']

每个字符串上的列表推导式和.replace:

>>> [e.replace('n','') for e in l]
['122,629,713']

Answer 5

我所理解的是你有一个列表a=['1n2n2n,n6n2n9n,n7n1n3']，你想从它删除n，你想要的结果是[122,629,713]。然而，list a中有一个错误，因为python将列表中的项目视为一个项目，a should be a=['1n2n2n','n6n2n9n','n7n1n3']

a=['1n2n2n','n6n2n9n','n7n1n3']
b=""
index=0
for i in a:
c=i.replace("n","")
b+=f"{c},"
if index==(len(a)-1):
b=b[0:len(b)-1]
index+=1
print(b)
#The outcome will be `122,629,713`

如果你想要[122,629,713]的结果，那么在for-loop后面加上

b=b.split(",")
# Now the outcome will be `[122,629,713]`