应用函数不起作用，类型错误为类型错误：类型"NoneType"的参数不可迭代

Question

我不知道为什么它不起作用，我想在使用apply的数据帧中使用该函数，但类型错误是TypeError："NoneType"类型的参数不可迭代

df[column1]  = ['https://www.some_text.com/ar?cc+cc&q={keyword}',
'https://www.some_text.com/ar?fsd+fsd&q={keyword}',
'https://www.some_text.com/ar?xc+sdc&q={keyword}']
url_s = 'some_text'
url_f = 'some_text_2'
def url_name(x):
if url_s in x:
a = x.replace(url_s,url_f)
return a
df['column1'] = df['column1'].apply(url_name)

Answer 1

IF循环中的缩进出现了一个小问题，这可能导致了错误。其他一切似乎都奏效了。

url_s = 'some_text'
url_f = 'some_text_2'
def url_name(x):
if url_s in x:
a = x.replace(url_s,url_f)
return a    

d = {'column1': ['https://www.some_text.com/ar?cc+cc&q={keyword}',
'https://www.some_text.com/ar?fsd+fsd&q={keyword}',
'https://www.some_text.com/ar?xc+sdc&q={keyword}']}
df = pandas.DataFrame(data=d)
df['column1'] = df['column1'].apply(url_name)

这就是结果。

df
Out[22]: 
column1
0   https://www.some_text_2.com/ar?cc+cc&q={keyword}
1  https://www.some_text_2.com/ar?fsd+fsd&q={keyw...
2  https://www.some_text_2.com/ar?xc+sdc&q={keyword}

希望这能有所帮助！