如何从熊猫数据框中的列中删除字符串值
我正在尝试编写一些代码,以逗号分隔数据框列中的字符串(因此它成为列表),并从该列表中删除某个字符串(如果存在)。删除不需要的字符串后,我想再次以逗号加入列表元素。我的数据框如下所示:
I am trying to write some code that splits a string in a dataframe column at comma (so it becomes a list) and removes a certain string from that list if it is present. after removing the unwanted string I want to join the list elements again at comma. My dataframe looks like this:
df:
Column1 Column2
0 a a,b,c
1 y b,n,m
2 d n,n,m
3 d b,b,x
所以基本上我的目标是从column2中删除所有b值,以便得到:
So basically my goal is to remove all b values from column2 so that I get:
df:
Column1 Column2
0 a a,c
1 y n,m
2 d n,n,m
3 d x
我编写的代码如下:
df=df['Column2'].apply(lambda x: x.split(','))
def exclude_b(df):
for index, liste in df['column2].iteritems():
if 'b' in liste:
liste.remove('b')
return liste
else:
return liste
第一行将列以逗号分隔的列表。使用现在的函数,我尝试遍历所有列表并删除b(如果存在),如果不存在则返回列表。如果我在末尾打印 liste,则仅返回Column2的第一行,而不返回其他行。我究竟做错了什么?并有一种方法可以将我的if条件实现为lambda函数吗?
The first row splits all the values in the column into a comma separated list. with the function now I tried to iterate through all the lists and remove the b if present, if it is not present return the list as it is. If I print 'liste' at the end it only returns the first row of Column2, but not the others. What am I doing wrong? And would there be a way to implement my if condition into a lambda function?
您只需应用正则表达式 b,?
即可替换 b
和中的任何值,在
如果存在 b
之后找到的
simply you can apply the regex b,?
, which means replace any value of b
and ,
found after the b
if exists
df['Column2'] = df.Column2.str.replace('b,?' , '')
Out[238]:
Column1 Column2
0 a a,c
1 y n,m
2 d n,n,m
3 d x