[英]compare two columns having list of strings in pandas
我在熊猫中有一个数据帧,该数据帧有两列,每一行是一个字符串列表,如何检查唯一行上的这两列中是否有单词匹配(标志列是所需的输出)
A B flag
hello,hi,bye bye, also 1
but, as well see, pandas 0
我努力了
df['A'].str.contains(df['B'])
但是我得到这个错误
TypeError: 'Series' objects are mutable, thus they cannot be hashed
您可以通过split和set
s将每个值转换为单独的单词,并通过&
检查交集,然后将值转换为boolean-将空集转换为False
,最后将其转换为int
s- Falses
为0
s和True
s为1
s 。
zipped = zip(df['A'], df['B'])
df['flag'] = [int(bool(set(a.split(',')) & set(b.split(',')))) for a, b in zipped]
print (df)
A B flag
0 hello,hi,bye bye,also 1
1 but,as well see,pandas 0
类似的解决方案:
df['flag'] = np.array([set(a.split(',')) & set(b.split(',')) for a, b in zipped]).astype(bool).astype(int)
print (df)
A B flag
0 hello,hi,bye bye, also 1
1 but,as well see, pandas 0
编辑:有可能是之前一些空格,
,所以添加map
与str.strip
并删除与空字符串filter
:
df = pd.DataFrame({'A': ['hello,hi,bye', 'but,,,as well'],
'B': ['bye ,,, also', 'see,,,pandas']})
print (df)
A B
0 hello,hi,bye bye ,,, also
1 but,,,as well see,,,pandas
zipped = zip(df['A'], df['B'])
def setify(x):
return set(map(str.strip, filter(None, x.split(','))))
df['flag'] = [int(bool(setify(a) & setify(b))) for a, b in zipped]
print (df)
A B flag
0 hello,hi,bye bye ,,, also 1
1 but,,,as well see,,,pandas 0
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.