Pandas Dataframe 检查列值是否在列列表中
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Pandas Dataframe Check if column value is in column list
提问by clg4
I have a dataframe df:
我有一个数据框df：
data = {'id':[12,112],
'idlist':[[1,5,7,12,112],[5,7,12,111,113]]
}
df=pd.DataFrame.from_dict(data)
which looks like this:
看起来像这样：
id idlist
0 12 [1, 5, 7, 12, 112]
1 112 [5, 7, 12, 111, 113]
I need to check and see if idis in the idlist, and select or flag it. I have tried variations of the following and receive the commented error:
我需要检查并查看是否id在 中idlist，然后选择或标记它。我尝试了以下变体并收到评论错误：
df=df.loc[df.id.isin(df.idlist),:] #TypeError: unhashable type: 'list'
df['flag']=df.where(df.idlist.isin(df.idlist),1,0) #TypeError: unhashable type: 'list'
Some possible other methods to a solution would be .applyin a list comprehension?
解决方案的一些可能的其他方法将.apply在列表理解中？
I am looking for a solution here that either selects the rows where idis in idlist, or flags the row with a 1 where idis in idlist. The resulting dfshould be either:
我在这里寻找一个解决方案，要么选择 where idis in的行，要么idlist用 1 where idis in标记该行idlist。结果df应该是：
id idlist
0 12 [1, 5, 7, 12, 112]
or:
或者：
flag id idlist
0 1 12 [1, 5, 7, 12, 112]
1 0 112 [5, 7, 12, 111, 113]
Thanks for the help!
谢谢您的帮助！
回答by jezrael
Use apply:
使用apply：
df['flag'] = df.apply(lambda x: int(x['id'] in x['idlist']), axis=1)
print (df)
id idlist flag
0 12 [1, 5, 7, 12, 112] 1
1 112 [5, 7, 12, 111, 113] 0
Similar:
相似的：
df['flag'] = df.apply(lambda x: x['id'] in x['idlist'], axis=1).astype(int)
print (df)
id idlist flag
0 12 [1, 5, 7, 12, 112] 1
1 112 [5, 7, 12, 111, 113] 0
With list comprehension:
与list comprehension：
df['flag'] = [int(x[0] in x[1]) for x in df[['id', 'idlist']].values.tolist()]
print (df)
id idlist flag
0 12 [1, 5, 7, 12, 112] 1
1 112 [5, 7, 12, 111, 113] 0
Solutions for filtering:
过滤解决方案：
df = df[df.apply(lambda x: x['id'] in x['idlist'], axis=1)]
print (df)
id idlist
0 12 [1, 5, 7, 12, 112]
df = df[[x[0] in x[1] for x in df[['id', 'idlist']].values.tolist()]]
print (df)
id idlist
0 12 [1, 5, 7, 12, 112]
回答by Aafaque Abdullah
You can use df.applyand process each row and create a new column flag that will check the condition and give you result as second output requested.
您可以使用df.apply和处理每一行并创建一个新的列标志，该标志将检查条件并在请求的第二个输出时为您提供结果。
df['flag'] = df.loc[:, ('id', 'idlist')].apply(lambda x: 1 if x[0] in x[1] else 0, axis=1)
print(df)
where x[0] is idand x[1] is idlist
哪里x[0] is id和x[1] is idlist
回答by YOBEN_S
By using issubset
通过使用 issubset
df.apply(lambda x : set([x.id]).issubset(x.idlist),1).astype(int)
Out[378]:
0 1
1 0
dtype: int32
By using np.vectorize
通过使用 np.vectorize
def myfun(x,y):
return np.in1d(x,y)
np.vectorize(myfun)(df.id,df.idlist).astype(int)
Timing :
时间：
%timeit np.vectorize(myfun)(df.id,df.idlist).astype(int)
10000 loops, best of 3: 92.3 μs per loop
%timeit df.apply(lambda x : set([x.id]).issubset(x.idlist),1).astype(int)
1000 loops, best of 3: 353 μs per loop
回答by rnso
Try simple forloop:
尝试简单的for循环：
flaglist = []
for i in range(len(df)):
if df.id[i] in df.idlist[i]:
flaglist.append(1)
else:
flaglist.append(0)
df["flag"] = flaglist
df:
df：
id idlist flag
0 12 [1, 5, 7, 12, 112] 1
1 112 [5, 7, 12, 111, 113] 0
To drop rows:
删除行：
flaglist = []
for i in range(len(df)):
if df.id[i] not in df.idlist[i]:
flaglist.append(i)
df = df.drop(flaglist)
df:
df：
id idlist flag
0 12 [1, 5, 7, 12, 112] 1
Above can be converted to list comprehension for creating a flag column:
以上可以转换为列表理解来创建标志列：
df["flag"] = [df.id[i] in df.idlist[i] for i in range(len(df))]
print(df)
# id idlist flag
# 0 12 [1, 5, 7, 12, 112] True
# 1 112 [5, 7, 12, 111, 113] False
or
或者
df["flag"] = [1 if df.id[i] in df.idlist[i] else 0 for i in range(len(df))]
print(df)
# id idlist flag
# 0 12 [1, 5, 7, 12, 112] 1
# 1 112 [5, 7, 12, 111, 113] 0
and for selecting out rows:
并选择行：
flaglist = [i for i in range(len(df)) if df.id[i] in df.idlist[i]]
print(df.iloc[flaglist])
# id idlist
# 0 12 [1, 5, 7, 12, 112]
