Check if column value is in other columns in pandas

Question

I have the following data file in pandas

  target   A       B      C
0 cat      bridge  cat    brush  
1 brush    dog     cat    shoe
2 bridge   cat     shoe   bridge

How to check if there is df.targetany column ['A','B','C', etc.]where there are many columns to check?

I tried merging A, B, and C into a string to use df.abcstring.str.contains(df.target), but this will not work.

Answer 1

You can use drop, isinand any.

• dropcolumn targethas df only to your column A, B,C
• check if the values ​​match the isintarget column
• and check for anyhits

What is it.

df["exists"] = df.drop("target", 1).isin(df["target"]).any(1)
print(df)

    target  A       B       C       exists
0   cat     bridge  cat     brush   True
1   brush   dog     cat     shoe    False
2   bridge  cat     shoe    bridge  True

Answer 2

eq, drop pop, neech :

mask = df.eq(df.pop('target'), axis=0)
print (mask)
       A      B      C
0  False   True  False
1  False  False  False
2  False  False   True

, True, any:

mask = df.eq(df.pop('target'), axis=0).any(axis=1)
print (mask)
0     True
1    False
2     True
dtype: bool

df['new'] = df.eq(df.pop('target'), axis=0).any(axis=1)
print (df)
        A     B       C    new
0  bridge   cat   brush   True
1     dog   cat    shoe  False
2     cat  shoe  bridge   True

, isin:

mask = df.isin(df.pop('target').values.tolist())
print (mask)
       A      B      C
0   True   True   True
1  False   True  False
2   True  False   True

, True all:

df['new'] = df.isin(df.pop('target').values.tolist()).all(axis=1)
print (df)
        A     B       C    new
0  bridge   cat   brush   True
1     dog   cat    shoe  False
2     cat  shoe  bridge  False

Answer 3

OneHotEncoder:

In [165]: x = pd.get_dummies(df.drop('target',1), prefix='', prefix_sep='')

In [166]: x
Out[166]:
   bridge  cat  dog  cat  shoe  bridge  brush  shoe
0       1    0    0    1     0       0      1     0
1       0    0    1    1     0       0      0     1
2       0    1    0    0     1       1      0     0

In [167]: x[df['target']].eq(1).any(1)
Out[167]:
0    True
1    True
2    True
dtype: bool

:

In [168]: x[df['target']]
Out[168]:
   cat  cat  brush  bridge  bridge
0    0    1      1       1       0
1    0    1      0       0       0
2    1    0      0       0       1

Answer 4

, , "target":

df["exist"] = df.apply(lambda row:row.value_counts()[row['target']] > 1 , axis=1)

, :

   b  c target
0  3  a      a
1  3  4      2
2  3  4      2
3  3  4      2
4  3  4      4

:

   b  c target  exist
0  3  a      a   True
1  3  4      2  False
2  3  4      2  False
3  3  4      2  False
4  3  4      4   True

Answer 5

, difference :

matches = df[df.columns.difference(['target'])].eq(df['target'], axis = 0)

#       A      B      C
#0  False   True  False
#1  False  False  False
#2  False  False   True

# Check if at least one match:
matches.any(axis = 1)

#Out[30]: 
#0     True
#1    False
#2     True

, , :

matches.apply(lambda x: ", ".join(x.index[np.where(x.tolist())]), axis = 1)

Out[53]: 
0    B
1     
2    C
dtype: object