If you have a df:
apple banana carrot
a 1 2 3
b 2 3 1
c 0 0 1
To find the index for the columns where a cell is equal to 0 is df[df['apple']==0].index
but can you transpose this so you find the index of row c where it is 0?
Basically I need to drop the columns where c==0 and would like to do this in one line by row rather than by each column.
If want test row c and get all columns if 0:
c = df.columns[df.loc['c'] == 0]
print (c)
Index(['apple', 'banana'], dtype='object')
If want test all rows:
c1 = df.columns[df.eq(0).any()]
print (c1)
Index(['apple', 'banana'], dtype='object')
If need remove columns if 0 in any row:
df = df.loc[:, df.ne(0).all()]
print (df)
carrot
a 3
b 1
c 1
Detail/explanation:
First compare all values of DataFrame by ne (!=):
print (df.ne(0))
apple banana carrot
a True True True
b True True True
c False False True
Then get all rows if all True rows:
print (df.ne(0).all())
apple False
banana False
carrot True
dtype: bool
Last filter by DataFrame.loc:
print (df.loc[:, df.loc['c'].ne(0)])
carrot
a 3
b 1
c 1
If need test only c row solution is similar, only first select c row by loc and omit all:
df = df.loc[:, df.loc['c'].ne(0)]
Yes you can, df.T[df.T['c']==0]
