Given a DataFrame with the columns xk and yk, we want to find the indexes of the DataFrame in which the values for xk and yk ==0.
I have it working perfectly fine for just the one column but I cant get it working for both
b = (df[df['xk'] ==0]).index.tolist()
How would I do it for xk and yk at the same time.
I think you can check if all values are True in compared subset ['xk', 'yk'] by all:
b = df[(df[['xk', 'yk']] == 0).all(1)].index.tolist()
Another solution is add second condition with &:
b = (df[(df['xk'] == 0) & (df['yk'] == 0)].index.tolist())
Sample:
df = pd.DataFrame({'xk':[0,2,3],
'yk':[0,5,0],
'aa':[0,1,0]})
print (df)
aa xk yk
0 0 0 0
1 1 2 5
2 0 3 0
b = df[(df[['xk', 'yk']] == 0).all(1)].index.tolist()
print (b)
[0]
b1 = (df[(df['xk'] == 0) & (df['yk'] == 0)].index.tolist())
print (b1)
[0]
Second solution is faster:
#length of df = 3k
df = pd.concat([df]*1000).reset_index(drop=True)
In [294]: %timeit df[(df[['xk', 'yk']] == 0).all(1)].index.tolist()
1000 loops, best of 3: 1.21 ms per loop
In [295]: %timeit (df[(df['xk'] == 0) & (df['yk'] == 0)].index.tolist())
1000 loops, best of 3: 828 µs per loop
df = pd.DataFrame({'xk':[0,2,0], 'yk':[0,5,0], 'aa':[0,1,0]}) print (df) and then you get DataFrame - a = (df.ix[(df['xk'] == 0) & (df['yk'] == 0)]). Do you need return e.g. first index or second index value by position?