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I have a pandas dataframe which contains 3 columns, each containing a site that a user has visited during a session.

In some cases, a user may have not visited 3 sites in a single session. This is shown by a 0, denoting that no site has been visited.

import pandas as pd

df = pd.DataFrame(data=[[5, 8, 1],[8,0,0],[1,17,0]], 
 columns=['site1', 'site2', 'site3'])
print(df)

 site1 site2 site3
0 5 8 1
1 8 0 0
2 1 17 0

In the example above, user 0 has visited sites 5, 8 and 1. User 1 has visited site 8 only, and user 2 has visited sites 1 and 17.

I would like to create a new column, last_site, which shows the last site visited by the user in that session.

The result I want is this:

 site1 site2 site3 last_site
0 5 8 1 1
1 8 0 0 8
2 1 17 0 17

How can I do this in a concise way using pandas?

Use forward filling of misisng values created by replacing 0 values and thenselect last column by iloc:

df['last'] = df.replace(0, np.nan).ffill(axis=1).iloc[:, -1].astype(int)
print (df)
 site1 site2 site3 last
0 5 8 1 1
1 8 0 0 8
2 1 17 0 17

If performance is important is possible use numpy:

a = df.values
m = a != 0

df['last'] = a[np.arange(m.shape[0]), m.shape[1]-m[:,::-1].argmax(1)-1]
print (df)
 site1 site2 site3 last
0 5 8 1 1
1 8 0 0 8
2 1 17 0 17

Code:

df['last_site'] = df.apply(lambda x: x.iloc[x.nonzero()].iloc[-1], axis=1)

Output:

 site1 site2 site3 last_site
0 5 8 1 1
1 8 0 0 8
2 1 17 0 17

mask + ffill

A "pure Pandas" solution:

df['last'] = df.mask(df.eq(0)).ffill(1).iloc[:, -1].astype(int)

numba

For efficiency over a large number of rows / columns, numba can help. To see why this works better than argmax, see Efficiently return the index of the first value satisfying condition in array.

from numba import njit

@njit
def get_last_val(A):
 m, n = A.shape
 res = A[:, -1]
 for i in range(m):
 for j in range(n):
 if A[i, j] == 0:
 res[i] = A[i, max(0, j-1)]
 break
 return res

df['last'] = get_last_val(df.values)