I'm trying to compare two data frames using an if statement, with the output being a new data frame. I'd like to compare data frame A to B and for each element in A that is larger than the corresponding element in B, return 1 else 0.
A = 5 3 2
4 7 1
1 9 5
B = 1 2 9
2 5 6
7 2 3
Return:
C = 1 1 0
1 1 0
0 1 1
You can use astype:
#create mask
print (A > B)
0 1 2
0 True True False
1 True True False
2 False True True
print (A > B).astype(int)
0 1 2
0 1 1 0
1 1 1 0
2 0 1 1
Next solution is use gt, but it is same:
print (A.gt(B)).astype(int)
0 1 2
0 1 1 0
1 1 1 0
2 0 1 1
In [13]: %timeit (A > B).astype(int)
The slowest run took 4.71 times longer than the fastest. This could mean that an intermediate result is being cached
1000 loops, best of 3: 908 µs per loop
In [14]: %timeit (A.gt(B)).astype(int)
The slowest run took 5.16 times longer than the fastest. This could mean that an intermediate result is being cached
1000 loops, best of 3: 901 µs per loop
You can compare directly which will generate a boolean mask and then cast the dtype to int using astype:
In [36]:
(A > B).astype(int)
Out[36]:
0 1 2
0 1 1 0
1 1 1 0
2 0 1 1
The boolean mask looks like:
In [37]:
A > B
Out[37]:
0 1 2
0 True True False
1 True True False
2 False True True
You can also just multiply Boolean True/False by one to get ones and zeros.
>>> (A > B) * 1
a b c
0 1 1 0
1 1 1 0
2 0 1 1
