Convert boolean DataFrame to binary number array

You can do:

aaa = pd.DataFrame([[False,False,False],
                      [True,True,True]])
aaa=aaa.astype(int)

Then aaa is

    0   1   2
0   0   0   0
1   1   1   1

If you want to get the array ['000','111'] you can do:

aaa = pd.DataFrame([[False,False,False],
                      [True,True,True]])
aaa=aaa.astype(int).astype(str)
[''.join(i) for i in aaa.values.tolist()]

You can convert after int and str to numpy array by values and then sum:

print (aaa.astype(int).astype(str).values.sum(axis=1))
['000' '111']

You can multiply by a bit shifted operator to simulate powers of two, sum, then convert to binary

aaa.mul(np.arange(3)[::-1] << 1).sum(1).apply(bin)

0      0b0
1    0b110
dtype: object

Notice how np.arange(3)[::-1] << 1 is successive powers of 2

array([4, 2, 0])

You can take this further by manipulating with str operations

aaa.mul(
    np.arange(3)[::-1] << 1
).sum(1).apply(bin).str.replace('0b', '').str.zfill(3)

0    000
1    110
dtype: object

I would do one of the following:

a.astype(int).astype(str).sum(axis=1).astype(int).astype(str)

but this is a bit too much retyping to my taste.

Another possibility is to use apply:

a.astype(int).astype(str).apply(lambda x: ''.join(list(x)))

But what seems cleanest to me is to obtain the desired number by multiplication, then convert it to binary:

a.dot([4, 2, 1]).map(lambda x: bin(x))

of course, if you don't want the '0b' on the beginning, you just use

a.dot([4, 2, 1]).map(lambda x: bin(x)[2:])