Converting one int to multiple bool columns in pandas

Question

Background

I got a data frame with integers. These integers represents a series of features that are either present or not present for that row.

I want these features to be named columns in my data frame.

Problem

My current solution explodes in memory and is crazy slow. How do I improve the memory efficiency of this?

import pandas as pd
df = pd.DataFrame({'some_int':range(5)})
df['some_int'].astype(int).apply(bin).str[2:].str.zfill(4).apply(list).apply(pd.Series).rename(columns=dict(zip(range(4), ["f1", "f2", "f3", "f4"])))

  f1 f2 f3 f4
0  0  0  0  0
1  0  0  0  1
2  0  0  1  0
3  0  0  1  1
4  0  1  0  0

It seems to be the .apply(pd.Series) that is slowing this down. Everything else is quite fast until I add this.

I cannot skip it because a simple list will not make a dataframe.

Answer 1

Here's a vectorized NumPy approach -

def num2bin(nums, width):
    return ((nums[:,None] & (1 << np.arange(width-1,-1,-1)))!=0).astype(int)

Sample run -

In [70]: df
Out[70]: 
   some_int
0         1
1         5
2         3
3         8
4         4

In [71]: pd.DataFrame( num2bin(df.some_int.values, 4), \
                    columns = [["f1", "f2", "f3", "f4"]])
Out[71]: 
   f1  f2  f3  f4
0   0   0   0   1
1   0   1   0   1
2   0   0   1   1
3   1   0   0   0
4   0   1   0   0

Explanation

1) Inputs :

In [98]: nums = np.array([1,5,3,8,4])

In [99]: width = 4

2) Get the 2 powered range numbers :

In [100]: (1 << np.arange(width-1,-1,-1))
Out[100]: array([8, 4, 2, 1])

3) Convert nums to a 2D array version as we later on want to do element-wise bit-ANDing between it and the 2-powered numbers in a vectorized mannner following the rules of broadcasting :

In [101]: nums[:,None]
Out[101]: 
array([[1],
       [5],
       [3],
       [8],
       [4]])

In [102]: nums[:,None] & (1 << np.arange(width-1,-1,-1))
Out[102]: 
array([[0, 0, 0, 1],
     [0, 4, 0, 1],
     [0, 0, 2, 1],
     [8, 0, 0, 0],
     [0, 4, 0, 0]])

To understand the bit-ANDIng, let's consider the number 5 from nums and its bit-ANDing for it against all 2-powered numbers [8,4,2,1] :

In [103]: 5 & 8    # 0101 & 1000
Out[103]: 0

In [104]: 5 & 4    # 0101 & 0100
Out[104]: 4

In [105]: 5 & 2    # 0101 & 0010
Out[105]: 0

In [106]: 5 & 1    # 0101 & 0001
Out[106]: 1

Thus, we see that there are no intersection against [8,2], whereas for others we have non-zeros.

4) In the final stage, look for matches (non-zeros) and simply convert those to 1s and rest to 0s by comparing against 0 resulting in a boolean array and then converting to int dtype :

In [107]: matches = nums[:,None] & (1 << np.arange(width-1,-1,-1))

In [108]: matches!=0
Out[108]: 
array([[False, False, False,  True],
       [False,  True, False,  True],
       [False, False,  True,  True],
       [ True, False, False, False],
       [False,  True, False, False]], dtype=bool)

In [109]: (matches!=0).astype(int)
Out[109]: 
array([[0, 0, 0, 1],
       [0, 1, 0, 1],
       [0, 0, 1, 1],
       [1, 0, 0, 0],
       [0, 1, 0, 0]])

Runtime test

In [58]: df = pd.DataFrame({'some_int':range(100000)})

# @jezrael's soln-1
In [59]: %timeit pd.DataFrame(df['some_int'].astype(int).apply(bin).str[2:].str.zfill(4).apply(list).values.tolist())
1 loops, best of 3: 198 ms per loop

# @jezrael's soln-2
In [60]: %timeit pd.DataFrame([list('{:20b}'.format(x)) for x in df['some_int'].values])
10 loops, best of 3: 154 ms per loop

# @jezrael's soln-3
In [61]: %timeit pd.DataFrame(df['some_int'].apply(lambda x: list('{:20b}'.format(x))).values.tolist())
10 loops, best of 3: 132 ms per loop

# @MaxU's soln-1
In [62]: %timeit pd.DataFrame([list(np.binary_repr(x, width=20)) for x in df.some_int.values])
1 loops, best of 3: 193 ms per loop

# @MaxU's soln-2
In [64]: %timeit df.some_int.apply(lambda x: pd.Series(list(np.binary_repr(x, width=20))))
1 loops, best of 3: 11.8 s per loop

# Proposed in this post
In [65]: %timeit pd.DataFrame( num2bin(df.some_int.values, 20))
100 loops, best of 3: 5.64 ms per loop

Answer 2

I think you need:

a = pd.DataFrame(df['some_int'].astype(int)
                               .apply(bin)
                               .str[2:]
                               .str.zfill(4)
                               .apply(list).values.tolist(), columns=["f1","f2","f3","f4"])
print (a)
  f1 f2 f3 f4
0  0  0  0  0
1  0  0  0  1
2  0  0  1  0
3  0  0  1  1
4  0  1  0  0

Another solution, thanks Jon Clements and ayhan:

a = pd.DataFrame(df['some_int'].apply(lambda x: list('{:04b}'.format(x))).values.tolist(), 
                 columns=['f1', 'f2', 'f3', 'f4'])
print (a)
  f1 f2 f3 f4
0  0  0  0  0
1  0  0  0  1
2  0  0  1  0
3  0  0  1  1
4  0  1  0  0

A bit changed:

a = pd.DataFrame([list('{:04b}'.format(x)) for x in df['some_int'].values], 
                  columns=['f1', 'f2', 'f3', 'f4'])
print (a)
  f1 f2 f3 f4
0  0  0  0  0
1  0  0  0  1
2  0  0  1  0
3  0  0  1  1
4  0  1  0  0

Timings:

df = pd.DataFrame({'some_int':range(100000)})

In [80]: %timeit pd.DataFrame(df['some_int'].astype(int).apply(bin).str[2:].str.zfill(20).apply(list).values.tolist())
1 loop, best of 3: 231 ms per loop

In [81]: %timeit pd.DataFrame([list('{:020b}'.format(x)) for x in df['some_int'].values])
1 loop, best of 3: 232 ms per loop

In [82]: %timeit pd.DataFrame(df['some_int'].apply(lambda x: list('{:020b}'.format(x))).values.tolist())
1 loop, best of 3: 222 ms per loop

In [83]: %timeit pd.DataFrame([list(np.binary_repr(x, width=20)) for x in df.some_int.values])
1 loop, best of 3: 343 ms per loop

In [84]: %timeit df.some_int.apply(lambda x: pd.Series(list(np.binary_repr(x, width=20))))
1 loop, best of 3: 16.4 s per loop

In [87]: %timeit pd.DataFrame( num2bin(df.some_int.values, 20))
100 loops, best of 3: 11.4 ms per loop

Answer 3

you can use numpy.binary_repr method:

In [336]: df.some_int.apply(lambda x: pd.Series(list(np.binary_repr(x, width=4)))) \
            .add_prefix('f')
Out[336]:
  f0 f1 f2 f3
0  0  0  0  0
1  0  0  0  1
2  0  0  1  0
3  0  0  1  1
4  0  1  0  0

or

In [346]: pd.DataFrame([list(np.binary_repr(x, width=4)) for x in df.some_int.values],
     ...:              columns=np.arange(1,5)) \
     ...:   .add_prefix('f')
     ...:
Out[346]:
  f1 f2 f3 f4
0  0  0  0  0
1  0  0  0  1
2  0  0  1  0
3  0  0  1  1
4  0  1  0  0