本文介绍了零和一间preT列二进制和存储作为一个整数列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我的零和一的数据帧。我想对待每列如果值是一个整数的二进制重新presentation。是什么让这个转换最简单的方法？
I have a dataframe of zeros and ones. I want to treat each column as if its values were a binary representation of an integer. What is easiest way to make this conversion?
我想这样的：
df = pd.DataFrame([[1, 0, 1], [1, 1, 0], [0, 1, 1], [0, 0, 1]])
print df
0 1 2
0 1 0 1
1 1 1 0
2 0 1 1
3 0 0 1
转换为：
0 12
1 6
2 11
dtype: int64
尽可能有效地
推荐答案
类似的解决方案，但更多更快的：
Similar solution, but more faster:
print (df.T.dot(1 << np.arange(df.shape[0] - 1, -1, -1)))
0 12
1 6
2 11
dtype: int64
时序
In [81]: %timeit df.apply(lambda col: int(''.join(str(v) for v in col), 2))
The slowest run took 5.66 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 264 µs per loop
In [82]: %timeit (df.T*(1 << np.arange(df.shape[0]-1, -1, -1))).sum(axis=1)
1000 loops, best of 3: 492 µs per loop
In [83]: %timeit (df.T.dot(1 << np.arange(df.shape[0] - 1, -1, -1)))
The slowest run took 6.14 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 204 µs per loop
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