如何将 pandas 中的浮点值二值化？

Question

我有这样的浮点数据，它是由 3 个神经元产生的神经网络输出。我想根据最大行值转换为二进制分类标签（互斥）。

0.423201  0.368718 0.338091
 0.246899  0.437535 0.000262
 0.978685 0.136219  0.027693

输出应该是

1 0 0
0 1 0
1 0 0

这意味着每一行可以有值 1 连续一次，其余全部为零（最大值变为 1）。

如何在pandas或python中完成？我知道 pandas 中的 get_dummies 是可行的方法，但它不起作用。

如果可以请帮忙

Answer 1

我想你可以使用 rank and then compare it with max value of df1. Last convert boolean DataFrame by astype 到 int:

print df
          0         1         2
0  0.423201  0.368718  0.338091
1  0.246899  0.437535  0.000262
2  0.978685  0.136219  0.027693

df1 = df.rank(method='max', axis=1)
print df1
   0  1  2
0  3  2  1
1  2  3  1
2  3  2  1

#get max value of df1
ma = df1.max().max()
print ma
3.0

print (df1 == ma)
       0      1      2
0   True  False  False
1  False   True  False
2   True  False  False

print (df1 == ma).astype(int)
   0  1  2
0  1  0  0
1  0  1  0
2  1  0  0

编辑:

我想你可以使用 eq for comparing by rows with max of df and last convert by astype 到 int:

print df.max(axis=1)
0    10
1     8
2     9
dtype: int64

print df.eq(df.max(axis=1), axis=0).astype(int)
   0  1  2
0  1  0  0
1  0  1  0
2  1  0  0

计时

len(df) = 3:

In [418]: %timeit df.eq(df.max(axis=1), axis=0).astype(int)
The slowest run took 5.44 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 334 µs per loop

In [419]: %timeit df.apply(lambda x: x == x.max(), axis='columns').astype(int)
The slowest run took 4.49 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 1.44 ms per loop

In [420]: %timeit (df.rank(method='max', axis=1) == df.rank(method='max', axis=1).max().max()).astype(int)
The slowest run took 4.83 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 656 µs per loop

len(df) = 3000:

In [426]: %timeit df.eq(df.max(axis=1), axis=0).astype(int)
The slowest run took 5.44 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 456 µs per loop

In [427]: %timeit df.apply(lambda x: x == x.max(), axis='columns').astype(int)
1 loops, best of 3: 496 ms per loop

In [428]: %timeit (df.rank(method='max', axis=1) == df.rank(method='max', axis=1).max().max()).astype(int)
The slowest run took 4.50 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 1.32 ms per loop

Answer 2

我认为这样会更简单、更快。

df.apply(lambda x: x == x.max(), axis='columns').astype(int)

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