python - 如何对 pandas 中的浮点值进行二值化？

Question

我有这样的 float 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

这意味着每行可以连续一次具有值 1，其余全部为零(最大值变为 1)。

如何在 pandas 或 python 中做到这一点？我知道 pandas 中的 get_dummies 是可行的方法，但它不起作用。

如果可以的话请帮忙。

Answer 1

我认为你可以使用rank，然后将其与df1的最大值进行比较。最后将 astype 的 bool DataFrame 转换为 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:

我认为您可以使用 eq 与 df 的 max 进行按行比较，最后由 astype 转换为 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

时间

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