给pd.DataFrame
和0.0 < values < 1.0
,我想根据定义的阈值0
将其转换为二进制值1
/ eps = 0.5
,
0 1 2
0 0.35 0.20 0.81
1 0.41 0.75 0.59
2 0.62 0.40 0.94
3 0.17 0.51 0.29
现在,我只有这个for loop
,对于大型数据集,它需要花费很长时间:
import numpy as np
import pandas as pd
data = np.array([[.35, .2, .81],[.41, .75, .59],
[.62, .4, .94], [.17, .51, .29]])
df = pd.DataFrame(data, index=range(data.shape[0]), columns=range(data.shape[1]))
eps = .5
b = np.zeros((df.shape[0], df.shape[1]))
for i in range(df.shape[0]):
for j in range(df.shape[1]):
if df.loc[i,j] < eps:
b[i,j] = 0
else:
b[i,j] = 1
df_bin = pd.DataFrame(b, columns=df.columns, index=df.index)
有人知道转换为二进制值的更有效方法吗?
0 1 2
0 0.0 0.0 1.0
1 0.0 1.0 1.0
2 1.0 0.0 1.0
3 0.0 1.0 0.0
谢谢
答案 0 :(得分:6)
df.round
>>> df.round()
np.round
>>> np.round(df)
astype
>>> df.ge(0.5).astype(int)
所有产生
0 1 2
0 0.0 0.0 1.0
1 0.0 1.0 1.0
2 1.0 0.0 1.0
3 0.0 1.0 0.0
注意:round
在这里起作用是因为它会自动在两个整数之间设置.5
的阈值。对于自定义阈值,请使用第三种解决方案
答案 1 :(得分:3)
答案 2 :(得分:3)
由于我们有很多答案,而且都使用不同的方法,所以我对速度比较感到好奇。以为我分享:
# create big test dataframe
dfbig = pd.concat([df]*200000, ignore_index=True)
print(dfbig.shape)
(800000, 3)
# pandas round()
%%timeit
dfbig.round()
101 ms ± 4.63 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# numpy round()
%%timeit
np.round(dfbig)
104 ms ± 2.71 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# pandas .ge & .astype
%%timeit
dfbig.ge(0.5).astype(int)
9.32 ms ± 170 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# numpy.where
%%timeit
np.where(dfbig<0.5, 0, 1)
21.5 ms ± 421 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
结局:
ge
和astype
np.where
np.round
round