我正在努力完成这样的任务:我需要从数据框中对列中的值进行离散化,并根据其他列中的值定义bin。
对于最小的工作示例,我们定义一个简单的数据框:
import pandas as pd
df = pd.DataFrame({'A' : ['one', 'one', 'two', 'three'] * 3,'B' : np.random.randn(12)})
数据框如下所示:
A B
0 one 2.5772143847077427
1 one -0.6394141654096013
2 two 0.964652049995486
3 three -0.3922889559403503
4 one 1.6903991754896424
5 one 0.5741442025742018
6 two 0.6300564981683544
7 three 0.9403680915507433
8 one 0.7044433078166983
9 one -0.1695006646595688
10 two 0.06376190217285167
11 three 0.277540580579127
现在我想介绍列C
,它将包含一个bin标签,列A
中的每个值都有不同的bin,即:
(-10,-1,0,1,10)
代表A == 'one'
,(-100,0,100)
代表A == 'two'
,(-999,0,1,2,3)
代表A == 'three'
。所需的输出是:
A B C
0 one 2.5772143847077427 (1, 10]
1 one -0.6394141654096013 (-1, 0]
2 two 0.964652049995486 (0, 100]
3 three -0.3922889559403503 (-999, 0]
4 one 1.6903991754896424 (1, 10]
5 one 0.5741442025742018 (0, 1]
6 two 0.6300564981683544 (0, 100]
7 three 0.9403680915507433 (0, 1]
8 one 0.7044433078166983 (0, 1]
9 one -0.1695006646595688 (-1, 0]
10 two 0.06376190217285167 (0, 100]
11 three 0.277540580579127 (0, 1]
我尝试将pd.cut
或np.digitize
与map
,apply
的不同组合一起使用,但没有成功。
目前我通过拆分框架并分别将pd.cut
应用于每个子集,然后合并以获取帧来实现结果,如下所示:
values_in_column_A = df['A'].unique().tolist()
bins = {'one':(-10,-1,0,1,10),'two':(-100,0,100),'three':(-999,0,1,2,3)}
def binnize(df):
subdf = []
for i in range(len(values_in_column_A)):
subdf.append(df[df['A'] == values_in_column_A[i]])
subdf[i]['C'] = pd.cut(subdf[i]['B'],bins[values_in_column_A[i]])
return pd.concat(subdf)
这很有效,但我认为它不够优雅,我还预计在生产中会出现一些速度或内存问题,因为我将拥有数百万行的帧。说到直接,我想这可以做得更好。
我很感激任何帮助或想法...
答案 0 :(得分:3)
这是否解决了您的问题?
df = pd.DataFrame({'A' : ['one', 'one', 'two', 'three'] * 3,
'B' : np.random.randn(12)})
bins = {'one': (-10,-1,0,1,10), 'two':(-100,0,100), 'three':(-999,0,1,2,3)}
def func(row):
return pd.cut([row['B']], bins=bins[row['A']])[0]
df['C'] = df.apply(func, axis=1)
这会返回一个DataFrame:
A B C
0 one 1.440957 (1, 10]
1 one 0.394580 (0, 1]
2 two -0.039619 (-100, 0]
3 three -0.500325 (-999, 0]
4 one 0.497256 (0, 1]
5 one 0.342222 (0, 1]
6 two -0.968390 (-100, 0]
7 three -0.772321 (-999, 0]
8 one 0.803178 (0, 1]
9 one 0.201513 (0, 1]
10 two 1.178546 (0, 100]
11 three -0.149662 (-999, 0]
更快版本的binnize:
def binize2(df):
df['C'] = ''
for key, values in bins.items():
mask = df['A'] == key
df.loc[mask, 'C'] = pd.cut(df.loc[mask, 'B'], bins=values)
%%timeit
df3 = binnize(df1)
10 loops, best of 3: 56.2 ms per loop
%%timeit
binize2(df2)
100 loops, best of 3: 6.64 ms per loop
这可能是由于它更改了DataFrame并且没有创建新的DataFrame。