Article Outline
情景描述：
最新，项目中涉及到评分卡操作，评分项目有大概几十项，每项基本都是按频次区间给一个分数，最后，累计所有项目的分数得出最后所需要的分数。
<!--more-->
demo1：
import pandas as pd
import numpy as np
df = pandas.DataFrame({"a": np.random.random(100), "b": np.random.random(100), "id": np.arange(100)})
# Bin the data frame by "a" with 10 bins...
bins = np.linspace(df.a.min(), df.a.max(), 10)
# array([ 0.00282977, 0.11097259, 0.2191154 , 0.32725822, 0.43540104,
0.54354385, 0.65168667, 0.75982948, 0.8679723 , 0.97611511])
# bins = np.linspace(0, 1, 11) # 优化版
# array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ])
groups = df.groupby(np.digitize(df.a, bins))
# Get the mean of each bin:
print groups.mean() # Also could do "groups.aggregate(np.mean)"
>>>output:
a b id
1 0.044003 0.525964 56.307692
2 0.167568 0.506078 55.454545
3 0.268109 0.510612 44.636364
4 0.375014 0.544154 69.833333
5 0.481702 0.590031 48.500000
6 0.599587 0.488921 38.076923
7 0.696548 0.643555 50.642857
8 0.830064 0.620650 50.571429
9 0.928396 0.545460 44.166667
10 0.976115 0.693051 28.000000
# Similarly, the median:
print groups.median()
>>>output:
a b id
1 0.028901 0.536857 61.0
2 0.167054 0.557716 49.0
3 0.267337 0.534911 43.0
4 0.374787 0.487063 73.0
5 0.480395 0.737007 49.5
6 0.603701 0.676479 42.0
7 0.695939 0.689144 57.5
8 0.836665 0.690757 41.0
9 0.924245 0.646487 47.0
10 0.976115 0.693051 28.0
# Apply some arbitrary function to aggregate binned data
print groups.aggregate(lambda x: np.mean(x[x > 0.5]))
>>>output:
a b id
1 NaN 0.671236 56.307692
2 NaN 0.704379 55.454545
3 NaN 0.768609 44.636364
4 NaN 0.804354 69.833333
5 0.514166 0.796151 48.500000
6 0.599587 0.755381 41.250000
7 0.696548 0.779524 50.642857
8 0.830064 0.766095 50.571429
9 0.928396 0.902529 44.166667
10 0.976115 0.693051 28.000000demo2 :
import numpy as np
import pandas
df = pandas.DataFrame({"a": np.random.random(100),
"b": np.random.random(100) + 10})
# Bin the data frame by "a" with 10 bins...
bins = np.linspace(df.a.min(), df.a.max(), 10)
groups = df.groupby(pandas.cut(df.a, bins))
# Get the mean of b, binned by the values in a
print groups.mean().b
>>>output:
a
(0.000351, 0.11] 10.596542
(0.11, 0.22] 10.690010
(0.22, 0.33] 10.250080
(0.33, 0.44] 10.546134
(0.44, 0.549] 10.471454
(0.549, 0.659] 10.455624
(0.659, 0.769] 10.501616
(0.769, 0.879] 10.588354
(0.879, 0.989] 10.461848
Name: b, dtype: float64
demo3:
In [144]: df = DataFrame({"a": np.random.random(100), "b": np.random.random(100), "id": np.arange(100)})
In [145]: bins = [0, .25, .5, .75, 1]
In [146]: a_bins = df.a.groupby(cut(df.a,bins))
In [147]: b_bins = df.b.groupby(cut(df.b,bins))
In [148]: a_bins.agg([mean,median])
Out[148]:
mean median
a
(0, 0.25] 0.124173 0.114613
(0.25, 0.5] 0.367703 0.358866
(0.5, 0.75] 0.624251 0.626730
(0.75, 1] 0.875395 0.869843
In [149]: b_bins.agg([mean,median])
Out[149]:
mean median
b
(0, 0.25] 0.147936 0.166900
(0.25, 0.5] 0.394918 0.386729
(0.5, 0.75] 0.636111 0.655247
(0.75, 1] 0.851227 0.838805demo4:
import pandas as pd
import numpy as np
In [168]: filter_values = [0, 5, 17, 33]
In [169]: df = pd.DataFrame(np.random.random(100)*100, columns = ['filtercol'])
In [170]: out = pd.cut(df.filtercol, bins = filter_values)
In [171]: counts = pd.value_counts(out)
Out[171]:
(17, 33] 16
(5, 17] 11
(0, 5] 5
# 排序
counts = counts.reindex(out.cat.categories)
counts = counts.sort_index()
In [172]: counts = counts.reindex(out.cat.categories)
In [173]: counts
Out[173]:
(0, 5] 5
(5, 17] 11
(17, 33] 16
Name: filtercol, dtype: int64
# 重置索引(reset index)
out = counts.reset_index(drop=True) # counts 不变
counts.reset_index(drop=True, inplace=True) # 直接改变counts
In [174]: out = counts.reset_index(drop=True)
In [175]: out
Out[175]:
0 5
1 11
2 16
Name: filtercol, dtype: int64
In [176]: counts
Out[176]:
(0, 5] 5
(5, 17] 11
(17, 33] 16
Name: filtercol, dtype: int64
In [177]: counts.reset_index(drop=True, inplace=True)
In [178]: counts
Out[178]:
0 5
1 11
2 16
Name: filtercol, dtype: int64总结：
第四个demo基本就可以完成当前目标了； 后续需要操作的是封装一个合适的通用方法，将每个项目评分标准代入即可。