big*_*bug 4 python normalization pandas
熊猫数据框:
>>> df
sales net_pft
STK_ID RPT_Date
002138 20140930 3.325 0.607
20150930 3.619 0.738
20160930 4.779 0.948
600004 20140930 13.986 2.205
20150930 14.226 3.080
20160930 15.499 3.619
600660 20140930 31.773 5.286
20150930 31.040 6.333
20160930 40.062 7.186
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只想知道如何获得输出,因为每行的值除以每组的第一行,如下所示:
sales net_pft
STK_ID RPT_Date
002138 20140930 1.000 1.000
20150930 1.088 1.216
20160930 1.437 1.562
600004 20140930 1.000 1.000
20150930 1.017 1.397
20160930 1.108 1.641
600660 20140930 1.000 1.000
20150930 0.977 1.198
20160930 1.261 1.359
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谢谢,
import pandas as pd
df = pd.DataFrame({'RPT_Date': ['20140930', '20150930', '20160930', '20140930', '20150930', '20160930', '20140930', '20150930', '20160930'], 'STK_ID': ['002138', '002138', '002138', '600004', '600004', '600004', '600660', '600660', '600660'], 'net_pft': [0.607, 0.738, 0.948, 2.205, 3.080, 3.619, 5.286, 6.333, 7.186], 'sales': [3.325, 3.619, 4.779, 13.986, 14.226, 15.499, 31.773, 31.040, 40.062]})
df = df.set_index(['STK_ID','RPT_Date'])
firsts = (df.groupby(level=['STK_ID']).transform('first'))
result = df / firsts
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产量
net_pft sales
STK_ID RPT_Date
002138 20140930 1.000000 1.000000
20150930 1.215815 1.088421
20160930 1.561779 1.437293
600004 20140930 1.000000 1.000000
20150930 1.396825 1.017160
20160930 1.641270 1.108180
600660 20140930 1.000000 1.000000
20150930 1.198070 0.976930
20160930 1.359440 1.260882
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上面的主要技巧是使用groupby/transform('first')创建一个与形状相同df但其值来自每个组的第一行的DataFrame :
firsts = df.groupby(level=['STK_ID']).transform('first')
# net_pft sales
# STK_ID RPT_Date
# 002138 20140930 0.607 3.325
# 20150930 0.607 3.325
# 20160930 0.607 3.325
# 600004 20140930 2.205 13.986
# 20150930 2.205 13.986
# 20160930 2.205 13.986
# 600660 20140930 5.286 31.773
# 20150930 5.286 31.773
# 20160930 5.286 31.773
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尽管这是对内存的大量使用,但这可能是获得所需结果的最快方法,因为它避免了在 Python 中循环遍历组。
如果上面的代码TypeError: Transform function invalid for data types在 Pandas 0.13 版本中引发了 a ,您可以尝试使用以下解决方法:
result = list()
for key, grp in df.groupby(level=['STK_ID']):
result.append(grp/grp.iloc[0])
result = pd.concat(result)
print(result)
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