
使用此设置:
import pandas as pdimport iotext = '''STK_ID RPT_Date sales cash000568 20120930 80.093 57.488000596 20120930 32.585 26.177000799 20120930 14.784 8.157'''df = pd.read_csv(io.BytesIO(text), delimiter = ' ', converters = {0:str})df.set_index(['STK_ID','RPT_Date'], inplace = True)df.index可以
MultiIndex像这样将索引重新分配给新的索引:
index = df.indexnames = index.namesindex = [('000999','20121231')] + df.index.tolist()[1:]df.index = pd.MultiIndex.from_tuples(index, names = names)print(df)# sales cash# STK_ID RPT_Date # 000999 20121231 80.093 57.488# 000596 20120930 32.585 26.177# 000799 20120930 14.784 8.157或者,可以将索引划分为列,然后可以重新分配列中的值,然后将列返回索引:
df.reset_index(inplace = True)df.ix[0, ['STK_ID', 'RPT_Date']] = ('000999','20121231')df = df.set_index(['STK_ID','RPT_Date'])print(df)# sales cash# STK_ID RPT_Date # 000999 20121231 80.093 57.488# 000596 20120930 32.585 26.177# 000799 20120930 14.784 8.157使用IPython进行基准测试
%timeit建议,重新分配索引(上面的第一种方法)比重置索引,修改列值然后再次设置索引(上面的第二种方法)要快得多:
In [2]: %timeit reassign_index(df)10000 loops, best of 3: 158 us per loopIn [3]: %timeit reassign_columns(df)1000 loops, best of 3: 843 us per loop
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