我有一个df
:
>>> df
sales cash
STK_ID RPT_Date
000568 20120930 80.093 57.488
000596 20120930 32.585 26.177
000799 20120930 14.784 8.157
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并希望将第一行的索引值更改('000568','20120930')
为('000999','20121231')
.最终结果将是:
>>> 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
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怎么做到这一点?
unu*_*tbu 20
使用此设置:
import pandas as pd
import io
text = '''\
STK_ID RPT_Date sales cash
000568 20120930 80.093 57.488
000596 20120930 32.585 26.177
000799 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)
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索引,df.index
可以重新分配给这样的新MultiIndex
:
index = df.index
names = index.names
index = [('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
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或者,可以将索引制作成列,然后可以重新分配列中的值,然后将列返回到索引:
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
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使用IPython进行基准测试%timeit
表明重新分配索引(上面的第一个方法)比重置索引,修改列值,然后再次设置索引要快得多(上面的第二种方法):
In [2]: %timeit reassign_index(df)
10000 loops, best of 3: 158 us per loop
In [3]: %timeit reassign_columns(df)
1000 loops, best of 3: 843 us per loop
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