如何更改 pandas 数据框索引值? [英] How to change Pandas dataframe index value?

问题描述

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

并且想要将第一行的索引值从('000568','20120930')更改为('000999','20121231').最终结果将是:

And want to change first row's index value from ('000568','20120930') to ('000999','20121231'). Final result will be:

>>> 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

如何实现?

推荐答案

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)

可以像这样将索引df.index重新分配给新的MultiIndex:

The index, df.index can be reassigned to a new MultiIndex like this:

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

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:

Or, the index could be made into columns, the values in the columns could be then reassigned, and then the columns returned to indices:

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 %timeit进行基准测试建议，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:

---

Benchmarking with IPython %timeit suggests reassigning the index (the first method, above) is significantly faster than resetting the index, modifying column values, and then setting the index again (the second method, above):

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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