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)
索引,df.index
可以MultiIndex
像这样重新分配给一个新的:
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,可以将索引设置为列,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
使用 IPython 进行基准测试%timeit
表明重新分配索引(上面的第一种方法)比重新设置索引、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 loop
In [3]: %timeit reassign_columns(df)
1000 loops, best of 3: 843 us per loop