我试图通过第一个索引组来移动Pandas数据帧列数据。这是演示代码：
In [8]: df = mul_df(5,4,3)
In [9]: df
Out[9]:
COL000 COL001 COL002
STK_ID RPT_Date
A0000 B000 -0.5505 0.7445 -0.3645
B001 0.9129 -1.0473 -0.5478
B002 0.8016 0.0292 0.9002
B003 2.0744 -0.2942 -0.7117
A0001 B000 0.7064 0.9636 0.2805
B001 0.4763 0.2741 -1.2437
B002 1.1563 0.0525 -0.7603
B003 -0.4334 0.2510 -0.0105
A0002 B000 -0.6443 0.1723 0.2657
B001 1.0719 0.0538 -0.0641
B002 0.6787 -0.3386 0.6757
B003 -0.3940 -1.2927 0.3892
A0003 B000 -0.5862 -0.6320 0.6196
B001 -0.1129 -0.9774 0.7112
B002 0.6303 -1.2849 -0.4777
B003 0.5046 -0.4717 -0.2133
A0004 B000 1.6420 -0.9441 1.7167
B001 0.1487 0.1239 0.6848
B002 0.6139 -1.9085 -1.9508
B003 0.3408 -1.3891 0.6739
In [10]: grp = df.groupby(level=df.index.names[0])
In [11]: grp.shift(1)
Out[11]:
COL000 COL001 COL002
STK_ID RPT_Date
A0000 B000 NaN NaN NaN
B001 -0.5505 0.7445 -0.3645
B002 0.9129 -1.0473 -0.5478
B003 0.8016 0.0292 0.9002
A0001 B000 NaN NaN NaN
B001 0.7064 0.9636 0.2805
B002 0.4763 0.2741 -1.2437
B003 1.1563 0.0525 -0.7603
A0002 B000 NaN NaN NaN
B001 -0.6443 0.1723 0.2657
B002 1.0719 0.0538 -0.0641
B003 0.6787 -0.3386 0.6757
A0003 B000 NaN NaN NaN
B001 -0.5862 -0.6320 0.6196
B002 -0.1129 -0.9774 0.7112
B003 0.6303 -1.2849 -0.4777
A0004 B000 NaN NaN NaN
B001 1.6420 -0.9441 1.7167
B002 0.1487 0.1239 0.6848
B003 0.6139 -1.9085 -1.9508
此处附有mul_df()代码：How to speed up Pandas multilevel dataframe sum?
现在我希望grp.shift(1)获得一个大数据帧。
In [1]: df = mul_df(5000,30,400)
In [2]: grp = df.groupby(level=df.index.names[0])
In [3]: timeit grp.shift(1)
1 loops, best of 3: 5.23 s per loop
5.23s太慢了。如何加快速度？
（我的电脑配置为：Pentium Dual-Core T4200@2.00GHZ，3.00GB RAM，WindowXP，Python 2.7.4，Numpy 1.7.1，Pandas 0.11.0，numexpr 2.0.1，Anaconda 1.5.0（32）位））
答案 0 :(得分：5)
如何移动整个DataFrame对象，然后将每个组的第一行设置为NaN？
dfs = df.shift(1)
dfs.iloc[df.groupby(level=0).size().cumsum()[:-1]] = np.nan
答案 1 :(得分：4)
问题是shift操作不是cython优化的，所以它涉及到python的回调。与此相比：
In [84]: %timeit grp.shift(1)
1 loops, best of 3: 1.77 s per loop
In [85]: %timeit grp.sum()
1 loops, best of 3: 202 ms per loop
答案 2 :(得分：0)
类似的问题和补充的答案适用于任何方向和幅度的转变：pandas: setting last N rows of multi-index to Nan for speeding up groupby with shift
代码（包括测试设置）是：
#
# the function to use in apply
#
def replace_shift_overlap(grp,col,N,value):
if (N > 0):
grp[col][:N] = value
else:
grp[col][N:] = value
return grp
length = 5
groups = 3
rng1 = pd.date_range('1/1/1990', periods=length, freq='D')
frames = []
for x in xrange(0,groups):
tmpdf = pd.DataFrame({'date':rng1,'category':int(10000000*abs(np.random.randn())),'colA':np.random.randn(length),'colB':np.random.randn(length)})
frames.append(tmpdf)
df = pd.concat(frames)
df.sort(columns=['category','date'],inplace=True)
df.set_index(['category','date'],inplace=True,drop=True)
shiftBy=-1
df['tmpShift'] = df['colB'].shift(shiftBy)
#
# the apply
#
df = df.groupby(level=0).apply(replace_shift_overlap,'tmpShift',shiftBy,np.nan)
# Yay this is so much faster.
df['newColumn'] = df['tmpShift'] / df['colA']
df.drop('tmpShift',1,inplace=True)
编辑：请注意，初始排序真的会影响到它的有效性。所以在某些情况下，原始答案更有效。
答案 3 :(得分：0)
尝试一下：
import numpy as np
import pandas as pd
df = pd.DataFrame({'A': [10, 20, 15, 30, 45,43,67,22,12,14,54],
'B': [13, 23, 18, 33, 48, 1,7, 56,66,45,32],
'C': [17, 27, 22, 37, 52,77,34,21,22,90,8],
'D': ['a','a','a','a','b','b','b','c','c','c','c']
})
df
#> A B C D
#> 0 10 13 17 a
#> 1 20 23 27 a
#> 2 15 18 22 a
#> 3 30 33 37 a
#> 4 45 48 52 b
#> 5 43 1 77 b
#> 6 67 7 34 b
#> 7 22 56 21 c
#> 8 12 66 22 c
#> 9 14 45 90 c
#> 10 54 32 8 c
def groupby_shift(df, col, groupcol, shift_n, fill_na = np.nan):
'''df: dataframe
col: column need to be shifted
groupcol: group variable
shift_n: how much need to shift
fill_na: how to fill nan value, default is np.nan
'''
rowno = list(df.groupby(groupcol).size().cumsum())
lagged_col = df[col].shift(shift_n)
na_rows = [i for i in range(shift_n)]
for i in rowno:
if i == rowno[len(rowno)-1]:
continue
else:
new = [i + j for j in range(shift_n)]
na_rows.extend(new)
na_rows = list(set(na_rows))
na_rows = [i for i in na_rows if i <= len(lagged_col) - 1]
lagged_col.iloc[na_rows] = fill_na
return lagged_col
df['A_lag_1'] = groupby_shift(df, 'A', 'D', 1)
df
#> A B C D A_lag_1
#> 0 10 13 17 a NaN
#> 1 20 23 27 a 10.0
#> 2 15 18 22 a 20.0
#> 3 30 33 37 a 15.0
#> 4 45 48 52 b NaN
#> 5 43 1 77 b 45.0
#> 6 67 7 34 b 43.0
#> 7 22 56 21 c NaN
#> 8 12 66 22 c 22.0
#> 9 14 45 90 c 12.0
#> 10 54 32 8 c 14.0