【python】pandas库常用函数之shift详解

- 首先看一下 df.shift(periods=1, freq=None, axis=0) 的源码解释：

df.shift?
Signature: df.shift(periods=1, freq=None, axis=0)
Docstring:
Shift index by desired number of periods with an optional time freq

Parameters
----------
periods : int
    Number of periods to move, can be positive or negative
freq : DateOffset, timedelta, or time rule string, optional
    Increment to use from the tseries module or time rule (e.g. 'EOM').
    See Notes.
axis : {0 or 'index', 1 or 'columns'}

Notes
-----
If freq is specified then the index values are shifted but the data
is not realigned. That is, use freq if you would like to extend the
index when shifting and preserve the original data.

- 注解：

• period：表示移动的幅度，可以是正数，也可以是负数，默认值是1,1就表示移动一次，注意这里移动的都是数据，而索引是不移动的，移动之后没有对应值的，就赋值为NaN。

• freq： DateOffset, timedelta, or time rule string，可选参数，默认值为None，只适用于时间序列，如果这个参数存在，那么会按照参数值移动时间索引，而数据值没有发生变化。
• axis： 轴向。

- 实例

1、设定 period 与 axis

df = pd.DataFrame(np.arange(16).reshape(4,4),columns=['AA','BB','CC','DD'],index =['a','b','c','d'])

df
Out[14]: 
   AA  BB  CC  DD
a   0   1   2   3
b   4   5   6   7
c   8   9  10  11
d  12  13  14  15
#当period为正时，默认是axis = 0轴的设定，向下移动
df.shift(2)
Out[15]: 
    AA   BB   CC   DD
a  NaN  NaN  NaN  NaN
b  NaN  NaN  NaN  NaN
c  0.0  1.0  2.0  3.0
d  4.0  5.0  6.0  7.0
#当axis=1，沿水平方向进行移动，正数向右移，负数向左移
df.shift(2,axis = 1)
Out[16]: 
   AA  BB    CC    DD
a NaN NaN   0.0   1.0
b NaN NaN   4.0   5.0
c NaN NaN   8.0   9.0
d NaN NaN  12.0  13.0
#当period为负时，默认是axis = 0轴的设定，向上移动
df.shift(-1)
Out[17]: 
     AA    BB    CC    DD
a   4.0   5.0   6.0   7.0
b   8.0   9.0  10.0  11.0
c  12.0  13.0  14.0  15.0
d   NaN   NaN   NaN   NaN

2、freq 参数实例

df = pd.DataFrame(np.arange(16).reshape(4,4),columns=['AA','BB','CC','DD'],index =pd.date_range('6/1/2012','6/4/2012'))

df
Out[38]: 
            AA  BB  CC  DD
2012-06-01   0   1   2   3
2012-06-02   4   5   6   7
2012-06-03   8   9  10  11
2012-06-04  12  13  14  15

df.shift(freq=datetime.timedelta(1))
Out[39]: 
            AA  BB  CC  DD
2012-06-02   0   1   2   3
2012-06-03   4   5   6   7
2012-06-04   8   9  10  11
2012-06-05  12  13  14  15

df.shift(freq=datetime.timedelta(-2))
Out[40]: 
            AA  BB  CC  DD
2012-05-30   0   1   2   3
2012-05-31   4   5   6   7
2012-06-01   8   9  10  11
2012-06-02  12  13  14  15