繁体   English   中英

获取连续两行之间的工作日

[英]Get business days between consecutive rows pandas

很简单的问题。

我知道:

df.diff()

给了我之间的日子,我知道我可以和

df.loc[df.Date.weekday == 4, 'Diff'] = 1

但这不是最佳选择。 我试过了

np.busday_count()

但是遇到一个我不太了解的错误。 这是带有该错误的示例代码:

In [36]: df = pd.DataFrame.from_dict({1: {'Date': '2016-01-01'}, 2: {'Date': '2016-01-02'}, 3: {'Date': '2016-01-03'}}, orient='index')

In [37]: df['Date'] = df.Date.astype('<M8[D]')

In [38]: np.busday_count(df.Date, df.Date.shift(1))
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-38-07a4ae9a16f6> in <module>()
----> 1 np.busday_count(df.Date, df.Date.shift(1))

TypeError: Iterator operand 0 dtype could not be cast from dtype('<M8[ns]') to dtype('<M8[D]') according to the rule 'safe'

In [39]: df = pd.DataFrame.from_dict({1: {'Date': '2016-01-01'}, 2: {'Date': '2016-01-02'}, 3: {'Date': '2016-01-03'}}, orient='index')

In [40]: np.busday_count(df.Date, df.Date.shift(1))
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-40-07a4ae9a16f6> in <module>()
----> 1 np.busday_count(df.Date, df.Date.shift(1))

TypeError: Iterator operand or requested dtype holds references, but the REFS_OK flag was not enabled

使用np.busday_count您也可以尝试:

x3 = [x.strftime('%Y-%m-%d') for x in df.Date]
x4 = [x.strftime('%Y-%m-%d') for x in df.Date.shift(1).fillna(0)]
np.busday_count(x4,x3)
array([12001,     1,     0])

%timeit np.busday_count(x4,x3)
The slowest run took 4.58 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 12.5 µs per loop

或者,如果您希望:

x1 = [x.date() for x in df.Date]
x2 = [x.date() for x in df.Date.shift(1).fillna(0)]
np.busday_count(x2,x1)
array([12001,     1,     0])

%timeit np.busday_count(x2,x1)
10000 loops, best of 3: 43.4 µs per loop

得到它了!

因此,我不知道这是否适合每个人的需求,但这是可行的:

np.busday_count(df.Date.values.tolist(), df.Date.shift(1).fillna(df.Date).values.tolist())

因此,添加tolist()和.fillna()部分都是必要的!

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM