pandas 累积和重置为 NaN
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/18196811/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Cumsum reset at NaN
提问by working4coins
If I have a pandas.core.series.Seriesnamed tsof either 1's or NaN's like this:
如果我有一个像这样的 1 或 NaN 的pandas.core.series.Series名字ts：
3382 NaN
3381 NaN
...
3369 NaN
3368 NaN
...
15 1
10 NaN
11 1
12 1
13 1
9 NaN
8 NaN
7 NaN
6 NaN
3 NaN
4 1
5 1
2 NaN
1 NaN
0 NaN
I would like to calculate cumsum of this serie but it should be reset (set to zero) at the location of the NaNs like below:
我想计算这个系列的 cumsum，但它应该在 NaN 的位置重置（设置为零），如下所示：
3382 0
3381 0
...
3369 0
3368 0
...
15 1
10 0
11 1
12 2
13 3
9 0
8 0
7 0
6 0
3 0
4 1
5 2
2 0
1 0
0 0
Ideally I would like to have a vectorized solution !
理想情况下，我想要一个矢量化的解决方案！
I ever see a similar question with Matlab : Matlab cumsum reset at NaN?
我曾经在 Matlab 中看到过类似的问题： Matlab cumsum reset at NaN？
but I don't know how to translate this line d = diff([0 c(n)]);
但我不知道如何翻译这一行 d = diff([0 c(n)]);
采纳答案by emprice
A simple Numpy translation of your Matlab code is this:
Matlab 代码的简单 Numpy 翻译是这样的：
import numpy as np
v = np.array([1., 1., 1., np.nan, 1., 1., 1., 1., np.nan, 1.])
n = np.isnan(v)
a = ~n
c = np.cumsum(a)
d = np.diff(np.concatenate(([0.], c[n])))
v[n] = -d
np.cumsum(v)
Executing this code returns the result array([ 1., 2., 3., 0., 1., 2., 3., 4., 0., 1.]). This solution will only be as valid as the original one, but maybe it will help you come up with something better if it isn't sufficient for your purposes.
执行此代码返回结果array([ 1., 2., 3., 0., 1., 2., 3., 4., 0., 1.])。此解决方案仅与原始解决方案一样有效，但如果它不足以满足您的目的，它可能会帮助您提出更好的解决方案。
回答by Phillip Cloud
Here's a slightly more pandas-onic way to do it:
这是一种稍微更像Pandas的方法：
v = Series([1, 1, 1, nan, 1, 1, 1, 1, nan, 1], dtype=float)
n = v.isnull()
a = ~n
c = a.cumsum()
index = c[n].index # need the index for reconstruction after the np.diff
d = Series(np.diff(np.hstack(([0.], c[n]))), index=index)
v[n] = -d
result = v.cumsum()
Note that either of these requires that you're using pandasat least at 9da899bor newer. If you aren't then you can cast the booldtypeto an int64or float64dtype:
请注意，其中任何一个都要求您pandas至少使用at9da899b或更新版本。如果不是，则可以将booldtype转换为int64or float64dtype：
v = Series([1, 1, 1, nan, 1, 1, 1, 1, nan, 1], dtype=float)
n = v.isnull()
a = ~n
c = a.astype(float).cumsum()
index = c[n].index # need the index for reconstruction after the np.diff
d = Series(np.diff(np.hstack(([0.], c[n]))), index=index)
v[n] = -d
result = v.cumsum()
回答by kadee
Even more pandas-onic way to do it:
更像Pandas的方式来做到这一点：
v = pd.Series([1., 3., 1., np.nan, 1., 1., 1., 1., np.nan, 1.])
cumsum = v.cumsum().fillna(method='pad')
reset = -cumsum[v.isnull()].diff().fillna(cumsum)
result = v.where(v.notnull(), reset).cumsum()
Contrary to the matlab code, this also works for values different from 1.
与 matlab 代码相反，这也适用于不同于 1 的值。
回答by Adam Fuller
If you can accept a similar boolean Series b, try
如果您可以接受类似的布尔系列b，请尝试
(b.cumsum() - b.cumsum().where(~b).fillna(method='pad').fillna(0)).astype(int)
Starting from your Series ts, either b = (ts == 1)or b = ~ts.isnull().
从系列开始ts，无论是b = (ts == 1)或b = ~ts.isnull()。
