考虑到 python 中的 pandas 数据帧中的组，如何计算累积和直到达到阈值并在达到阈值后重置它？

Question

我有一个这样的数据框：

import pandas as pd
import numpy as np

data={'trip':[1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3],
      'timestamps':[1235471761, 1235471763, 1235471765, 1235471767, 1235471770, 1235471772, 1235471776, 1235471779, 1235471780, 1235471789,1235471792,1235471793,1235471829,1235471833,1235471835,1235471838,1235471844,1235471847,1235471848,1235471852,1235471855,1235471859,1235471900,1235471904,1235471911,1235471913]}

df = pd.DataFrame(data)
df['TimeDistance'] = df.groupby('trip')['timestamps'].diff(1)
df

我正在寻找的是从“TimeDistance”列中的第一行（将其视为原点）开始，对其值进行累积求和，每当此求和达到 10 时，重新启动 cumsum 并继续此过程直到结束行程（正如您在此数据框中看到的，我们在“行程”列中有 3 次行程）。 我想要一个新列中的所有累计总和，比方说，“cumu”列。 另一个重要问题是，在达到阈值后，“cumu”列中阈值之后的下一行必须为零，并且求和再次从这个新原点重新开始。

Answer 1

我希望我已经理解你的问题了。您可以将生成器与

.send()

一起使用：

def my_accumulate(maxval):
    val = 0
    yield
    while True:
        if val < maxval:
            val += yield val
        else:
            yield val
            val = 0


def fn(x):
    a = my_accumulate(10)
    next(a)
    x["cumu"] = [a.send(v) for v in x["TimeDistance"]]
    return x


df = df.groupby("trip").apply(fn)
print(df)

印花：

    trip  timestamps  TimeDistance  cumu
0      1  1235471761           NaN   0.0
1      1  1235471763           2.0   2.0
2      1  1235471765           2.0   4.0
3      1  1235471767           2.0   6.0
4      1  1235471770           3.0   9.0
5      1  1235471772           2.0  11.0
6      1  1235471776           4.0   0.0
7      1  1235471779           3.0   3.0
8      1  1235471780           1.0   4.0
9      1  1235471789           9.0  13.0
10     1  1235471792           3.0   0.0
11     1  1235471793           1.0   1.0
12     2  1235471829           NaN   0.0
13     2  1235471833           4.0   4.0
14     2  1235471835           2.0   6.0
15     2  1235471838           3.0   9.0
16     2  1235471844           6.0  15.0
17     2  1235471847           3.0   0.0
18     2  1235471848           1.0   1.0
19     2  1235471852           4.0   5.0
20     2  1235471855           3.0   8.0
21     2  1235471859           4.0  12.0
22     3  1235471900           NaN   0.0
23     3  1235471904           4.0   4.0
24     3  1235471911           7.0  11.0
25     3  1235471913           2.0   0.0

---

另一个解决方案：

df = df.groupby("trip").apply(
    lambda x: x.assign(
        cumu=(
            val := 0,
            *(
                val := val + v if val < 10 else (val := 0)
                for v in x["TimeDistance"][1:]
            ),
        )
    ),
)
print(df)

Answer 2

Andrej 的回答更好，因为我的可能效率不高，这取决于

df

按行程订购，而

TimeDistance

是

nan

作为每次行程的第一个值。

cummulative_sum = 0
df['cumu'] = 0

for i in range(len(df)):
    if np.isnan(df.loc[i,'TimeDistance']) or cummulative_sum >= 10:
        cummulative_sum = 0
        df.loc[i, 'cumu'] = 0
    else:
        cummulative_sum += df.loc[i,'TimeDistance']
        df.loc[i, 'cumu'] = cummulative_sum

print(df)

输出：

    trip  timestamps  TimeDistance  cumu
0      1  1235471761           NaN     0
1      1  1235471763           2.0     2
2      1  1235471765           2.0     4
3      1  1235471767           2.0     6
4      1  1235471770           3.0     9
5      1  1235471772           2.0    11
6      1  1235471776           4.0     0
7      1  1235471779           3.0     3
8      1  1235471780           1.0     4
9      1  1235471789           9.0    13
10     1  1235471792           3.0     0
11     1  1235471793           1.0     1
12     2  1235471829           NaN     0
13     2  1235471833           4.0     4
14     2  1235471835           2.0     6
15     2  1235471838           3.0     9
16     2  1235471844           6.0    15
17     2  1235471847           3.0     0
18     2  1235471848           1.0     1
19     2  1235471852           4.0     5
20     2  1235471855           3.0     8
21     2  1235471859           4.0    12
22     3  1235471900           NaN     0
23     3  1235471904           4.0     4
24     3  1235471911           7.0    11
25     3  1235471913           2.0     0