[英]reset cumulative sum base on condition pandas and return other cumulative sum
我有这个 dataframe -
counter duration amount
0 1 0.08 1,235
1 2 0.36 1,170
2 3 1.04 1,222
3 4 0.81 1,207
4 5 3.99 1,109
5 6 1.20 1,261
6 7 4.24 1,068
7 8 3.07 1,098
8 9 2.08 1,215
9 10 4.09 1,043
10 11 2.95 1,176
11 12 3.96 1,038
12 13 3.95 1,119
13 14 3.92 1,074
14 15 3.91 1,076
15 16 1.50 1,224
16 17 3.65 962
17 18 3.85 1,039
18 19 3.82 1,062
19 20 3.34 917
我想根据以下逻辑创建另一列:
对于每一行,我想计算“持续时间”的运行总和,但它应该是当前行下方的行的运行总和(领先而不是滞后)。 我想在运行总和达到 5 时停止计算 -> 当它达到 5 时,我想返回“金额”的运行总和(使用相同的逻辑)。
例如,对于“计数器”1,它应该取前 4 行 (0.08+0.36+1.04+0.81<5),然后返回 1,235+1,170+1,222+1,207=4834
对于“计数器”2,它应该只需要 0.36 + 1.04 + 0.81<5 并返回 1,170+1,222+1,207=3599
将不胜感激任何帮助!
让我们尝试使用循环构建您自己的逻辑
c = df.duration.values
v=df.amount.values
result = []
lim=5
check = []
for i in range(len(c)):
total = 0
value = 0
for x, y in zip(v[i:],c[i:]):
total += y
value += x
if total >= lim:
result.append(value-x)
print(total)
break
#result
#[4834, 3599, 2429, 2316, 1109, 1261, 1068, 1098, 1215, 1043, 1176, 1038, 1119, 1074, 1076, 1224, 962, 1039, 1062]
我将首先通过 2 列 go 一次以获得它们的累积总和。
cum_amount = df['amount'].cumsum()
cum_duration = df['duration'].cumsum()
为结果准备好清单
results = []
然后循环遍历每个索引(相当于计数器)
for idx in cum_duration.index:
# keep only rows within `5` and the max. index is where the required numbers are located
wanted_idx = (cum_duration[cum_duration<5]).index.max()
# read those numbers with the wanted index
results.append({'idx': idx, 'cum_duration': cum_duration[wanted_idx], 'cum_amount': cum_amount[wanted_idx]})
# subtract the lag (we need only the leads not the lags)
cum_amount -= cum_amount[idx]
cum_duration -= cum_duration[idx]
最后的结果是 DataFrame。
pd.DataFrame(results)
idx cum_duration cum_amount
0 0 2.29 4834.0
1 1 2.21 3599.0
2 2 1.85 2429.0
3 3 4.80 2316.0
4 4 3.99 1109.0
5 5 1.20 1261.0
6 6 4.24 1068.0
7 7 3.07 1098.0
8 8 2.08 1215.0
9 9 4.09 1043.0
10 10 2.95 1176.0
11 11 3.96 1038.0
12 12 3.95 1119.0
13 13 3.92 1074.0
14 14 3.91 1076.0
15 15 1.50 1224.0
16 16 3.65 962.0
17 17 3.85 1039.0
18 18 3.82 1062.0
19 19 3.34 917.0
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.