[英]If row below duplicate, use value from other columns until a new value is found
我有一个棘手的数据操作问题。 基本上，我有一个日期列表。 每天都会统计有多少未解决的问题。 我想创建一个新列ideal_issues_left ，它使用np.linspace来计算理想的剩余问题数，如果它们每天都以稳定的速度完成，在日期范围结束时为零。
我设法从每个起点创建了每天 dataframe 的估计值，但我现在要做的是根据以下逻辑用估计值填充ideal_issues_left列：
ideal_issues_left 。ideal_issues_left ，直到达到新的open_issues数量。例如，假设这是日期范围和未解决的问题：
import pandas as pd
chart_data = pd.DataFrame({
'date': pd.date_range('2018-08-19', '2018-09-01', freq='d'),
'open_issues': [23.0, 25.0, 26.0, 26.0, 28.0, 36.0, 33.0, 39.0, 39.0, 38.0, 38.0, 38.0, 38.0, 38.0]
})
chart_data
date open_issues
0 2020-08-19 23.0
1 2020-08-20 25.0
2 2020-08-21 26.0
3 2020-08-22 26.0
4 2020-08-23 28.0
5 2020-08-24 36.0
6 2020-08-25 33.0
7 2020-08-26 39.0
8 2020-08-27 39.0
9 2020-08-28 38.0
10 2020-08-29 38.0
11 2020-08-30 38.0
12 2020-08-31 38.0
13 2020-09-01 38.0
p = []
for day, val in enumerate(chart_data.loc[:, 'open_issues']):
days_left = 14 - day
p.append(np.linspace(start=val, stop=0, num=days_left))
estimates = pd.DataFrame(p)
estimates
0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 23.0 21.230769 19.461538 17.692308 15.923077 14.153846 12.384615 10.615385 8.846154 7.076923 5.307692 3.538462 1.769231 0.0
1 25.0 22.916667 20.833333 18.750000 16.666667 14.583333 12.500000 10.416667 8.333333 6.250000 4.166667 2.083333 0.000000 NaN
2 26.0 23.636364 21.272727 18.909091 16.545455 14.181818 11.818182 9.454545 7.090909 4.727273 2.363636 0.000000 NaN NaN
3 26.0 23.400000 20.800000 18.200000 15.600000 13.000000 10.400000 7.800000 5.200000 2.600000 0.000000 NaN NaN NaN
4 28.0 24.888889 21.777778 18.666667 15.555556 12.444444 9.333333 6.222222 3.111111 0.000000 NaN NaN NaN NaN
5 36.0 31.500000 27.000000 22.500000 18.000000 13.500000 9.000000 4.500000 0.000000 NaN NaN NaN NaN NaN
6 33.0 28.285714 23.571429 18.857143 14.142857 9.428571 4.714286 0.000000 NaN NaN NaN NaN NaN NaN
7 39.0 32.500000 26.000000 19.500000 13.000000 6.500000 0.000000 NaN NaN NaN NaN NaN NaN NaN
8 39.0 31.200000 23.400000 15.600000 7.800000 0.000000 NaN NaN NaN NaN NaN NaN NaN NaN
9 38.0 28.500000 19.000000 9.500000 0.000000 NaN NaN NaN NaN NaN NaN NaN NaN NaN
10 38.0 25.333333 12.666667 0.000000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
11 38.0 19.000000 0.000000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
12 38.0 0.000000 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
13 38.0 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
期望的最终结果应该是：
chart_data
date open_issues ideal_issues_left
0 2018-08-19 23.0 23.0
1 2018-08-20 25.0 25.0
2 2018-08-21 26.0 26.0 # <- this value is from estimates row 2 col 0
3 2018-08-22 26.0 23.6 # <- this value is from estimates row 2 col 1
4 2018-08-23 28.0 28.0
5 2018-08-24 36.0 36.0
6 2018-08-25 33.0 33.0
7 2018-08-26 39.0 39.0 # <- this value is from estimates row 7 col 0
8 2018-08-27 39.0 32.5 # <- this value is from estimates row 7 col 1
9 2018-08-28 38.0 38.0 # <- this value is from estimates row 9 col 0
10 2018-08-29 38.0 28.5 # <- this value is from estimates row 9 col 1
11 2018-08-30 38.0 19.0 # <- this value is from estimates row 9 col 2
12 2018-08-31 38.0 9.5 # <- this value is from estimates row 9 col 3
13 2018-09-01 38.0 0.0 # <- this value is from estimates row 9 col 4
谢谢！
如果存在相同数量的问题，则从累积总数的总和中获取累积计数。 该值使用在参考数据框中相同数量的问题中提取的数据进行更新。
chart_data['flg'] = chart_data['open_issues'].groupby(((chart_data['open_issues'] != chart_data['open_issues'].shift())).cumsum()).cumcount()
chart_data
date open_issues flg
0 2018-08-19 23.0 0
1 2018-08-20 25.0 0
2 2018-08-21 26.0 0
3 2018-08-22 26.0 1
4 2018-08-23 28.0 0
5 2018-08-24 36.0 0
6 2018-08-25 33.0 0
7 2018-08-26 39.0 0
8 2018-08-27 39.0 1
9 2018-08-28 38.0 0
10 2018-08-29 38.0 1
11 2018-08-30 38.0 2
12 2018-08-31 38.0 3
13 2018-09-01 38.0 4
for i,issues in enumerate(chart_data['open_issues']):
k = chart_data.loc[i,'flg']
df = estimates[estimates[0] == issues]
l = df.iloc[:1, k].values
# print(l)
chart_data.loc[i,'idea_issues_left'] = l
chart_data
date open_issues flg idea_issues_left
0 2018-08-19 23.0 0 23.000000
1 2018-08-20 25.0 0 25.000000
2 2018-08-21 26.0 0 26.000000
3 2018-08-22 26.0 1 23.636364
4 2018-08-23 28.0 0 28.000000
5 2018-08-24 36.0 0 36.000000
6 2018-08-25 33.0 0 33.000000
7 2018-08-26 39.0 0 39.000000
8 2018-08-27 39.0 1 32.500000
9 2018-08-28 38.0 0 38.000000
10 2018-08-29 38.0 1 28.500000
11 2018-08-30 38.0 2 19.000000
12 2018-08-31 38.0 3 9.500000
13 2018-09-01 38.0 4 0.000000
如果您的数据集很大并且您想避免循环，则可以改用merge 。
chart_data["prev_day_open_issues"] = chart_data["open_issues"].shift(1)
chart_data["no match"] = chart_data["open_issues"] != chart_data["prev_day_open_issues"]
# same idea as in r-beginners code
chart_data["ideal_pos"] = (chart_data["open_issues"]
.groupby(chart_data["no match"].cumsum())
.cumcount())
# tidy up and remove temp columns
new_chart_data = chart_data[["date", "open_issues", "ideal_pos"]]
# make your estimates dataframe into a one-to-one lookup in long format
estimates["open_issues"] = estimates[0]
new_estimates = (estimates
.drop_duplicates(subset=["open_issues"])
.melt(id_vars="open_issues", var_name="ideal_pos",
value_name="ideal_issues_left"))
# join
final = new_chart_data.merge(new_estimates, how="left", on=["open_issues", "ideal_pos"])
print(final[["date", "open_issues", "ideal_issues_left"]])
date open_issues ideal_issues_left
2018-08-19 23.0 23.000000
2018-08-20 25.0 25.000000
2018-08-21 26.0 26.000000
2018-08-21 26.0 23.636364
2018-08-23 28.0 28.000000
2018-08-24 36.0 36.000000
2018-08-25 33.0 33.000000
2018-08-26 39.0 39.000000
2018-08-26 39.0 32.500000
2018-08-28 38.0 38.000000
2018-08-28 38.0 28.500000
2018-08-28 38.0 19.000000
2018-08-28 38.0 9.500000
2018-08-28 38.0 0.000000
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