我有一个包含项目,开始日期和结束日期的数据框。对于每一行,我想返回项目开始时正在处理的其他项目的数量。使用df.apply()
时如何嵌套循环?我尝试过使用for循环,但是我的数据帧很大,并且花费的时间太长。
import datetime as dt
data = {'project' :['A', 'B', 'C'],
'pr_start_date':[dt.datetime(2018, 9, 1), dt.datetime(2019, 4, 1), dt.datetime(2019, 6, 8)],
'pr_end_date': [dt.datetime(2019, 6, 15), dt.datetime(2019, 12, 1), dt.datetime(2019, 8, 1)]}
df = pd.DataFrame(data)
def cons_overlap(start):
overlaps = 0
for i in df.index:
other_start = df.loc[i, 'pr_start_date']
other_end = df.loc[i, 'pr_end_date']
if (start > other_start) & (start < other_end):
overlaps += 1
return overlaps
df['overlap'] = df.apply(lambda row: cons_overlap(row['pr_start_date']), axis=1)
这是我正在寻找的输出:
pr pr_start_date pr_end_date overlap
0 A 2018-09-01 2019-06-15 0
1 B 2019-04-01 2019-12-01 1
2 C 2019-06-08 2019-08-01 2
答案 0 :(得分:3)
我建议您利用numpy broadcasting:
ends = df.pr_start_date.values < df.pr_end_date.values[:, None]
starts = df.pr_start_date.values > df.pr_start_date.values[:, None]
df['overlap'] = (ends & starts).sum(0)
print(df)
输出
project pr_start_date pr_end_date overlap
0 A 2018-09-01 2019-06-15 0
1 B 2019-04-01 2019-12-01 1
2 C 2019-06-08 2019-08-01 2
开头和结尾都是3x3的矩阵,当满足条件时,它们就是真值:
# ends
[[ True True True]
[ True True True]
[ True True True]]
# starts
[[False True True]
[False False True]
[False False False]]
然后找到与逻辑&
的交点,并求和成列(sum(0)
)。
答案 1 :(得分:2)
答案 2 :(得分:0)
我假设这些行按开始日期排序,然后检查以前启动的尚未完成的项目。 df.index.get_loc(r.name)产生正在处理的行的索引。
df["overlap"]=df.apply(lambda r: df.loc[:df.index.get_loc(r.name),"pr_end_date"].gt(r["pr_start_date"]).sum()-1, axis=1)