按行列出的重叠时间范围的长度

Question

我正在使用 pandas 版本 1.0.5

下面的示例数据框列出了三天内记录的时间间隔，我寻找每天都有哪些时间间隔重叠的地方。

例如，所有三个日期（黄色突出显示）的重叠时间之一是 1:16 - 2:13。另一个（蓝色突出显示）为 18:45 - 19:00

所以我的预期输出是这样的：[57,15]因为

• 57 - 1:16 - 2:13 之间的分钟。
• 18:45 - 19:00 之间 15 分钟

请使用输入数据帧的生成器：

import pandas as pd
dat1 = [
    ['2023-12-27','2023-12-27 00:00:00','2023-12-27 02:14:00'],
    ['2023-12-27','2023-12-27 03:16:00','2023-12-27 04:19:00'],
    ['2023-12-27','2023-12-27 18:11:00','2023-12-27 20:13:00'],
    ['2023-12-28','2023-12-28 01:16:00','2023-12-28 02:14:00'],
    ['2023-12-28','2023-12-28 02:16:00','2023-12-28 02:28:00'],
    ['2023-12-28','2023-12-28 02:30:00','2023-12-28 02:56:00'],
    ['2023-12-28','2023-12-28 18:45:00','2023-12-28 19:00:00'],
    ['2023-12-29','2023-12-29 01:16:00','2023-12-29 02:13:00'],
    ['2023-12-29','2023-12-29 04:16:00','2023-12-29 05:09:00'],
    ['2023-12-29','2023-12-29 05:11:00','2023-12-29 05:14:00'],
    ['2023-12-29','2023-12-29 18:00:00','2023-12-29 19:00:00']
       ]
df = pd.DataFrame(dat1,columns = ['date','Start_tmp','End_tmp'])
df["Start_tmp"] = pd.to_datetime(df["Start_tmp"])
df["End_tmp"] = pd.to_datetime(df["End_tmp"])

Answer 1

该解决方案使用：

• numpy，没有不常见的Python模块，所以使用Python 1.0.5你应该，希望，是清楚的，
• 没有嵌套循环来处理不断增长的数据集的速度问题，

方法：

• 绘制重叠的景观
• 然后选择与记录天数相对应的重叠，
• 最后用长度来描述重叠部分

记录的天数：（如Python 中：将 dataframe 中的 timedelta 转换为 int）

n = 1 + ( max(df['End_tmp']) - min(df['Start_tmp']) ).days
n
3

加法景观：

# initial flat whole-day landcape (height: 0)
L = np.zeros(24*60, dtype='int')
# add up ranges: (reused @sammywemmy's perfect formula for time of day in minutes)
for start, end in zip(df['Start_tmp'].dt.hour.mul(60) + df['Start_tmp'].dt.minute,  # Start_tmp timestamps expressed in minutes
                      df['End_tmp'].dt.hour.mul(60)   + df['End_tmp'].dt.minute):   # End_tmp timestamps expressed in minutes
    L[start:end+1] += 1

plt.plot(L)
plt.hlines(y=[2,3],xmin=0,xmax=1400,colors=['green','red'], linestyles='dashed')
plt.xlabel('time of day (minutes)')
plt.ylabel('time range overlaps')

（请原谅拼写错误：这显然是分钟，而不是秒）

仅保留所有日期的重叠：（红线，n=3）

# Reduce heights <n to 0 because not overlaping every day
L[L<n]=0
# Simplify all greater values to 1 because only their presence matters
L[L>0]=1
# Now only overlaps are highlighted
# (Actually this latest line is disposable, provided we filter all but the overlaps of rank n. Useful only if you were to include lower overlaps)

提取重叠范围及其长度

# Highlight edges of overlaping intervals
D = np.diff(L)
# Describe overlaps as ranges
R = list(zip([a[0]   for a in np.argwhere(D>0)],  # indices where overlaps *begin*, with scalar indices instead of arrays
             [a[0]-1 for a in np.argwhere(D<0)])) # indices where overlaps *end*, with scalar indices instead of arrays
R
[(75, 132), (1124, 1139)]

# Finally their lengths
[b-a for a,b in R]

最终输出：[57, 15]