检查一行上的日期是否早于下一行的另一个日期

Question

我在 Python 中有以下代码：

import pandas as pd
import numpy as np
date_rng = pd.date_range(start='5/18/2019', end='7/22/2020', freq='S')

df = pd.DataFrame(date_rng, columns=['start_timestamp'])
df['end_timestamp'] = date_rng
df['start_timestamp'] = np.random.randint(1589760000,1595376000,size=(len(date_rng)))
df['end_timestamp'] = np.random.randint(1589760000,1595376000,size=(len(date_rng)))
df = df[(df.end_timestamp/df.start_timestamp<=1.000009)&(df.end_timestamp/df.start_timestamp>=1.000001)]
df = df.sort_values(by=['start_timestamp','end_timestamp'])
df['start_timestamp'] = pd.to_datetime(df['start_timestamp'],unit='s')
df['end_timestamp'] = pd.to_datetime(df['end_timestamp'],unit='s')

因此，我有以下数据框：

  start_timestamp     end_timestamp
2020-05-18 00:00:30 2020-05-18 00:54:07
2020-05-18 00:01:40 2020-05-18 03:50:39
2020-05-18 00:02:08 2020-05-18 02:39:41
2020-05-18 00:04:01 2020-05-18 00:47:25
2020-05-18 00:04:01 2020-05-18 02:26:50
2020-05-18 00:04:44 2020-05-18 02:17:53

                .
                .
                .

我应该怎么做才能确保在我的数据集中每个end_timestamp 都是在其下一行的start_timestamp 之前的日期时间？

已实施的解决方案

我基本上将数据集转换为数组，将其按升序排列，然后将其转换回数据框。它可能不是最优雅的解决方案，但它可以正常工作并为我打算使用的内容生成一致的数据。

import pandas as pd
import numpy as np
date_rng = pd.date_range(start='7/22/2019', end='7/22/2020', freq='S')

df = pd.DataFrame(date_rng, columns=['start_timestamp'])
df['end_timestamp'] = date_rng
df['start_timestamp'] = np.random.randint(1563753600,1595376000,size=(len(date_rng)))
df['end_timestamp'] = np.random.randint(1563753600,1595376000,size=(len(date_rng)))
df = df[(df.end_timestamp/df.start_timestamp<=1.0000009)&(df.end_timestamp/df.start_timestamp>=1.0000001)]
df = df.to_numpy()
df = df.reshape(df.shape[0]*2,1)
df = np.sort(df,axis=0)
df = df.reshape(int(df.shape[0]/2),2)
df = pd.DataFrame(df,columns=['start_timestamp','end_timestamp'])
df['start_timestamp'] = pd.to_datetime(df['start_timestamp'],unit='s')
df['end_timestamp'] = pd.to_datetime(df['end_timestamp'],unit='s')

Answer 1

编写你的逻辑，一切都很好

1. freq='S' 毫无意义，您将生成与开始日期和结束日期之间的秒数一样多的行
2. 在开始时间随机化后，使用当前行和下一行作为结束时间的随机函数的种子。这样做是为了理解列表
3. 在获取范围开始和结束时的 UTC 秒数方面更聪明了

import pandas as pd
import numpy as np
from datetime import datetime
# date_rng = pd.date_range(start='5/18/2019', end='7/22/2020', freq='S')
date_rng = pd.date_range(start='5/18/2019', end='5/19/2019', freq='min')

sec = [(date_rng.min() - datetime(1970, 1, 1)).total_seconds(),
       (date_rng.max() - datetime(1970, 1, 1)).total_seconds() ]
df = pd.DataFrame(date_rng, columns=['start_timestamp'])
df['start_timestamp'] = np.random.randint(sec[0],sec[1],size=(len(date_rng)))
df = df.sort_values(by="start_timestamp")
l = df["start_timestamp"].tolist()  # get randomised start times
l[-1] = sec[1] # set last time to end of range
# randomise end time between two start times
df['end_timestamp'] = [np.random.randint(l[i], l[i+1]) if i<len(l)-1  and l[i]<l[i+1] else l[i] for i, s in enumerate(l)]
df['start_timestamp'] = pd.to_datetime(df['start_timestamp'],unit='s')
df['end_timestamp'] = pd.to_datetime(df['end_timestamp'],unit='s')