在 Python Pandas 中将时间分组为时间段

Question

我想编写一个将时间分组为时间段的代码。我有两列from 和to，我有列表periods。根据来自两列的值，我需要将新列插入到名为 periods 的数据框中，这将代表时间段。 这是代码：

import pandas as pd

df = pd.DataFrame({"from":['08:10', '14:00', '15:00', '17:01', '13:41'],
                   "to":['10:11', '15:32', '15:35' , '18:23', '16:16']})
print(df)

periods = ["00:01-06:00", "06:01-12:00", "12:01-18:00", "18:01-00:00"]
#if times are between two periods, for example '17:01' and '18:23', it counts as first period ("12:01-18:00") 

结果应如下所示：

    from     to       period
0  08:10  10:11  06:01-12:00
1  14:00  15:32  12:01-18:00
2  15:00  15:35  12:01-18:00
3  17:01  18:03  18:01-00:00
4  18:41  19:16  18:01-00:00

两列中的值是日期时间。

Answer 1

这是一种方法（我假设“18:00”属于“12:01-18:00”期间）：

results = [0 for x in range(len(df))]
for row in df.iterrows():
    item = row[1]
    start = item['from']
    end = item['to']

    for ind, period in enumerate(periods):
        per_1, per_2 = period.split("-")
        if start.split(":")[0] >= per_1.split(":")[0]:            #hours
            if start.split(":")[0] == per_1.split(":")[0]:
                if start.split(":")[1] >= per_1.split(":")[1]:    #minutes
                    if start.split(":")[1] == per_1.split(":")[1]:
                        results[row[0]] = period
                        break
                    #Wrap around if you reach the end of the list
                    index = ind+1 if ind<len(periods) else 0
                    results[row[0]] = periods[index]
                    break
                index = ind-1 if ind>0 else len(periods)-1
                results[row[0]] = periods[index]
                break

            if start.split(":")[0] <= per_2.split(":")[0]:
                if start.split(":")[0] == per_2.split(":")[0]:
                    if start.split(":")[1] == per_2.split(":")[1]:
                        results[row[0]] = period
                        break
                    #If anything else, then its greater, so in next period
                    index =  ind+1 if ind<len(periods) else 0
                    results[row[0]] = periods[index]
                    break
                results[row[0]] = period
                break

print(results)
df['periods'] = results
['06:01-12:00', '12:01-18:00', '12:01-18:00', '12:01-18:00', '18:01-00:00']

df['periods'] = results
df
    from     to      periods
0  08:10  10:11  06:01-12:00
1  14:00  15:32  12:01-18:00
2  15:00  15:35  12:01-18:00
3  17:01  18:23  12:01-18:00
4  18:41  16:16  18:01-00:00

这应该涵盖所有场景。但是您应该在可能的时间的每个边缘情况下对其进行测试以确保。

Answer 2

下面

import pandas as pd
from datetime import datetime

df = pd.DataFrame({"from": ['08:10', '14:00', '15:00', '17:01', '13:41'],
                   "to": ['10:11', '15:32', '15:35', '18:23', '16:16']})
print(df)

periods = ["00:01-06:00", "06:01-12:00", "12:01-18:00", "18:01-00:00"]
_periods = [(datetime.strptime(p.split('-')[0], '%H:%M').time(), datetime.strptime(p.split('-')[1], '%H:%M').time()) for
            p in periods]


def match_row_to_period(row):
    from_time = datetime.strptime(row['from'], '%H:%M').time()
    to_time = datetime.strptime(row['to'], '%H:%M').time()
    for idx, p in enumerate(_periods):
        if from_time >= p[0] and to_time <= p[1]:
            return periods[idx]
    for idx, p in enumerate(_periods):
        if idx > 0:
            prev_p = _periods[idx - 1]
            if from_time <= prev_p[1] and to_time >= p[0]:
                return periods[idx - 1]


df['period'] = df.apply(lambda row: match_row_to_period(row), axis=1)
print('-----------------------------------')
print('periods: ')
for _p in _periods:
    print(str(_p[0]) + ' -- ' + str(_p[1]))
print('-----------------------------------')

print(df)

输出

    from     to
0  08:10  10:11
1  14:00  15:32
2  15:00  15:35
3  17:01  18:23
4  13:41  16:16
-----------------------------------
periods: 
00:01:00 -- 06:00:00
06:01:00 -- 12:00:00
12:01:00 -- 18:00:00
18:01:00 -- 00:00:00
-----------------------------------
    from     to       period
0  08:10  10:11  06:01-12:00
1  14:00  15:32  12:01-18:00
2  15:00  15:35  12:01-18:00
3  17:01  18:23  12:01-18:00
4  13:41  16:16  12:01-18:00

Answer 3

不确定，如果有更好的解决方案，但这里有一种使用 apply 和 assign pandas 方法的方法，这通常比迭代 DataFrame 更 Pythonic，因为 pandas 针对完整的 df ix 分配操作进行了优化，而不是逐行更新（请参阅这个很棒的博客post）。

附带说明，我在这里使用的数据类型是datetime.time 实例，而不是您的示例中的字符串。在处理时间时，最好使用适当的时间库而不是字符串表示。

from datetime import time

df = pd.DataFrame({
    "from": [
        time(8, 10),
        time(14, 00),
        time(15, 00),
        time(17, 1),
        time(13, 41)
    ],
    "to": [
        time(10, 11),
        time(15, 32),
        time(15, 35),
        time(18, 23),
        time(16, 16)
    ]
})

periods = [{
    'from': time(00, 1),
    'to': time(6, 00),
    'period': '00:01-06:00'
}, {
    'from': time(6, 1),
    'to': time(12, 00),
    'period': '06:01-12:00'
}, {
    'from': time(12, 1),
    'to': time(18, 00),
    'period': '12:01-18:00'
}, {
    'from': time(18, 1),
    'to': time(0, 00),
    'period': '18:01-00:00'
}]


def find_period(row, periods):
    """Map the df row to the period which it fits between"""
    for ix, period in enumerate(periods):
        if row['to'] <= periods[ix]['to']:
            if row['from'] >= periods[ix]['from']:
                return periods[ix]['period']

# Use df assign to assign the new column to the df
df.assign(
    **{
        'period':
            df.apply(lambda row: find_period(row, periods), axis='columns')
    }
)

Out:
       from        to       period
0  08:10:00  10:11:00  06:01-12:00
1  14:00:00  15:32:00  12:01-18:00
2  15:00:00  15:35:00  12:01-18:00
3  17:01:00  18:23:00         None
4  13:41:00  16:16:00  12:01-18:00

注意ix 3 处的行正确显示 None，因为它不准确地适合您定义的两个时期中的任何一个（而是桥接 12:00-18:00 和 18:00-00:00）