如何在给定值的 n 分钟内找到列表中的所有日期时间？

Question

我有两个时间列表（Hour:Min:Sec 格式），我一直在努力将 list_a 中的每个条目与 list_b 中的所有条目进行比较，以确定 30 分钟内的值：

list_a = ["10:26:42", "8:55:43", "7:34:11"]
list_b = ["10:49:20", "8:51:10", "10:34:35", "8:39:47", "7:11:49", "7:42:10"]

预期输出：

10:26:42 is within 30m of 10:49:20, 10:34:35
8:55:43 is within 30m of 8:51:10, 8:39:47
7:34:11 is within 30m of 7:11:49, 7:42:10

到目前为止，我一直在做的是：

import datetime

# Convert the Lists to Datetime Format

for data in list_a:
    convert = datetime.datetime.strptime(data,"%H:%M:%S")
    list_a_times.append(convert)

for data in list_b:
    convert = datetime.datetime.strptime(data,"%H:%M:%S")
    list_b_times.append(convert)

# Using a Value of List A, Find the Closest Value in List B

for data in list_a_times:
     closest_to_data = min(list_b_times, key=lambda d: abs(d - data))

     print(data, closest_to_data)

这种方法有效，但它只能找到一个最接近的值！只要值在所需的 30 分钟或更短的时间内，我如何操纵 min() 函数来继续提供值？

Answer 1

IIUC，你想比较所有的组合，所以你需要检查所有。

请阅读答案末尾的datetime/timedelta。

使用itertools.product：

list_a = ['10:26:42', '8:55:43', '7:34:11']
list_b = ['10:49:20', '8:51:10', '10:34:35', '8:39:47', '7:11:49', '7:42:10']

import datetime
from itertools import product

str2time = lambda s: datetime.datetime.strptime(s, "%H:%M:%S")

for a,b in product(map(str2time, list_a), map(str2time, list_b)):
    if abs(a-b).total_seconds() <= 1800:
        print(f'{a:%H:%M:%S} is within 30m of {b:%H:%M:%S}')

输出：

10:26:42 is within 30m of 10:49:20
10:26:42 is within 30m of 10:34:35
08:55:43 is within 30m of 08:51:10
08:55:43 is within 30m of 08:39:47
07:34:11 is within 30m of 07:11:49
07:34:11 is within 30m of 07:42:10

使用嵌套的 for 循环：

import datetime

str2time = lambda s: datetime.datetime.strptime(s, "%H:%M:%S")

for a in map(str2time, list_a):
    start = f'{a:%H:%M:%S} is within 30m of'
    for b in map(str2time, list_b):
        if abs(a-b).total_seconds() <= 1800:
            print(f'{start} {b:%H:%M:%S}', end='')
            start = ','
    if start == ',':
        print()

输出：

10:26:42 is within 30m of 10:49:20, 10:34:35
08:55:43 is within 30m of 08:51:10, 08:39:47
07:34:11 is within 30m of 07:11:49, 07:42:10

注意datetime

使用不带日期的 datetime 将默认为 1900-01-01，这可能会在接近午夜时产生边缘效应。相反，您可以使用 timedelta 对象。使用我的代码，您需要将 str2time 函数更改为：

def str2time(s):
    h,m,s = map(int, s.split(':'))
    return datetime.timedelta(hours=h, minutes=m, seconds

并稍微修改一下代码，以便能够转换为字符串：

z = datetime.datetime(1900,1,1)

for a in map(str2time, list_a):
    start = f'{z+a:%H:%M:%S} is within 30m of'
    for b in map(str2time, list_b):
        if abs(a-b).total_seconds() <= 1800:
            print(f'{start} {z+b:%H:%M:%S}', end='')
            start = ','
    if start == ',':
        print()

Answer 2

你循环和掠夺所有元素的绝对时间差异，而不是使用min：

list_a = ["10:26:42", "8:55:43", "7:34:11"]
list_b = ["10:49:20", "8:51:10", "10:34:35", "8:39:47", "7:11:49", "7:42:10"]

import datetime
import datetime

# Convert the Lists to Datetime Format
list_a = [datetime.datetime.strptime(d,"%H:%M:%S") for d in list_a]
list_b = [datetime.datetime.strptime(d,"%H:%M:%S") for d in list_b]

for value in list_a:
    for v in list_b:
        if abs(value-v) < datetime.timedelta(minutes=30):
            print (value, "=>", v, "diff: ", (value-v).total_seconds() // 60)
    print()
            

输出：

1900-01-01 10:26:42 => 1900-01-01 10:49:20 diff:  -23.0
1900-01-01 10:26:42 => 1900-01-01 10:34:35 diff:  -8.0

1900-01-01 08:55:43 => 1900-01-01 08:51:10 diff:  4.0
1900-01-01 08:55:43 => 1900-01-01 08:39:47 diff:  15.0

1900-01-01 07:34:11 => 1900-01-01 07:11:49 diff:  22.0
1900-01-01 07:34:11 => 1900-01-01 07:42:10 diff:  -8.0

这对于像 0:05:00 和 23:55:00 这样的日期时间会出错，因为它们位于不同的日期。

您可以通过自己编写的增量计算来解决这个问题：

def abs_time_diff(dt1, dt2, *, ignore_date = False):
    if not ignore_date:
        return abs(dt1-dt2)
    # use day before, this day and day after, report minimum
    return min ( (abs(dt1 + datetime.timedelta(days = delta) - dt2) 
                  for delta in range(-1,2)))

list_a = ["0:5:0"]
list_b = ["0:20:0", "23:55:0"]

list_a = [datetime.datetime.strptime(d,"%H:%M:%S")  for d in list_a]
list_b = [datetime.datetime.strptime(d,"%H:%M:%S")  for d in list_b]

for value in list_a:
    for v in list_b:
        print (value, v, abs_time_diff(value,v))
        print (value, v, abs_time_diff(value,v, ignore_date = True))

输出：

1900-01-01 00:05:00 1900-01-01 00:20:00 0:15:00
1900-01-01 00:05:00 1900-01-01 00:20:00 0:15:00

1900-01-01 00:05:00 1900-01-01 23:55:00 23:50:00 # with date
1900-01-01 00:05:00 1900-01-01 23:55:00 0:10:00  # ignores date

Answer 3

我会为此提出使用pandas 的建议：

# Convert to pandas datetime series
import pandas as pd
dt_a = pd.Series(list_a, dtype='datetime64[ns]')
dt_b = pd.Series(list_b, dtype='datetime64[ns]')

# Comparison loop
interv_size = '30m'   # Thirty minutes
for el in dt_a:
    hits = df_b.loc[ abs(el - df_b) < interv_size ].dt.time
    print(f'{el.time()} is within {interv_size} of', *hits) 

优势？ 你让 python 处理你的日期格式

Answer 4

from datetime import datetime, timedelta

list_a = ["10:26:42", "8:55:43", "7:34:11"]
list_b = ["10:49:20", "8:51:10", "10:34:35", "8:39:47", "7:11:49", "7:42:10"]

time_format = "%H:%M:%S"


def convert_to_datetime(time_str):
    return datetime.strptime(time_str, time_format)


# Overriding list_a and list_ to avoid polluting the namespace
# Sorting for simple optimization
list_a = sorted([convert_to_datetime(time_str) for time_str in list_a])
list_b = sorted([convert_to_datetime(time_str) for time_str in list_b])

time_range_limit_in_seconds = timedelta(minutes=30).total_seconds()

result = []
for list_a_datetime in list_a:
    with_in_time_limit = []
    for list_b_datetime in list_b:
        difference_in_seconds = (
            list_a_datetime-list_b_datetime).total_seconds()

        if difference_in_seconds <= time_range_limit_in_seconds:
            # Convert back to string
            with_in_time_limit.append(
                list_b_datetime.strftime(time_format)
            )

        # Since the list is sorted, all the rest don't fall in time range
        if difference_in_seconds < 0:
            break

    print(list_a_datetime.strftime(time_format), with_in_time_limit)