Pythonic的方式来寻找日期时间最小的时间差异目标从列表？

Question

我有一个日期时间对象target_time。我也有一个列表，例如times_list[i]['time']是日期时间对象。我想遍历这个列表并找到与target_time具有最短时间距离的项目的索引。这些差异将以毫秒为单位，列表中涵盖的时间范围将以分钟为单位。我目前的解决方案是相当笨拙的：

best_index = -1 
best_diff = 999999999999.0 
for i in range(len(times_List)): 
    cur_time = times_list[i]['time'] 
    diff = abs((cur_time-target_time).total_seconds())) 
    if diff < best_diff: 
     best_index = i 
     best_diff = diff 
if best_index > -1: 
    result = times_list[best_index]['time'] 

是否有一个更笨拙和更pythonic这样做呢？

Answer 1

import operator 
deltas = [abs(ti['time'] - target_time) for ti in times_list] 
min_index, min_delta = min(enumerate(deltas), key=operator.itemgetter(1)) 

Answer 2

替代方法：

temp = sorted(times_list, key=lambda x: abs(abs((x-target_time).total_seconds()))) 
result = times_list.index(temp[0]) 

Answer 3

该功能可以找到最接近的：

from datetime import datetime 

def closest(target, dates):  
    return min(dates, key=lambda date: abs(date-target)) 


print(closest(target1, dates)) 
#2014-12-31 16:00:00 

print(closest(target2, dates)) 
#2015-01-01 12:00:00 
print(abs(target1 - target2)) 

Answer 4

使用min()找到差异最小。再加上enumerate()保持指数的轨迹：

min_pos, min_diff = min(enumerate(times_list), key=lambda t: abs(target_time - t[1]['time'])) 

无需产生分歧的中间列表。