Question

我有一个以这种形式输入的列表：

open_info = ['Cube 1, 9:30am to 10:00am, Thursday, March 3, 2016', 'Cube 2, 5:00pm to 5:30pm, Thursday, March 3, 2016']

我想解析这些信息，以此形式创建一个新列表：

open_times = [[9, 30, 'am'],[5, 0, 'pm']]

以第一个索引为小时，第二个为分钟，第三个索引为上午/下午。我只记录每个列表元素的第一次值，因为我处理的间隔总是30分钟。

我通过使用以下python列表推导完成了这个：

open_times = [x.split(",")[1].replace(" ","").split("to") for x in open_info]
open_times = [x[0].split(":")+x[1].split(":") for x in open_times]
open_times = [[int(x[0]),int(x[1][:2]),x[1][2:]] for x in open_times]

我想知道是否要在所有这些中创建嵌套列表解析。我已经查看了python文档并阅读了一些关于这个主题的博客，但是我仍然无法完成这个。

Answer 1

要回答如何“嵌套”列表推导的问题，你可以这样做来组合第1行和第2行....

open_times = [y[0].split(":")+y[1].split(":") for y in [x.split(",")[1].replace(" ","").split("to") for x in open_info]]

......但这真是一团糟。这三条线在这里更容易理解和清洁。你可能还会考虑把它写成一系列循环，因为在理解中有很多东西会在一个循环之外更清晰。

Answer 2

您可以使用正则表达式来提取时间：

>>> import re
>>> 
>>> [[int(val) if val.isdigit() else val for val in re.search(r'(\d+):(\d+)(am|pm)',item, re.I).groups()] for item in open_info]
[[9, 30, 'am'], [5, 00, 'pm']]

但如果它不能匹配正则表达式，那么它可能会引发AttributeError，所以如果您不确定，可以使用try-except表达式来处理错误。

times = []
for item in open_info:
    match = re.search(r'(\d+):(\d+)(am|pm)',item, re.I)
    try:
        h, m, b = match.groups()
    except (AttributeError, ValueError):
        pass # or append a proper value to times, instead.
    else:
        times.append([int(h), int(m), b])
   times.append(match)

Answer 3

您可以使用以下内容：

open_info = ['Cube 1, 9:30am to 10:00am, Thursday, March 3, 2016', 'Cube 2, 5:00pm to 5:30pm, Thursday, March 3, 2016']

answer = [[int(s.split(':',1)[0][-2:]), int(s.split(':')[1][:2]), 
           s.split(':')[1][2:4]] for s in open_info]
print(answer)

<强>输出

[[9, 30, 'am'], [5, 0, 'pm']]

但是，在这些情况下，使用map代替list理解可能更具可读性：

def func(s):
    hour = int(s.split(':')[0][-2:])
    minute = int(s.split(':')[1][:2])
    suffix = s.split(':')[1][2:4]
    return [hour, minute, suffix]

answer = map(func, open_info)
print(answer)

<强>输出

[[9, 30, 'am'], [5, 0, 'pm']]

Answer 4

您可以简单地创建处理函数，而不是将所有逻辑推送到列表理解表达式。

为了更好的可读性，我重命名了一些值。

def extract(s):
    time_from, time_to = s.split(",")[1].replace(" ", "").split("to")
    hour, min_am_pm = time_from.split(":")
    min = min_am_pm[:2]
    am_pm = min_am_pm[2:]
    return [int(hour), int(min), am_pm]


open_info = ['Cube 1, 9:30am to 10:00am, Thursday, March 3, 2016', 'Cube 2, 5:00pm to 5:30pm, Thursday, March 3, 2016']
open_times = [extract(x) for x in open_info]

Answer 5

from csv import reader
answer = [[int(a), int(b[:2]), c[2:]] for a, b, c in (inf[1].split(":")
             for inf in reader(open_info, skipinitialspace=True))]

实际上符合您的预期输出：

[[9, 30, 'am'], [5, 0, 'pm']]

一个简单的函数是一个更好的主意，也不需要不断地重复分割同一行：

def spl(l):
    for inf in l:
        a, b, c = inf.split(", ", 2)[1].split(":", 2)
        yield [int(a), int(b[:2]), c[2:]]

print(list(spl(open_info)))

输出：

[[9, 30, 'am'], [5, 0, 'pm']]

或者让csv lib解析项目：

from csv import reader

def spl(l):
    for inf in reader(l, skipinitialspace=True):
        a, b, c = inf[1].split(":", 2)
        yield [int(a), int(b[:2]), c[2:]]
print(list(spl(open_info)))

冷凝多列表理解

5 个答案: