如何在列表理解中添加额外的中间步骤?

可以使用内部生成器表达式进行拆分:

[int(hh)*60 + int(mm) for hh, mm in (ts.split(':') for ts in timestamps)]

虽然就我个人而言,我更愿意使用助手函数:

def timestamp_to_minutes(timestamp: str) -> int:
    hh, mm = timestamp.split(":")
    return int(hh)*60 + int(mm)

[timestamp_to_minutes(ts) for ts in timestamps]

# Alternative
list(map(timestamp_to_minutes, timestamps))

如果不想将字符串拆分两次,可以使用 := 分配操作员:

timestamps = [int((s := t.split(":"))[0]) * 60 + int(s[1]) for t in timestamps]
print(timestamps)

印刷品:

[1378, 191, 741]

备选方案:

print([int(h) * 60 + int(m) for h, m in (t.split(":") for t in timestamps)])

印刷品:

[1378, 191, 741]

晚会迟到了。。但是为什么不使用datetime/timedelta来转换时间呢?

对于“hh:mm”,这可能有些过分,但您可以轻松地将其调整为更复杂的时间字符串:

from datetime import datetime as dt
import typing

def timestamps_to_minutes(timestamps: typing.List[str]) -> typing.List[any]:
    """Uses datetime.strptime to parse a datetime string and return
    minutes spent in this day."""
    return [int(((p:=dt.strptime(t,"%H:%M")) - dt(p.year,p.month, p.day)
                 ).total_seconds()//60) for t in timestamps]

timestamps = ["22:58", "03:11", "12:21"]

print(timestamps_to_minutes(timestamps))

产出:

[1378, 191, 741]

只是为了好玩,我们也可以 operator.methodcaller :

from operator import methodcaller
out = [int(h) * 60 + int(m) for h, m in map(methodcaller("split", ":"), timestamps)]

输出:

[1378, 191, 741]