字典列表集理解计算

Question

我的数据结构是一个字典列表。我想对某些键的值运行一个函数，然后只输出一定数量的字典作为结果。

from datetime import datetime
from dateutil.parser import parse

today = '05/17/18'
adict = [{'taskid':1,'desc':'task1','complexity':5,'dl':'05/28/18'},{'taskid':2,'desc':'task2','complexity':3,'dl':'05/20/18'},
         {'taskid':3,'desc':'task3','complexity':1,'dl':'05/25/18'}]

def conv_tm(t):
    return datetime.strptime(t,'%m/%d/%y')

def days(obj):
    day = conv_tm(today)
    dl = conv_tm(obj)
    dur = (dl-day).days
    if dur <0:
        dur = 1
    return dur

我发现处理“dl”键日期的最简单方法是运行这个字典理解：

vals = [days(i['dl']) for i in adict]

#this also worked, but I didn't like it as much
vals = list(map(lambda x: days(x['dl']), adict))

现在，我需要做 2 件事：1) 将此列表压缩回 'dl' 键，以及 2) 返回或打印一组（随机）2 个不改变原始字典的字典，也许像所以：

{'taskid':1,'desc':task1,'dl':8,'complexity':5}
{'taskid':3,'desc':task3,'dl':8,'complexity':1}

干杯

Answer 1

你可以像这样直接生成新的字典：

new_dicts = [{**d, 'dl': days(d['dl'])} for d in adict]

如果您需要单独的 val，您也可以使用它来执行此操作：

new_dicts = [{**d, 'dl': v} for d, v in zip(adict, vals)]