Python映射dicts
假设我有一些dict,其中包含几个月的短名称和一些数据Python映射dicts,python,dictionary,Python,Dictionary,假设我有一些dict,其中包含几个月的短名称和一些数据 data_in = {"Jan":2.0, "Feb":5.5} 和一些包含映射的dict month_names = {"Jan":"January", "Feb":"February"} day_month = {day:"Jan" for day in range(1,32)} day_month.update({day:"Feb" for day in range(32,60)}) 如何获得以下dict,其中包含以元组形式表示
data_in = {"Jan":2.0, "Feb":5.5}
和一些包含映射的dict
month_names = {"Jan":"January", "Feb":"February"}
day_month = {day:"Jan" for day in range(1,32)}
day_month.update({day:"Feb" for day in range(32,60)})
如何获得以下dict,其中包含以元组形式表示的长名称和天数的原始数据
{("January", 1):2.0, ("January", 2):2.0 ...}
似乎您要做的是迭代
day\u month
字典的键,并为每个键取month\u names
和中的data\u中的值的匹配项,因此它应该类似于:
result = {}
for day in day_month.keys():
short_month = day_month[day]
data = data_in[short_month]
name = month_names[short_month]
new_key = (name, day)
result[new_key] = data
这是我自己的尝试-
from itertools import count
c = count(1,1)
days = {"Feb" : 28, "Jan" : 31}
mapping = {(month_names[i], next(c)) : data_in[i] for i in data_in.keys() for _ in range(days[i])}
print (mapping)
或使用您的日/月
dict
mapping = {(month_names[day_month[i]], i) : data_in[day_month[i]] for i in day_month.keys()}
print mapping
输出-
{('January', 1): 2.0,
('January', 2): 2.0,
('January', 3): 2.0,
('January', 4): 2.0,
('January', 5): 2.0,
('January', 6): 2.0,
('January', 7): 2.0,
('January', 8): 2.0,
('January', 9): 2.0,
('January', 10): 2.0,
('January', 11): 2.0,
('January', 12): 2.0,
('January', 13): 2.0,
('January', 14): 2.0,
('January', 15): 2.0,
('January', 16): 2.0,
('January', 17): 2.0,
('January', 18): 2.0,
('January', 19): 2.0,
('January', 20): 2.0,
('January', 21): 2.0,
('January', 22): 2.0,
('January', 23): 2.0,
('January', 24): 2.0,
..... }
请注意,update
修改dict,但返回None
。因此,day\u-month={…}.update(…)
使day\u-month
等于None
。您的更新
缺少结尾)
@depperm不要介意其他无效语法,对吗?:)day\u month错误已修复。“day number”是Julian day吗?@depperm-需要详细说明吗?将其放在他的代码后面,它会给出错误AttributeError:“NoneType”对象没有属性“keys”
@depperm-是的,我看到了帖子本身的评论,明白了这一点。但这与答案无关。我同意使用正确的语法更好,但这不是问题所在。