从列表中查询，从字典的字典中返回

Question

一个列表和一个字典，如下，我想做一些查询。

列表中的每个元素都是人名和查询日期（字符串格式）

字典有人名key，value是一个以announcement_date（字符串格式）为key，bonus为value的字典。

从给定人员姓名和查询日期的列表中，我想找出人员的奖金，即公告日期最接近（并且早于）查询日期的时间。

例如，“Mike 20191022”将返回 Mike 在 20190630 宣布的奖金（即 105794.62）。

我尝试的是找出每个announcement_date的差异，并将它们与query_date进行比较。从最小的差异，我得到一个索引，并使用该索引返回对应的奖金。

import numpy as np

to_do = ["Mike 20191022"]

bonus = {'Mike': {'20200630': '105794.62', '20191231': '105794.62', '20190630': '105794.62', '20181231': '105794.62', '20180630': '95122.25', '20171231': '95122.25', '20170630': '95122.25'}}

for ox in to_do:
    to_do_people = ox.split(' ')[0]"
    to_do_date = ox.split(' ')[1]"
    for key, s_dict in bonus.items():"
        if to_do_people == key:"
            tem_list = []"
            for k, v in (s_dict.items()):"
                tem_list.append(int(to_do_date) - int(k))"
            idx = np.argmin(tem_list)"
            print (ox + '@Bonus@'  + list(s_dict.keys())[idx] + '@' + list(s_dict.values())[idx])"

但是它不起作用。输出是：

Mike 20191022@Bonus@20200630@105794.62

出了什么问题，我该如何纠正？谢谢。

Answer 1

实际上，您应该附加abs 值。

因为min(-1000, 100, 0) 是-1000，但你想要的是0。

tem_list.append(abs(int(to_do_date) - int(k)))

输出为

Mike 20191022@Bonus@20191231@105794.62

但这并不能解决您的结果日期应该在查询日期之前的问题。通过将查询日期之后的所有日期设置为一个非常大的数字，可以轻松解决此问题。

tem_list.append(abs(int(to_do_date) - int(k)) if (int(to_do_date)>=int(k)) else float('inf'))

它的作用是为查询日期之后的所有日期设置infinity。

所以，解决方案变成了

import numpy as np

to_do = ["Mike 20191022"]

bonus = {'Mike': {'20200630': '105794.62', '20191231': '105794.62', '20190630': '105794.62', '20181231': '105794.62', '20180630': '95122.25', '20171231': '95122.25', '20170630': '95122.25'}}

for ox in to_do:
    to_do_people = ox.split(' ')[0]
    to_do_date = ox.split(' ')[1]
    for key, s_dict in bonus.items():
        if to_do_people == key:
            tem_list = []
            for k, v in (s_dict.items()):
                tem_list.append(abs(int(to_do_date) - int(k)) if (int(to_do_date)>=int(k)) else float('inf'))
            idx = np.argmin(tem_list)
            print (ox + '@Bonus@'  + list(s_dict.keys())[idx] + '@' + list(s_dict.values())[idx])

输出

Mike 20191022@Bonus@20190630@105794.62

Answer 2

由于您要导入numpy，请使用numpy.searchsorted：

>>> for query in to_do:
       name, date = query.split()
       keys = sorted(map(int,bonus[name].keys()))
       ix = np.searchsorted(keys, int(date), side='right') - 1
       print(f"{name} {date}@bonus@{keys[ix]}@{bonus[name][str(keys[ix])]}")

Mike 20191022@bonus@20190630@105794.62

解释：

For first iteration, query == "Mike 20191022"
>>> name, date = query.split()
>>> name
"Mike"
>>> date
"20191022"
# then we sort the keys of the dict:
>>> sorted(map(int,bonus[name].keys()))
[20170630, 20171231, 20180630, 20181231, 20190630, 20191231, 20200630]
# Now we search for the index of the first date >= query_date
>>> np.searchsorted(keys, int(date), side='right')
5
# So 5th index is first date >= query_date
# If we check, we will find: keys[5] == 20191231 > 20191022

# However, we want the date before that, so subtract 1 from index
>>> ix = np.searchsorted(keys, int(date), side='right') - 1
>>> ix
4
# Now we first get the required date:
>>> keys[ix] # == keys[4]
20190630
# And then we access the bonus:
>>> bonus[name][str(keys[ix])]
"105794.62"

现在我们使用f-strings 打印它