我有一个像这样的字典列表:
{
"Player_Name":"Byeong-Hun An",
"Tournament":[
{
"Name":"Arnold Palmer Invitational presented by Mastercard",
"Points":"32.80",
"Salary":"10300.00"
}
]
},
{
"Player_Name":"Byeong-Hun An",
"Tournament":[
{
"Name":"Different",
"Points":"18.80",
"Salary":"10400.00"
}
]
}
我想要这个:
[
{
"Player_Name":"Byeong-Hun An",
"Tournament":[
{
"Name":"Arnold Palmer Invitational presented by Mastercard",
"Points":"32.80",
"Salary":"10300.00"
},
{
"Name":"Different",
"Points":"18.80",
"Salary":"10400.00"
}
]
}
]
我已经尝试过收藏,但是它并不能完全满足我的需求。我本质上是想让每一个玩家都将所有锦标赛对象组合为一个,以便每个玩家只有一个对象,而不是每个事件都有自己的对象。
这是我的代码
import json
import numpy as np
import pandas as pd
from collections import Counter
# using json open the player objects file and set it equal to data
with open('PGA_Player_Objects.json') as json_file:
data = json.load(json_file)
points = []
players = []
for a in data:
for b in a['Tournament']:
points.append(int(float(b['Points'])))
for x in data:
players.append(x['Player_Name'])
def Average(lst):
unrounded = sum(lst) / len(lst)
return round(unrounded,2)
result = Counter()
for d in data:
for b in d['Tournament']:
result[d['Player_Name']] += int(float(b['Points']))
我怎样才能做到这一点?
如果您的清单在l
:
l = [{'Player_Name': 'Byeong-Hun An', 'Tournament': [{'Name': 'Arnold Palmer Invitational presented by Mastercard', 'Points': '32.80', 'Salary': '10300.00'}]},
{'Player_Name': 'Byeong-Hun An', 'Tournament': [{'Name': 'Different', 'Points': '18.80', 'Salary': '10400.00'}]},]
尝试这个:
from itertools import groupby
result = []
for k,g in groupby(sorted(l, key=lambda x:x['Player_Name']), lambda x:x['Player_Name']):
result.append({'Player_Name':k, 'Tournament':[i['Tournament'][0] for i in g]})
那么结果将是:
[{'Player_Name': 'Byeong-Hun An',
'Tournament': [
{'Name': 'Arnold Palmer Invitational presented by Mastercard',
'Points': '32.80',
'Salary': '10300.00'},
{'Name': 'Different',
'Points': '18.80',
'Salary': '10400.00'}]}]
本文收集自互联网,转载请注明来源。
如有侵权,请联系 [email protected] 删除。
我来说两句