在python中将多列表转换为字典

我正在尝试从列表列表创建词典。我有一个姓名和标记列表，如：

str=[
["Amit", 23],
["Amit", 127],
["Rahul", 3],
["Rahul", 71],
["Deepak", 31],
["Rahul", 22],
["Amit", 81]
]

使用键作为名称，值作为平均标记的字典，如：

{"Amit":77, "Rahul":32, "Deepak":31}

我试图获取每个列表，然后将其转换为字典值。但是我犯了一个错误

def highest_avg(str):
mydict={}
for each in str:
    my_dict=dict.fromkeys(each[0],each[1])
    print(my_dict)

一,。使用

键存储
编号的
列表

#驱动程序值：

IN :s = [['Amit', 23], ['Amit', 127], ['Rahul', 3], ['Rahul', 71],
         ['Deepak', 31], ['Rahul', 22], ['Amit', 81]]

我会使用列表的
OrderedDict
，因为看起来您希望维护顺序

from collections import OrderedDict

o = collections.OrderedDict()
for i in data:
     o.setdefault(i[0], []).append(i[1])

现在已经完成了向字典中添加数据，请按键计算平均值

for k in o:
     o[k] = sum(o[k]) / len(o[k])

如果顺序在这里不是最重要的，那么一本普通的
{}
字典也可以


使用列表理解：

{x[0]:x[1] for x in str}

或将数据分为两个列表键和值：

keys = [x[0] for x in str]
values = [x[1] for x in str]
dict(zip(keys, values))

输出：

{'Deepak': 31, 'Amit': 81, 'Rahul': 22}

使用标准Python方法：

myList=[
["Amit", 23],
["Amit", 127],
["Rahul", 3],
["Rahul", 71],
["Deepak", 31],
["Rahul", 22],
["Amit", 81]
]

myDict = dict()
for key, value in myList:
    if key not in myDict:
        myDict[key] = list()
    myDict[key].append(value)

for key, values in myDict.items():
    myDict[key] = sum(values) / len(values)

print(myDict)

首先，我得到属于某个键的所有值，然后计算平均值。还要注意str是一个内置变量，所以我用myList替换了它。
您可以使用
itertools.groupby
：

from itertools import groupby

k = lambda x: x[0]
{k: sum(l)//len(l) for k, l in ((k, [x[1] for x in g]) for k, g in groupby(sorted(str, key=k), key=k))}
# {'Amit': 77, 'Deepak': 31, 'Rahul': 32}

希望这有帮助
这不是OP的要求。即使期望的输出也不匹配。请把这个问题再复习一遍。
keys = [x[0] for x in str]
values = [x[1] for x in str]
dict(zip(keys, values))

{'Deepak': 31, 'Amit': 81, 'Rahul': 22}

myList=[
["Amit", 23],
["Amit", 127],
["Rahul", 3],
["Rahul", 71],
["Deepak", 31],
["Rahul", 22],
["Amit", 81]
]

myDict = dict()
for key, value in myList:
    if key not in myDict:
        myDict[key] = list()
    myDict[key].append(value)

for key, values in myDict.items():
    myDict[key] = sum(values) / len(values)

print(myDict)

from itertools import groupby

k = lambda x: x[0]
{k: sum(l)//len(l) for k, l in ((k, [x[1] for x in g]) for k, g in groupby(sorted(str, key=k), key=k))}
# {'Amit': 77, 'Deepak': 31, 'Rahul': 32}

    data =[
["Amit", 23],
["Amit", 127],
["Rahul", 3],
["Rahul", 71],
["Deepak", 31],
["Rahul", 22],
["Amit", 81]
]

import numpy as np
import pandas as pd

data_df = pd.DataFrame(data) #convert your list into a Pandas DataFrame

data_df

       0    1
0    Amit   23
1    Amit  127
2   Rahul    3
3   Rahul   71
4  Deepak   31
5   Rahul   22
6    Amit   81

agg_data = data_df.groupby(data_df[0]).agg(np.mean) #Group data on names & aggregate using mean of scores(values)

agg_data
         1
0         
Amit    77
Deepak  31
Rahul   32

agg_data.to_dict()[1] #Convert your Pandas DataFrame into dict using 'dict' orientation
#the [1] is the key that represents your column. Dict is created like this: {column -> {index -> value}}

{'Amit': 77, 'Deepak': 31, 'Rahul': 32} #This is the output