遍历python中的字典列表以查找所有出现的一对值

Question

我搜索了一段时间，但找不到解决问题的确切方法。我有一个格式的字典列表：

d = [{"sender": "a", "time": 123, "receiver": "b", "amount": 2}, {"sender": "c", "time": 124, "receiver": "b", "amount": 10}, {"sender": "a", "time": 130, "receiver": "b", "amount": 5}]

我想找到迭代所有字典并计算给定的发送者-接收者对出现的次数以及总数的最佳方法。

所以我想得到：

result = [{"sender": "a", "receiver":b, "count": 2, "total_amount":7}, {"sender": "c", "receiver":b, "count": 1, "total_amount":10}]

我很确定我可以通过逐个迭代列表中的所有字典来完成这项工作，将信息保存在临时字典中，但这会导致大量嵌套 if 循环。我希望有一种更清洁的方法来做到这一点。

我知道我可以使用Counter 来计算唯一值的出现次数：

from collections import Counter
Counter(val["sender"] for val in d)

这会给我：

>>> ({"a":2, "c":1})

但我怎样才能为一对值执行此操作并为每个值分别设置字典？

提前谢谢你，我希望我的问题足够清楚

Answer 1

这是一种使用dict 方法的简单迭代的方法。

例如：

d = [{"sender": "a", "time": 123, "reciever": "b", "amount": 2}, {"sender": "c", "time": 124, "reciever": "b", "amount": 10}, {"sender": "a", "time": 130, "reciever": "b", "amount": 5}]
result = {}
for i in d:
    key = (i['sender'], i['reciever'])
    # del i['time']  # if you do not need `time` key
    if key not in result:
        i.update({'total_amount': i.pop('amount'), 'count': 1})
        result[key] = i
    else:
        result[key]['total_amount'] += i['amount']
        result[key]['count'] += 1

print(list(result.values()))

输出：

[{'count': 2, 'reciever': 'b', 'sender': 'a', 'time': 123, 'total_amount': 7},
 {'count': 1, 'reciever': 'b', 'sender': 'c', 'time': 124, 'total_amount': 10}]

Answer 2

纯python方式是创建一个sender:reciever对的新哈希表

我更新了它以按要求计算total amount。

d = [{"sender": "a", "time": 123, "reciever": "b", "amount": 2},
     {"sender": "c", "time": 124, "reciever": "b", "amount": 10},
     {"sender": "a", "time": 130, "reciever": "b", "amount": 5}]

nd = {}

for o in d:
  sender = o['sender']
  recv = o['reciever']
  amount = o['amount']

  k = sender + ":" + recv
  if k not in nd:
    nd[k] = (0, 0)

  nd[k] = (nd[k][0] + 1, nd[k][1] + amount)

print nd

导致{'c:b': (1, 10), 'a:b': (2, 7)}

Answer 3

您可以使用 pandas 将字典列表解析为数据框。
数据帧将允许您轻松总结某些发送方接收方对的 amount 字段。

import pandas as pd

dict = [{"sender": "a", "time": 123, "receiver": "b", "amount": 2},   
        {"sender": "c", "time": 124, "receiver": "b", "amount": 10},   
        {"sender": "a", "time": 130, "receiver": "b", "amount": 5}]

df = pd.DataFrame.from_records(dict)
group = df.groupby(by=['sender', 'receiver'])

result = group.sum()
result['occurrences'] = group.size()
print(result)

会输出

                 time  amount  occurrences
sender receiver
a      b          253       7            2
c      b          124      10            1

Answer 4

Max Crous's 答案比这更优雅，但如果你想避免额外的库：这是一种纯 python 方式：

import collections
result = collections.defaultdict(lambda : [0,0])
for e in d: 
    result[(e['sender'],e['reciever'])][0]+=e['amount']
    result[(e['sender'],e['reciever'])][1]+=1 

Result 现在是一个字典，其中发送者和接收者的元组作为键，2 元素列表 [total_amount, count] 作为值

Answer 5

Imo 最简单和最干净的解决方案是使用 defaultdict:

from collections import defaultdict

dct = [{"sender": "a", "time": 123, "reciever": "b", "amount": 2},
       {"sender": "c", "time": 124, "reciever": "b", "amount": 10},
       {"sender": "a", "time": 130, "reciever": "b", "amount": 5}]

result = defaultdict(int)
for item in dct:
    key = "{}:{}".format(item["sender"], item["reciever"])
    result[key] += item["amount"]

print(result)

结果

defaultdict(<class 'int'>, {'a:b': 7, 'c:b': 10})

此外，不要调用你的变量dict 或list。

Answer 6

使用字典，您可以将发送者设置为键，将值设置为接收者和数量，然后递增/添加接收者，数量

dict = [{"sender": "a", "time": 123, "reciever": "b", "amount": 2}, {"sender": "c", "time": 124, "reciever": "b", "amount": 10}, {"sender": "a", "time": 130, "reciever": "b", "amount": 5}]
dict1={}
for eachitem in dict:
        if(eachitem["sender"] in dict1.keys()):
             dict1[eachitem["sender"]]["amount"]=dict1[eachitem["sender"]]["amount"]+eachitem["amount"]
             dict1[eachitem["sender"]]["reciever"]+=1

        else:
            dict1[eachitem["sender"]]={"reciever":1,"amount":eachitem["amount"]}

print(dict1)

输出

{'a': {'reciever': 2, 'amount': 7}, 'c': {'reciever': 1, 'amount': 10}}