Question

所以我有两个字典，每个字典都有以下方式的键值对：

firstDict = {
     'HImJYsulal': 0
     'psNnxwFVmv': 0
     '4B0IaN1P5x': 0
     'MxZzGOlefq': 0
}

我想用我的代码实现的目标如下：

循环遍历第一个和第二个字典。如果第一个字典的值在第二个字典中，则将restaurant_key的int值（在firstDict中）增加1
做同样的事情，除了主要的for循环是secondDict而内部循环是firstDict（其中有一个原因）。 restaurant_key（of secondDict）的int值增加1

代码运行全部，但我没有得到我想要的东西，我得到了：

(u'HImJYsulal', 0)
(u'jXDXpoeuWY', 1)
(u'ctNyMKpCoE', 2)
(u'vWsFNwTnz1', 3)
(u'0zcfI67S6X', 4)
(u'aGjUr2Ittw', 5)
(u'eQ5rGvpoRs', 6)
(u'6Q96oO26ua', 7)
...and so on and so forth

这不是我想要的。 int值应该有所不同。理想情况下它看起来应该是这样的：

(u'HImJYsulal', 4)
(u'jXDXpoeuWY', 0)
(u'ctNyMKpCoE', 1)
(u'vWsFNwTnz1', 5)
(u'0zcfI67S6X', 2)

以下是代码：

import read_spins

firstDict = {}

# initialSpots
    read_spins.read_json('initialSpots.json', firstDict)

#for obj in firstDict:
    #print(obj,firstDict[obj])


#chosenSpots

secondDict = {}

read_spins.read_json('chosenSpots.json', secondDict)


#for all merchants in the initial spot
for k, v in firstDict.iteritems():
    #for all merchants in the chosen spot
    for k2, v2 in secondDict.iteritems():

            #if the merchant appears in the initial spot, and also in the chosen spot, 
            #end the loop and go to the next one. We're only interested in those that aren't in the chosen spot. 
            #This means that they were dropped. 

            if k == k2:

                print("broke with: " + k)
                break

            else:

                #the merchant isn't in the chosen spots,so therefore the merchant was dropped. 
                firstDict[k] = firstDict[k] + 1

#for all merchants in the chosen spot
for k, v in secondDict.iteritems():
    #for all merchants in the initial spot
    for k2, v2 in firstDict.iteritems():

        #if the merchant appears in the chosen spot, but also in the initial spot,
        #end the loop and go to the next merchant. THis means the merchant was
        #originally selected. 

        if k == k2:

            print("broke with: " + k)
            break

        else:

            #the merchant isn't in the initial spot, thus the merchant was added. 
            secondDict[k] = secondDict[k] + 1


for obj in firstDict:
    print(obj, firstDict[obj])

print(" ")
print("CHOSEN SPOTS")
print("++++++++++++")
print(" ")

for obj in secondDict:
    print(obj, secondDict[obj])

非常感谢您提前。

Answer 1

如果密钥互相存在，则将两个字典中的int值增加1

通过set intersection方法从两个字典中获取公用密钥。
对常用密钥进行迭代。
为各个字典的公共密钥增加1。

<强>演示：

>>> a = {"a": 1, "b":2, "c":0}
>>> b = {"d": 1, "b":2, "c":0}
>>> ab = set(b.keys()).intersection(set(a.keys()))
>>> ab
set(['c', 'b'])
>>> for i in ab:
...     a[i] = a[i] + 1
...     b[i] = b[i] + 1
... 
>>> a
{'a': 1, 'c': 1, 'b': 3}
>>> b
{'c': 1, 'b': 3, 'd': 1}

您的代码问题：

只有在k==k2时才需要增加。因此，当此条件为false时，每次代码进入else loop时，否则我们将值递增1。

只需在if循环中增加值。

尝试1 ：

for k, v in secondDict.iteritems():
    increment = False        
    for k2, v2 in firstDict.iteritems():
        if k == k2:
            secondDict[k] = secondDict[k] + 1
            print("broke with: " + k)
            break

尝试2 ：

for k, v in secondDict.iteritems():
    if k in firstDict:
        secondDict[k] = secondDict[k] + 1

Answer 2

您不需要执行嵌套迭代来执行您所声明的操作：

for key in set(firstDict.keys()) & set(secondDict.keys()):
    firstDict[key] += 1
    secondDict[key] += 1

关键是要注意你的两个操作都是在dicts共有的键上操作的，即交叉点。然后你可以使用内置的set数据类型，与嵌套循环相比，速度会非常快 - 更不用说你的意图会更清晰： - ）

Python：增加字典中的int值

2 个答案: