Python:在字典中递增int值

Question

所以我有两个字典，每个字典都有如下方式的键值对：

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

我想用我的代码实现以下目标：

1. 循环查看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（在firstDict中）的int值增加1

2. 做同样的事情，除了主for循环是secondDict，内部循环是firstDict（这是有原因的）。将餐厅键（secondDict的）的int值增加1

代码运行了，但是我得到的不是我想要的，而是：

^{pr2}$

这不是我想要的。int值应该不同。理想情况下，它应该看起来像这样：

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

代码如下：

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])

事先非常感谢。在

Answer 1

如果两个字典中都有键，则两个字典中的int值递增1

1. 通过set intersection方法从这两个字典中获取公共密钥。在
2. 迭代公共键。在
3. 每个常用字典的键值按1递增。在

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}

---

代码问题：

仅当k==k2时才需要递增。所以每当代码进入else loop时，当这个条件为false时，在其他情况下，我们将值递增1。在

只需在if循环中增加值。在

尝试1：

^{pr2}$

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所述操作：

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有共同点的键上操作的，即交叉点。然后您可以使用内置的set数据类型，与嵌套循环相比，它的速度将惊人地快，更不用说您的意图会更清楚：-）