流行与新鲜

>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>> 

流行与新鲜

>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>> 

对于熊猫，你可以有这样的东西，

from pandas import DataFrame
df = DataFrame([{"fruit":"apple", "colour":"red"}])
df.rename(columns = {'fruit':'fruit_name'}, inplace = True)
df.to_dict('records')[0]
>>> {'fruit_name': 'apple', 'colour': 'red'}

我只是要帮我妻子做一些python类的事情，所以我写了这段代码来告诉她如何做。正如标题所示，它只替换键名。这是非常罕见的，你必须替换一个键名，并保持字典的顺序完整，但我还是想分享，因为这篇文章是当你搜索它时谷歌返回的，即使它是一个非常老的线程。

代码:

dictionary = {
    "cat": "meow",
    "dog": "woof",
    "cow": "ding ding ding",
    "goat": "beh"
}


def countKeys(dictionary):
    num = 0
    for key, value in dictionary.items():
        num += 1
    return num


def keyPosition(dictionary, search):
    num = 0
    for key, value in dictionary.items():
        if key == search:
            return num
        num += 1


def replaceKey(dictionary, position, newKey):
    num = 0
    updatedDictionary = {}
    for key, value in dictionary.items():
        if num == position:
            updatedDictionary.update({newKey: value})
        else:
            updatedDictionary.update({key: value})
        num += 1
    return updatedDictionary


for x in dictionary:
    print("A", x, "goes", dictionary[x])
    numKeys = countKeys(dictionary)

print("There are", numKeys, "animals in this list.\n")
print("Woops, that's not what a cow says...")

keyPos = keyPosition(dictionary, "cow")
print("Cow is in the", keyPos, "position, lets put a fox there instead...\n")
dictionary = replaceKey(dictionary, keyPos, "fox")

for x in dictionary:
    print("A", x, "goes", dictionary[x])

输出:

A cat goes meow
A dog goes woof
A cow goes ding ding ding
A goat goes beh
There are 4 animals in this list.

Woops, that's not what a cow says...
Cow is in the 2 position, lets put a fox there instead...

A cat goes meow
A dog goes woof
A fox goes ding ding ding
A goat goes beh

没有直接的方法做到这一点，但你可以删除然后分配

d = {1:2,3:4}

d[newKey] = d[1]
del d[1]

或者做大量的键改变:

d = dict((changeKey(k), v) for k, v in d.items())

d = {1:2,3:4}

假设我们想要改变列表元素p=['a'， 'b']的键值。 下面的代码可以做到:

d=dict(zip(p,list(d.values()))) 

我们得到

{'a': 2, 'b': 4}