add value to dictionary key as list loop for-CodePudding

I am trying to add a list of value for a key. I use the following lines :

d = dict()

for i in range(5):
    d['key'].append(i)

But I get the error : KeyError: 'key'

The expected output is : {'key' : [0,1,2,3,4]}

How to fix this error ?

CodePudding user response：

You can't append to a nonexistent list. Initialize d['key'] first.

d = {'key': []}
for i in range(5):
    d['key'].append(i)

CodePudding user response：

You can use defaultdict to create automatically an object for a missing key, in this case an empty list.

from collections import defaultdict

d = defaultdict(list)

for i in range(5):
    d['key'].append(i)

print(d)
# defaultdict(<class 'list'>, {'key': [0, 1, 2, 3, 4]})

CodePudding user response：

Python has no way to know that the values for the key key are a list. First you should initialize the key and then append the values like:

d = dict()
d['key'] = []

for i in range(5):
    d['key'].append(i) 
    
print(d)
# {'key': [0, 1, 2, 3, 4]}

CodePudding user response：

Given a dict that might contain 'key', I would write this with a try/except over KeyError and start the list if it doesn't exist

for i in range(5):
    try:
        d['key'].append(i)
    except KeyError:    # key not in d (yet)
        d['key'] = [i]  # start a new list

CodePudding user response：

First of all what are you trying to accomplish here? If you want to create a list value for that key you should do it like

d['key'] = [0, 1, 2, 3, 4]

Anyway, the problem in your code is that you are trying to append a value in a list that does not exist. A solution to your problem could be

d = dict()

for i in range(5):
    if 'key' not in d.keys():
        d['key'] = []
    else:
        d['key'].append(i)

CodePudding user response：

Just another unmentioned way, for the one liners:

d = {'key' : [i for i in range(5)]}

This is called comprehension in python: python comprehension

CodePudding user response：

d = {}
xkey =1
i=[]
for x in range(5):
  i.append(x)
d[xkey] = i
print(d)

The result

{1: [0, 1, 2, 3, 4]}