如何使用Python在此字典中使用空字符串将值添加为缺少的键

Question

the dictionary sample is as below :

d = { 1:'',2:'',3:'',5:'',6:'2',7:'',9:'',10:'6',11:'7',13:'9',14:'',15:'11'}

and i want to add key 4 with empty string as value after key 3, key 8 with empty string as value after key 7 and so on ....I want the simplest code in python.

Answer 1

>>> d = { 1:'',2:'',3:'',5:'',6:'2',7:'',9:'',10:'6',11:'7',13:'9',14:'',15:'11'}
>>> d.update(dict.fromkeys(set(range(16)).difference(d), ''))
>>> d
{0: '', 1: '', 2: '', 3: '', 4: '', 5: '', 6: '2', 7: '', 8: '', 9: '', 10: '6', 11: '7', 12: '', 13: '9', 14: '', 15: '11'}

Note that the dict is unordered, even though it may look ordered in this example!

Answer 2

Python dictionaries do not maintain an order. You cannot add anything 'after' another key.

Just add the keys you are missing:

d.update((missing, '') for missing in set(range(1, max(d) + 1)).difference(d))

which is a compact way of saying:

for index in range(1, max(d) + 1):  # all numbers from 1 to the highest value in the dict
    if index not in d:              # index is missing
        d[index] = ''               # add an empty entry

However, it looks more like you need a list instead:

alist = [None, '', '', '', '', '', '2', '', '', '', '6', '7', '', '9', '', '11']

where alist[15] returns '11' just like your d . The None at the start makes it easier to treat this whit 1-based indexing instead of 0-based, you could adjust your code for that otherwise.