How to use a dictionary to append values to a list?-CodePudding

Trying to solve a leetcode problem right now and think I have an idea for a solution but need some help. I am given 2 lists: 1 is a list of values, and the other is the index value where I should place such values. Such as below:

nums = [1, 5, 3]
index = [0, 1, 2]

So, I would have to append number 1 at index 0, 5 at index 1, etc.

I have solved the problem with simple while loop but now I want to solve the problem with a dictionary, so I have made this so far:

 list2 = []

 ans = dict(zip(nums, index))
        
 list2.insert(someindex, somenumber)

My question is how I can access the data in the dictionary to finish my insert command. how can I replace the code at 'someindex' and 'somenumber' to represent the index and corresponding value stored in my dictionary? Is this possible? Thanks for the help

CodePudding user response：

if that is the case, you could order your code as follow and get a list from the dictionary values you are building there, something like this might help you solve the issue, if I understood what you are trying to accomplish with this:

nums = [1, 5, 3]
index = [0, 1, 2]
list2 = list(dict(zip(index,nums)).values())
print(list2)

CodePudding user response：

You could do something like this:

nums = [1, 5, 3]
index = [0, 1, 2]
ans = dict(zip(index, nums))
list2 = [0 for _ in range(len(ans))]

for index, value in ans.items():
    list2[index] = value

Note that you must instantiate list2 the way described because you'll get index out of range exception otherwise.

CodePudding user response：

This is a very convoluted way of doing things. (keep it simple. for loop is much better) but here are a few ways.

list(map(lambda x: x[0], sorted(ans.items(), key=lambda x: x[1])))


[x[0] for x in sorted(ans.items(), key=lambda x: x[1])]

Need sorted here because you never mentioned that indices list is sorted

Going one step back you do not need to create a dict. zip object would suffice

[x[0] for x in sorted(zip(nums, index), key=lambda x: x[1])]

this would save unnecessary transformation to dict