将列表元素映射到它们的位置（Python）

Question

I have a list of URLs. (We can assume that a given URL is met in the list no more than once.)

I need a fast way to determine which of two URLs is before in the list.

I think, I should create the dict from URL to its position in the list.

What is the easy way (without writing a for loop with manual increasing of the counter) to map elements of a list into their positions in the list?

The best thing I conceived is:

order = {}
i = 0
for item in list:
    order[item] = i
    i += 1

Now to check if url1 is before url2 , I check order[url1] < order[url2] .

Can this code be shortened?

Answer 1

This creates your order

order = {k: v for v, k in enumerate(list)}

Example:

L = list('abc')

Your version:

order1 = {}
i = 0
for item in L:
    order1[item] = i
    i += 1
print(order1)

My version:

order2 = {k: v for v, k in enumerate(L)}
print(order2)

Output:

{'a': 0, 'b': 1, 'c': 2}
{'a': 0, 'b': 1, 'c': 2}

Better don't use list for your variable name because it is a built-in.

enumerate provides an iterate that gives you the index and the value for each iteration through your list.

Answer 2

If you want to know which comes first for a specific pair of items, you can use the index method on the list:

a = ['cat', 'dog', 'fish']

a.index('cat') < a.index('dog') # True
a.index('fish') < a.index('dog') # False

Answer 3

List of URLs:

urls = ['A', 'B', 'C', 'D']

List of indices:

index = range(len(urls))

Create the dict:

order = dict(zip(urls, index))

Test:

print(order['A'] < order['B'])  # True

Demo