Get New List Based on Dictionary Key

[leoahum]

I'm new to Python. If I have a list indicate the order for a new list with the key in a dictionary, how could I get the new list with the values in the dictionary.

For instance:

l = ['a','b','a','c','b']
D = {'a':[3,2],'b':[4,5],'c':[1]}

How can I get a new list as:
L = [3,4,2,1,5]


A lot of thanks!

[ichabod801]

What have you tried? We're not big on writing code for people, but we're big on helping people with code they've written.

[leoahum]

Here is my code. It gives me the wanted list, but the two "for" loops is a little redundant. I'm wondering if there is simpler way to code it out.

l = ['a','b','a','c','b']
D = {'a':[3,2],'b':[4,5],'c':[1]}


L = []

for i in range(len(l)):
    for k in D:
        if l[i] == k:
            L.append(D[k][0])
            D[k].remove(D[k][0])

Thanks!

[buran]

something like

my_list= ['a','b','a','c','b']
my_dict = {'a':[3,2],'b':[4,5],'c':[1]}
result = [my_dict[key].pop(0) for key in my_list]
print(result)

but it assumes all elements in my_list are present in my_dict.keys and there are enough elements in the respective list in the dict.
Otherwise

my_list= ['a','b','a','c','b']
my_dict = {'a':[3],'b':[4,5],'c':[1]}
result = []
for key in my_list:
    try:
        result.append(my_dict[key].pop(0))
    except (KeyError, IndexError):
        continue
print(result) 

[leoahum]

Awesome, thanks!