How do I find the corresponding value when iterating two lists in Python?

Question

I would like to find the corresponding item in a list but the code below seems to find the corresponding item and then moves on item further as shown:

a = [1,2,3,4]
b = [5,6,7,8]
for i in a:
    if i == 1:
       print b[i]

Which gives me:

When I thought it should print:

In which case I'm now having to write:

print b[i-1]

I stated that when looking for "1" it should print "5" but it gives me "6" instead. The question and code seems clear enough to understand for those who have already provided an answer @DimKoim — python_starter, Dec 09 '15 at 12:08
@python_starter take it easy, I am here to help you. It is nice that you find your answer though. — DimKoim, Dec 09 '15 at 12:16
hahah sorry @DimKoim too much coffee and too little sleep : ) — python_starter, Dec 09 '15 at 12:20

score 3 · Accepted Answer · answered Dec 09 '15 at 12:05

python list index starts from 0 .

when you try to iterate a list you will be getting each and every element in the list

for example:

a = [1,2,3,4]
for element in a:
    print element

will be printing [1,2,3,4]

incase if you want to get the index along with the element you should use enumerate function

for example:

a = [1,2,3,4]
for index, element in enumerate(a):
    print index, element

will be printing 0,1 where 0 is the index of 1 in list 1,2 2,3 3,4

in you program you can use enumerate to achieve what you want

a = [1,2,3,4]
b = [5,6,7,8]
for index, i in enumerate(a):
    if i == 1:
       print b[index]

will yield you the result you want.

score 1 · Answer 2 · answered Dec 09 '15 at 12:08

1

Your condition is i==1, so when this is satisfied it naturally always looks up the second element (counting from 0) of b. (If the value 1 appeared at some other location in a, the workaround b[i-1] you are using would fail.)

I suspect what you want is

a = [1,2,3,4]
b = [5,6,7,8]
print b[a.index(1)]

(note that this avoids the unnecessary for loop). But your question isn't clear.

answered Dec 09 '15 at 12:08

snadathur

11
1

Not really a general answer and clearly not what the OP wanted. – erip Dec 09 '15 at 12:15
1

Assuming that `a` contains the value being searched for (in this case, `1`) my answer achieves exactly the same thing as that in the accepted answer, but it will be faster and is more Pythonic. So I don't understand your comment. – snadathur Dec 09 '15 at 14:35
`a = [1,2,3,4,1,2]; b = [5,6,7,8,10000,6,3]; print b[a.index(1)]` Doesn't work. – erip Dec 09 '15 at 20:51

score 1 · Answer 3 · answered Dec 09 '15 at 12:59

In your code, i==1 corresponds to the 0th element of a.

a.index(i) gives us the index with which we can print the corresponding element of b.

a = [1,2,3,4]
b = [5,6,7,8]

for i in a:
    if i == 1:
       print b[a.index(i)]

This would print the first element of b i.e 5

score 0 · Answer 4 · answered Dec 09 '15 at 12:06

The reason, as everyone has mentioned, is because Python lists are 0-based.

When i == 1, you then print b[i] == b[1] == 2

To get the index, you need to use enumerate, which returns the index then the element at that index.

Here's how you should use it.

   for idx, elem in enumerate(a):
     if elem == 1:
       print b[idx]

How do I find the corresponding value when iterating two lists in Python?

4 Answers4