如何在一个循环中比较两个列表的值是否相等？

Question

So I want to make this so I can check if another code is working, but I keep getting this error:

'list index out of range'

For the following code:

for L1[i] in range(0, len(L1)):
    if L1[i]==L2[i]:
        L1[i]='ok'

What is going wrong?

Answer 1

You probably are looking for something more like this. I recommend prevalidating your list lengths are equal so that your loop doesn't fall over.

assert len(L1) == len(L2)

for i in range(len(L1)):
    if L1[i] == L2[i]:
        L1[i] = 'ok'

Alternately, if it is acceptable for your lists to be of different lengths, simply take the minimum of the two lengths as your exclusive upper bound.

upper_bound = min(len(L1), len(L2))
for i in range(upper_bound):

Answer 2

How about:

import itertools
zipped_pairs = itertools.izip_longest(L1, L2, fillvalue=object()) # generator of pairs (L1[n],L2[n])
equals_tests = (a == b for a,b in zipped_pairs) # perform equality test on each pair
all_equal = all(equals_tests) # True if all of the equals_tests items are True

Or in one line:

all((a == b for a,b in itertools.izip_longest(L1, L2, fillvalue=object()))

Note that this will ultimately return false whenever L1 and L2 are of different lengths. If you want to have something else, alter the equals_tests step to use the test you prefer. You may also need to use a different fillvalue parameter for izip_longest .

In fact, here's a version which will treat an empty place as equal to anything:

nonce = object()
all(((a == b) or (nonce in (a,b)) for a,b in itertools.izip_longest(L1, L2, fillvalue=nonce))

Answer 3

Assuming this is Python, there are two problems:

1. You only want to specify i in the beginning of the for -loop.
2. L2 may not have as many items as L1 .

---

for i in range(0, len(L1)):
    try:
        if L1[i] == L2[i]:
            L1[i] = 'ok'
    except IndexError:
        break

As Frederik points out, you could use enumerate as well:

for i, l1 in enumerate(L1):
    try:
        if L[i] == L2[i]:
            L1[i] = 'ok'
    except:
        break

In my opinion, the increase in readability of enumerate over range is mostly offset by defining an extra variable ( l1 ) which you never use. But it is just my opinion.

---

One last option, which may be best, is to use zip to merge the two lists ( zip truncates the longer of the two):

for i, l1, l2 in enumerate( zip(L1, L2) ):
    if l1 == l2:
        L1[i] = 'ok'