在 python 中不使用嵌套 for 循环访问来自两个子列表的元素

Question

for example:

list = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]

and I want to compare the first element of two consecutive sublists in every iteration of for loop.
How should I write the for loop?

I have tried this following code:

for x in list:
    print(x[0])
    print(x+1[0])

I know the use of ( x+1[0] ) is completely illogical. But what should I use at this place? Any suggestions? Like comparing(any type of mathematical comparison) 1st elements of list1 & list2, list3 & list4.

Answer 1

Try:

for x in list1:
    for y in list1:
        if x[0] == y[0]:
            print(x[0],y[0])

Now if you don't want to compare the same elements (which the above code does):

for x in list1:
    for y in [i for i in list1 if i!=x]:
        if x[0] == y[0]:
            print(x[0],y[0])

Or you could print like:

for x in list1:
    for y in [i for i in list1 if i!=x]:
        if x[0] == y[0]:
            print(x ,"and", y , 'have the same elements first')

Here:

list1 = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

Answer 2

You can use zip() to pair off consecutive lists by passing slices to zip. To pair them off at an offset like: [0, 1], [1, 2], [2, 3] you could use:

zip(l, l[1:])

To pair them off as pairs (rather than repeating the last one like [0, 1], [2, 3] ) you could use:

zip(l[::], [1:1])

In your example that would look like:

l = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]

for l1, l2 in zip(l[::2], l[1::2]):
    print(l1[0], l2[0])

Prints:

1 5
9 13

Answer 3

following code will genrate first element from every sublist. Tweak it according to your need

lst = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

for i in range(len(lst[0])):
    for j in range(len(lst)):
        print(lst[j][i], end=" ")
    print()