使用for循环python创建简单表

Question

Im just trying to create simple table with following this code

n = 0
x = ["list1","list2","list3","list4"]
for y in x:
    n+=1
    h="{}. {}".format(n,y) + '   ' + "{}. {}".format(n+n,y)
    print(h)

And output i got

1. list1  5. list1
2. list2 6. list2
3. list3  7. list3
4. list4  8. list4

What should i do to got output like this

1. list1  3. list3
2. list2  4. list4

Im just beginning and so confused to solved it,thanks for who can answer my issue.

Answer 1

The below should give you what you are looking for, no matter how long the list is. There may be a more efficient way than this but this way breaks it down into its parts, so essentially you need to find the halfway point in the list and print the respective value on the right.

x = ["list1","list2","list3","list4", "list5",]

if len(x) % 2 == 1 :
    half = int((len(x) + 1) / 2)
else:
    half = int(len(x) / 2)

for i in range(half):
    if half > len(x) / 2:
        if i == half - 1:
            h = "{}. {}".format (i + 1 , x[i])
        else:
            h = "{}. {}".format (i + 1 , x[i]) + '   ' + "{}. {}".format (half + (i + 1) , x[int (len (x) / 2) + i + 1])
    else:
        h = "{}. {}".format (i + 1 , x[i]) + '   ' + "{}. {}".format (half + (i + 1), x[int (len (x) / 2) + i])
    print(h)

I have also taken into account lists that have an odd number of elements like the above example for x , which will give you the following output

1. list1   4. list4
2. list2   5. list5
3. list3

Answer 2

Something like this?

import math
x = ["list1","list2","list3","list4", "list5", "list6", "list7"]
y = math.ceil(len(x)/2)
for i, j in enumerate(x[:math.ceil(len(x)/2)], start=1):
    try:
        print("{}. {}".format(i, j) + '     '"{}. {}".format(i+math.ceil(len(x)/2), x[y]))
    except:
        print("{}. {}".format(i, j))
    y +=1

Output:

1. list1     5. list5
2. list2     6. list6
3. list3     7. list7
4. list4