如何并排打印两个不同大小的列表？

Question

I have two lists of different length that I want to print side by side, separated by a tab. As an example:

a_list = ["a","b","c","d","e"]
b_list = ["f","g","h","i","j","k","l","m","n"]

I tried:

print('A-list:'+ '\t' + 'B-list:')
for i in range(len(b_list)):
    print(a_list[i] + '\t' + b_list[i])

I off course get an "out of range-trace" back because one list is shorter. I do not wish to use zip.

Answer 1

your code is rasing IndexError because the length of b_list is greater than the length of a_list , for this, you could use a try except statement:

for i in range(max(len(b_list), len(a_list))):
    try:
        print(f"{a_list[i]}\t", end="")
    except IndexError:
        print(f" \t", end="")
    try:
        print(f"{b_list[i]}\t")
    except IndexError:
        print(f" \t")

output:

a   f
b   g
c   h
d   i
e   j
    k
    l
    m
    n

---

Answer 2

This is a possible solution. It doesn't matter if a_list is longer or shorter than b_list .

def get(lst, idx):
    try:
        return lst[idx]
    except:
        return " "

a_list = ['a','b','c','d','e'] 
b_list = ['f','g','h','i','j','k','l','m','n']

result = []

for i in range(min(len(a_list), len(b_list))):
    result.append(get(a_list, i))
    result.append(get(b_list, i))

for i in range(min(len(a_list), len(b_list)), max(len(a_list), len(b_list))):
    result.append(get(a_list, i))
    result.append(get(b_list, i))

print('\n'.join('\t'.join((result[i], result[i+1]))
                          for i in range(0, len(result), 2)))

This prints the expected output:

a   f
b   g
c   h
d   i
e   j
    k
    l
    m
    n

Answer 3

without any extra loop:

list_a = ["A", "B", "C"]
list_b = ["1", "2", "3", "4"]
len_a = len(list_a)
len_b = len(list_b)
i = 0
while True:
    if i < len_a:
        print(list_a[i], end="")
    if i < len_b:
        print("\t", list_b[i], end="")
    print()
    i += 1
    if i > max(len_a, len_a):
        break

Answer 4

def iterate(a_list, b_list):
  # iterate through short list
  min_length = min(((len(a_list), len(b_list)))) 

  for i in range(min_length):
    print('{}\t{}'.format(a_list[i], b_list[i]))

  # print remaining items accordingly
  if i + 1 == len(a_list):
    for item in b_list[i+1:]:
      print('\t{}'.format(item))  
  else:
    for item in a_list[i+1:]:
      print(item) 

I advise use the string method format which allows you to have elements of any kind in your lists.

Answer 5

In all solutions, the trick is to go through the longest list and use an empty space for missing entries in the shorter list.

For example:

a_list = ['a','b','c','d','e'] 
b_list = ['f','g','h','i','j','k','l','m','n']

for i in range(max(len(a_list),len(b_list))):
    a = a_list[i:i+1] or [""]
    b = b_list[i:i+1] or [""]
    print(a[0]+"\t"+b[0])

If you're not concerned with memory usage, you could also pad the lists with blanks in order to have a simpler printing process that assumes lists of same sizes:

padding  = len(b_list)-len(a_list)
a_padded = a_list + [""]*padding
b_padded = b_list + [""]*-padding
for i in range(len(a_padded)):
    print(a_padded[i]+"\t"+b_padded[i])

Answer 6

You could try this:

test1 = [1, 2, 3, 4]
test2 = [1, 2, 3, 4, 5]
test3 = [1, 2]

while test or test2 or test3:
    print(test.pop() if test else '',\
          test2.pop() if test2 else '',\
          test3.pop()if test3 else '')