Python中轮流使用time.sleep打印两个列表

Question

I have two lists: [1,2,3,4,5] and ['a','b','c','d','e','f'] . (Note that one has 5 items, the other 6). What I want is to print them, item-by-item, taking turns, with a one-second delay, each list on its own line. I would imagine the code for printing one list to be something like this:

import time
for x in [1,2,3,4,5]:
    print(x, end=' ')
    time.sleep(1)

But, after each digit appears in the output, I want a letter to appear underneath it, with a 1 second delay, so that the printing order would be: 1-a-2-b-3-c... , and the end result would look as follows:

1 2 3 4 5 
a b c d e f 

Is there a way to make it happen in Python?

Answer 1

import time
# import os
from IPython.display import clear_output
a = [1, 2, 3, 4, 5]
b = ['a', 'b', 'c', 'd', 'e', 'f']

line_one = ''
line_two = ''
for i1, i2 in zip(a, b):
    # os.system('CLS')
    clear_output(wait=True)
    line_one += (str(i1)+" ")
    print(line_one, end="\n")
    print(line_two, end="", flush=True)
    line_two += (i2+" ")
    time.sleep(1)
    print(i2, end=" ", flush=True)
    time.sleep(1)
for i in range(len(b)-len(a)):
    print(b[len(a)+i],end=" ",flush=True)
    time.sleep(1)

This code should work in jupyter notebook. for windows command line, you can use os.system('CLS') instead of clear_output() .