如何检查整数序列是否在列表中

Question

I want the program to continue executing until every element on the list is a string.
That is

li = [1, 2, 6, 'h', 'y', 'h', 'y', 4]

should stop when

li = [all elements type are strings]

li = [1, 2, 6, 'h', 'y', 'h', 'y', 4]

while '''there are still numbers in the list''':
    #code keeps on executing and updating li[] with string input
else:
    # output the list with no numbers

This is what I have tried, but if the first[0] and and last[7] element becomes a string, the while loop goes to the last else condition even with int type present in the list. if done sequentially, it works fine.

li = [8, 2, 6, 'h', 'y', 'h', 'y', 4]

for a in li:
    while a in li:
        if type(a) == int:
            x = int(input('Position: '))
            entry = input('Enter: ')
            li.pop(x)
            li.insert(x, entry)
            print(li) # to see what's happening
            li = li
    else:
        print('Board is full')

print(li)

But, I don't want it sequentially.
Therefore, it should continue if

li = [c, 2, 6, 'h', 'y', 'h', 'y', f]

and stop when

li = [a, b, c, 'h', 'y', 'h', 'y', d]

all strings

Answer 1

You can use all or any in conjunction with string.isnumeric() to check this

li = ['a', 1, 2]
while any(str(ele).isnumeric() for ele in li):
    modify li ...

Answer 2

The solution you asked for is possible. But one piece of advice while writing in python, is to avoid doing type checking explicitly (type(a) == int). below is a simple solution.

import string
import random
letter = random.choice(string.ascii_letters) #swapping data
xe = [8, 2, 6, 'h', 'y', 'h', 'y', 4]
for i in range(len(xe)):
    try:
        xe[i].upper()
    except:
        xe.pop(i)
        xe.insert(i,letter)
print(xe)

Answer 3

I think you can do this:

li = [1, 2, 3, h, y, 5, g, 3]
length = len(li)
while count != length:
    for row in li:
    if type(row) == int:
        li.pop(row)
        count += 1

Answer 4

Geckos gave me a clue. Thank you for that. I found out that using isinstance with any() solves the problem

li = ['a', 1, 2]
while any(isinstance(element, int) for element in li):
       #code goes here...