How can I find out if a list contains letters without using any libraries?-CodePudding

In a list of lists:

list=[['3','4','5'],['6','3','5'],['hello','goodbye','something56']]

I want to get rid of the one that has letters. My attempt is:

for i in sub_list:
    if '.*[a-z] .*' in i:
        continue
    else:
        print(i)

However, this is not working.

CodePudding user response：

Try to use str.isalpha() to check if string contains alphabet

list_of_lists = [['3','4','5'], ['6','3','5'], ['hello','goodbye','something56']]
for sub_list in list_of_lists:
    if any(x.isalpha() for x in sub_list):
        continue    #this is what you are looking for
    else:
        print(sub_list)

Output

['3', '4', '5']
['6', '3', '5']

CodePudding user response：

Using just basic Python (no libraries)
It's Bad form to name a variable list since it hides the built-in function list.

Code

numbers = "0123456789"  # list of digits
lst = [['3','4','5'],['6','3','5'],['hello','goodbye','something56'], ['2', '4', 'today']]

new_lst = []
for sublist in lst:
    new_sublist = []
    for item in sublist:
        for c in item:
            if not c in numbers:
                break
        else:
          # no break encountered so only numbers in for c in item
          continue
        break   # break in for c initem, so issue break in for item in sublist
    else:
      # no break, so all items where numbers
      new_lst.append(sublist)  # sublist only had numbers
        
print(new_lst)
# Output: [['3', '4', '5'], ['6', '3', '5']]