如何在不使用任何库的情况下找出列表是否包含字母？

Question

在列表列表中：

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

我想摆脱那个有字母的人。 我的尝试是：

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

但是，这是行不通的。

Answer 1

尝试使用 str.isalpha() 检查字符串是否包含字母

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']

Answer 2

• 仅使用基本的 Python（无库）
• 命名变量列表的形式很糟糕，因为它隐藏了内置的 function 列表。

代码

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']]