Python从列表中删除项目，其中仅部分项目已知

Question

I have a list of things like this:

My_list = [['A1'],['B2'],['C3']]

I am asking the user to enter a LETTER (only) indicating which element to delete. So, for example, if the user entered 'B', the 2nd entry should be deleted and the list should look like this:

My_list = [['A1'],['C3']]

I have tried several solutions, but below is my best attempt at this...and i believe i'm not too far away from the final answer.

My_list = [['A1','B2','C3']]

print('What LETTER pairing would you like to delete?')
get_letter = input()
delete_letter = get_letter.upper()

for each_element in My_list:
    for each_pair in each_element:
        if each_pair[0] == delete_letter:
            location = My_list.index(each_pair)
            clues_list.pop(location)
            print('done')

when i run and enter 'B' the Compile error says...

ValueError: 'B2' is not in list

...but B2 is in list!

Where am i going wrong?

Answer 1

The error message is correct: 'B2' is not in My_list , although ['B2'] is!

You need to remove each_element , not each_pair , and should not do so while iterating over the list. For example, try:

location = None
for index, each_element in enumerate(My_list):
    for each_pair in each_element:
        if each_pair[0] == delete_letter:
            location = index
            break
if location is not None:
    My_list.pop(location)

However, it seems likely that a different data structure would be more useful to you, eg a dictionary:

data = {'A': 1, 'B': 2, 'C': 3}

where the code becomes as simple as:

if delete_letter in data: 
    del data[delete_letter]

Answer 2

You can have it done in a much more concise way (using list comprehension ):

My_list = [['A1','B2','C3']]

get_letter = raw_input('What LETTER pairing would you like to delete? ')
delete_letter = get_letter.upper()
print delete_letter
new_list = [var for var in My_list[0] if delete_letter not in var]
print new_list

A few caveats: not in will try to find delete_letter anywhere in the variable, so if you want it to have more exact matches you can look into .startswith . Then, my piece of code blindly follows your list of lists.