How to find position of char in Two-dimensional lists?-CodePudding

I need to find position of specific char in two-dimensional list

I tired

list=[["a","b","c"],["d","e","f"],["g","h","i"]]
print(list.index("c"))

and

print(list.index[0]("c"))

But it does not work:
print(list.index("c"))
builtins.ValueError: 'c' is not in list

CodePudding user response：

You have to loop over the rows. Then check whether the searched character is in the row before trying to get its index.

for i, row in enumerate(list):
    if 'c' in row:
        print(i, row.index('c'))

CodePudding user response：

You could try this way:

Explain: your list is a nested list containing sublists, so you have to go one level deeper to get the value that's you're interested.

Note - please avoid using built-in list as your variable name -that's not a good practice.

# L is your list...
# val = 'c'

for idx, lst in enumerate(L):
    if val in lst:
        print(f' which sub-list: {idx}, the value position: {lst.index(val)}')

# Output:
# which sub-list: 0, the value position: 2

# running with val is `g`
# >>> which sub-list: 2, the value position: 0

CodePudding user response：

print(list.index("c")) does not work, as list as a list containing lists, not chars. I assume you are looking for coordinates as an answer; something like

for i in range(len(list)):
    for j in range(len(list[i])):
        if (list[i][j] == 'c'):
            print(f"c at list[{i}][{j}]")

will give you that.

As for your second suggestion, you need to call index to list[0], so:

list[0].index('c')

CodePudding user response：

You could do something like this [enter image description here][1]

def find_char(list, char):
  for i in range(len(list)):
    for j in range(len(list[i])):
      if list[i][j] == char:
        return (i, j)
  return None

list = [["a","b","c"],["d","e","f"],["g","h","i"]]
char = "c"
position = find_char(list, char)
found = ""
if position:
  found = list[position[0]][position[1]]
else:
  print(f"Character {char} not found in list")
  
print(found)


  [1]: https://i.stack.imgur.com/iDnI1.png