My recursive function always returns None - list

Question

This is a recursive function that attempts to return how many times an item appears in a list. Following is the code:
def count (item, l_ist2): # function that returns the number of times item appears in l_ist
# Base case: item appears 0 times
print (item in l_ist2)
if item in l_ist2 == False:
return 0
# Recursive case
elif item in l_ist2 == True:
l_ist2.remove(item)
print (l_ist2, "\n")
return count (item, l_ist2) + 1
list_for_task_2 = [5,56,2,57,2,46,5,345,5,56,2,56,56,345]
item = int (input ("Choose an item in the list [5,56,2,57,2,46,5,345,5,56,2,56,56,345]: "))
print ("The item", item, "appears", count(item,list_for_task_2), "time(s) in the list above.\n")
Somehow, the output is always: "The item 5 appears None time(s) in the list above."
Thanks in advance to anyone who can help me answer this question!
P.S. This assignment of mine specifically ask for the code to be written recursively.

Answer 1

There is one case in your recursive function that silently returns None; this is what is creating the issue.
Further, your recursion uses python in that scans the entire sequence each time the function is called... this is probably not what you intended to do.
I replaced this with a mechanism that pops an element from the sequence and compares it to the item to be counted. This implies that the recursive case is now whether or not the sequence is empty or not.
def count (item, seq):
"""returns the number of times item appears in seq
"""
if not seq: # is the sequence empty or not
return 0
elif seq.pop() == item: # if not, pop an element and compares it to the item
return count(item, seq) + 1 # add 1 if found
else:
return count(item, seq) # do not add 1 if not found
list_for_task_2 = [5, 56, 2, 57, 2, 46, 5, 345, 5, 56, 2, 56, 56, 345]
item = int(input ("Choose an item in the list [5, 56, 2, 57, 2, 46, 5, 345, 5, 56, 2, 56, 56, 345]: "))
print("The item", item, "appears", count(item,list_for_task_2), "time(s) in the list above.\n")