Home/ Questions/Typeerror argument of type nonetype is not iterable - simple way to solve it
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Dakota Miller
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Asked: 2022-05-18T15:47:08+00:00 In:

Typeerror argument of type nonetype is not iterable – simple way to solve it

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I encountered the following problem in completing my work:

File "main.py", line 42, in <module>
  main()
  File "main.py", line 32, in main
  diff()
  File "main.py", line 17, in diff
  gamePlay(difficulty)
  File "/Users/Nathan/Desktop/Hangman/gameplay.py", line 9, in gamePlay
  getInput(word)
  File "/Users/Nathan/Desktop/Hangman/gameplay.py", line 25, in getInput
  if guess in wordInput:

Below is the code I ran:

import random
 
 easyWords = ["car", "dog", "apple", "door", "drum"]
 
 mediumWords = ["airplane", "monkey", "bananana", "window", "guitar"]
 
 hardWords = ["motorcycle", "chuckwalla", "strawberry", "insulation", "didgeridoo"]
 
 wordCount = []
 
 #is called if the player chooses an easy game. 
 #The words in the array it chooses are the shortest.
 #the following three functions are the ones that
 #choose the word randomly from their respective arrays.
 def pickEasy():
  word = random.choice(easyWords)
  word = str(word)
  for i in range(1, len(word) + 1):
  wordCount.append("_")
 
 #is called when the player chooses a medium game.
 def pickMedium():
  word = random.choice(mediumWords)
  for i in range(1, len(word) + 1):
  wordCount.append("_")
 
 #is called when the player chooses a hard game. 
 def pickHard():
  word = random.choice(hardWords)
  for i in range(1, len(word) + 1):
  wordCount.append("_")
from words import *
 from art import *
 
 def gamePlay(difficulty):
  if difficulty == 1:
  word = pickEasy()
  print start
  print wordCount
  getInput(word)
 
  elif difficulty == 2:
  word = pickMedium()
  print start
  print wordCount
 
  elif difficulty == 3:
  word = pickHard()
  print start
  print wordCount
 
 def getInput(wordInput):
  wrong = 0
  guess = raw_input("Type a letter to see if it is in the word: \n").lower()
 
  if guess in wordInput:
  print "letter is in word"
 
  else:
  print "letter is not in word"

What’s causing it, and how can it be resolved in the “typeerror argument of type nonetype is not iterable“ in the python?

typeerror
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2 Answers

  1. Best Answer
    Expert

    The cause: If a function returns nothing, for example:

    def test():
    pass

    It has a return value of None implicitly.

    As your pick* methods don’t return anything, for example:

    def pickEasy():
    word = random.choice(easyWords)
    word = str(word)
    for i in range(1, len(word) + 1):
        wordCount.append("_")

    the lines that call them, for example:

    word = pickEasy()

    word has been set to None, so wordInput in getInput is also None. That is to say:

    if guess in wordInput:

    has the same meaning as

    if guess in None:

    Because None is an instance of NoneType, which lacks iterator/iteration capabilities, the type error occurs.

    The solution: You can fix this error by adding the return type:

    def pickEasy():
    word = random.choice(easyWords)
    word = str(word)
    for i in range(1, len(word) + 1):
        wordCount.append("_")
    return word
    • 18

  2. The python error wordInput says wordInput is an iterable. It is actually of NoneType.

    You will notice wordInput if you print wordInput prior to the line in question.

    wordInput is None. This means that the argument passed on to the function is also None. This is word. The pickEasy result is assigned to word.

    Your pickEasy function doesn’t return anything. A Python method that doesn’t return any results in a NoneType.

    You may have wanted to return word. This will suffice. 

    def pickEasy():
 word = random.choice(easyWords)
 word = str(word)
 for i in range(1, len(word) + 1):
 wordCount.append("_")
 return word
    • 11

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