Python: Got an Error Statement for "local variable 'idx' referenced before assignment"

Question

def removeNonAlpha(myString, key):
      from string import ascii_lowercase
      alphabet = ascii_lowercase + ' '
      myString = myString.lower()
      cipherText = ' '
      for ch in myString:
        if ch == 1 or ch == 2 or ch == 3 or ch == 4 or ch == 5 or ch == 6 or ch == 7 or ch == 8 or ch == 9 :
          idx = alphabet.find(ch)
        cipherText = cipherText + key[idx] 
      return cipherText

Running my script with:

removeNonAlpha('My favorite Cat was named Patches', alphabet)

returns the following error

          7     if ch == 1 or ch == 2 or ch == 3 or ch == 4 or ch == 5 or ch == 6 or ch == 7 or ch == 8 or ch == 9 :
          8       idx = alphabet.find(idx)
    ----> 9     cipherText = cipherText + key[idx]
         10   return cipherText

UnboundLocalError: local variable 'idx' referenced before assignment

What is wrong with my code?

Think about what happens when your `if` condition is false on the first iteration, then you *don't assign anything to `idx`* — juanpa.arrivillaga, Sep 30 '20 at 04:33
Note also, if `myString` is a `str`, then `ch` will *never* equal an `int` as you are testing in your condition — juanpa.arrivillaga, Sep 30 '20 at 04:35
See [UnboundLocalError on local variable when reassigned after first use](https://stackoverflow.com/questions/370357/unboundlocalerror-on-local-variable-when-reassigned-after-first-use) — Shivam Jha, Sep 30 '20 at 04:37

score 2 · Answer 1 · answered Sep 30 '20 at 04:37

2

It means you tried to use the variable 'idx' before creating. 'idx' is only created inside the if, but the line that uses 'idx' is outside the 'if' statement, so it runs even when the 'if' fails

answered Sep 30 '20 at 04:37

João Vitor Barbosa

204
1
9

score 0 · Answer 2 · answered Sep 30 '20 at 04:41

just change the function to following:

def removeNonAlpha(myString, key):
  from string import ascii_lowercase
  alphabet = ascii_lowercase + ' '
  myString = myString.lower()
  cipherText = ' '
  for ch in myString:
    if ch == 1 or ch == 2 or ch == 3 or ch == 4 or ch == 5 or ch == 6 or ch == 7 or ch == 8 or ch == 9 :
      idx = alphabet.find(ch)
      cipherText = cipherText + key[idx] #edits
  return cipherText

score 0 · Answer 3 · answered Sep 30 '20 at 10:03

its because your idx is not getting initialised if the condition does not pass, moreover there are some changes you could make with regards to that massive conditional, try this:

def removeNonAlpha(myString, key):
      from string import ascii_lowercase
      alphabet = ascii_lowercase + ' '
      myString = myString.lower()
      cipherText = ' '
      for ch in myString:
        idx = #make some default value 
        if ch in ['123456789']:
          idx = alphabet.find(ch)
        cipherText = cipherText + key[idx] 
      return cipherText

You have to initialise idx with some default value and maybe add a check if idx was found or not.

score 0 · Answer 4 · answered Apr 19 '21 at 16:19

Update: Completely changed my code but here's the working results:

`def removeNonAlpha(myString):
  from string import ascii_lowercase
  alphabet = ascii_lowercase + ' '
  myString = myString.lower()
  cipherText = ' '
  

  return myString


def getFreq(tuples): 
  return tuples[1]
  # this returns the second element of a tuple #
countDict = {}
with open('wells.txt', 'r', encoding='utf-8') as wells:
  text = wells.read()
  for word in text.split():
    word = removeNonAlpha(word)
    if word in countDict:
      countDict[word] += 1
    else: 
      countDict[word] = 1


wordList = list(countDict.items())

wordList.sort(key = getFreq)

for wordTuple in wordList[-10:]:
  print(wordTuple)`

Python: Got an Error Statement for "local variable 'idx' referenced before assignment"

4 Answers4