NameError occurring after calling nested function











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1
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So I split a .txt file into a list of lists (shown below). However, when I try to run print(splitKeyword(keywords[1][0])) to try and print the first element of the second list/element within keywordList, I get the error: NameError: name 'keywordList' is not defined. How can I fix this?



def functionOne(textFile):
textFileVar = open(textFile, 'r')

def splitKeyword(argument):
keywordList =
for line in argument:
keywordList.append(line.strip().split(','))
return keywordList

splitKeyword(textFileVar)
print(keywordList[1][0])

results = functionOne("text1.txt")
print(results)


This is the text1.txt/textFile/textFileVar contents




hello,world



123,456




This is what keywordList looks like when printed:



[[hello, world], [123, 456]]









share|improve this question


















  • 1




    you are returning from the function but not catching. so try to keywordList = splitKeyword(textFileVar) as keywordList is local to that function.
    – Jay Parikh
    Nov 11 at 5:41















up vote
1
down vote

favorite












So I split a .txt file into a list of lists (shown below). However, when I try to run print(splitKeyword(keywords[1][0])) to try and print the first element of the second list/element within keywordList, I get the error: NameError: name 'keywordList' is not defined. How can I fix this?



def functionOne(textFile):
textFileVar = open(textFile, 'r')

def splitKeyword(argument):
keywordList =
for line in argument:
keywordList.append(line.strip().split(','))
return keywordList

splitKeyword(textFileVar)
print(keywordList[1][0])

results = functionOne("text1.txt")
print(results)


This is the text1.txt/textFile/textFileVar contents




hello,world



123,456




This is what keywordList looks like when printed:



[[hello, world], [123, 456]]









share|improve this question


















  • 1




    you are returning from the function but not catching. so try to keywordList = splitKeyword(textFileVar) as keywordList is local to that function.
    – Jay Parikh
    Nov 11 at 5:41













up vote
1
down vote

favorite









up vote
1
down vote

favorite











So I split a .txt file into a list of lists (shown below). However, when I try to run print(splitKeyword(keywords[1][0])) to try and print the first element of the second list/element within keywordList, I get the error: NameError: name 'keywordList' is not defined. How can I fix this?



def functionOne(textFile):
textFileVar = open(textFile, 'r')

def splitKeyword(argument):
keywordList =
for line in argument:
keywordList.append(line.strip().split(','))
return keywordList

splitKeyword(textFileVar)
print(keywordList[1][0])

results = functionOne("text1.txt")
print(results)


This is the text1.txt/textFile/textFileVar contents




hello,world



123,456




This is what keywordList looks like when printed:



[[hello, world], [123, 456]]









share|improve this question













So I split a .txt file into a list of lists (shown below). However, when I try to run print(splitKeyword(keywords[1][0])) to try and print the first element of the second list/element within keywordList, I get the error: NameError: name 'keywordList' is not defined. How can I fix this?



def functionOne(textFile):
textFileVar = open(textFile, 'r')

def splitKeyword(argument):
keywordList =
for line in argument:
keywordList.append(line.strip().split(','))
return keywordList

splitKeyword(textFileVar)
print(keywordList[1][0])

results = functionOne("text1.txt")
print(results)


This is the text1.txt/textFile/textFileVar contents




hello,world



123,456




This is what keywordList looks like when printed:



[[hello, world], [123, 456]]






python list function file






share|improve this question













share|improve this question











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asked Nov 11 at 5:38









CosmicCat

775




775








  • 1




    you are returning from the function but not catching. so try to keywordList = splitKeyword(textFileVar) as keywordList is local to that function.
    – Jay Parikh
    Nov 11 at 5:41














  • 1




    you are returning from the function but not catching. so try to keywordList = splitKeyword(textFileVar) as keywordList is local to that function.
    – Jay Parikh
    Nov 11 at 5:41








1




1




you are returning from the function but not catching. so try to keywordList = splitKeyword(textFileVar) as keywordList is local to that function.
– Jay Parikh
Nov 11 at 5:41




you are returning from the function but not catching. so try to keywordList = splitKeyword(textFileVar) as keywordList is local to that function.
– Jay Parikh
Nov 11 at 5:41












3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










Try this:



def functionOne(textFile):
textFileVar = open(textFile, 'r')

def splitKeyword(argument):
keywordList =
for line in argument:
keywordList.append(line.strip().split(','))
return keywordList

output = splitKeyword(textFileVar)
print(output[1][0])
return output

results = functionOne("text1.txt")
print(results)


look at return keywordList in splitKeyword function. it returns the value(keywordList). but in other scopes you can not access that variable, so you need to store that in something.






share|improve this answer





















  • is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
    – CosmicCat
    Nov 11 at 6:23










  • I think it is because of another line of your code. but remove it and check if you want.
    – mehrdad-pedramfar
    Nov 11 at 6:30










  • when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
    – CosmicCat
    Nov 11 at 6:37










  • that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
    – mehrdad-pedramfar
    Nov 11 at 6:44


















up vote
0
down vote













Your keywordList is local to the function splitKeyword(), not to the function functionOne(). That's why you're getting a NameError.






share|improve this answer





















  • Is there a workaround?
    – CosmicCat
    Nov 11 at 5:49










  • @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
    – Arjofocolovi
    Nov 11 at 5:51


















up vote
0
down vote













keywordlist is a local variable to the function splitKeyword which return it so you can directly use this function and reduce the code.



def functionOne(textFile):
textFileVar = open(textFile, 'r')
def splitKeyword(argument):
keywordList =
for line in argument:
keywordList.append(line.strip().split(','))
return keywordList

print(splitKeyword(textFileVar))

results = functionOne("text1.txt")
print(results)





share|improve this answer





















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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






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    active

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    up vote
    1
    down vote



    accepted










    Try this:



    def functionOne(textFile):
    textFileVar = open(textFile, 'r')

    def splitKeyword(argument):
    keywordList =
    for line in argument:
    keywordList.append(line.strip().split(','))
    return keywordList

    output = splitKeyword(textFileVar)
    print(output[1][0])
    return output

    results = functionOne("text1.txt")
    print(results)


    look at return keywordList in splitKeyword function. it returns the value(keywordList). but in other scopes you can not access that variable, so you need to store that in something.






    share|improve this answer





















    • is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
      – CosmicCat
      Nov 11 at 6:23










    • I think it is because of another line of your code. but remove it and check if you want.
      – mehrdad-pedramfar
      Nov 11 at 6:30










    • when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
      – CosmicCat
      Nov 11 at 6:37










    • that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
      – mehrdad-pedramfar
      Nov 11 at 6:44















    up vote
    1
    down vote



    accepted










    Try this:



    def functionOne(textFile):
    textFileVar = open(textFile, 'r')

    def splitKeyword(argument):
    keywordList =
    for line in argument:
    keywordList.append(line.strip().split(','))
    return keywordList

    output = splitKeyword(textFileVar)
    print(output[1][0])
    return output

    results = functionOne("text1.txt")
    print(results)


    look at return keywordList in splitKeyword function. it returns the value(keywordList). but in other scopes you can not access that variable, so you need to store that in something.






    share|improve this answer





















    • is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
      – CosmicCat
      Nov 11 at 6:23










    • I think it is because of another line of your code. but remove it and check if you want.
      – mehrdad-pedramfar
      Nov 11 at 6:30










    • when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
      – CosmicCat
      Nov 11 at 6:37










    • that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
      – mehrdad-pedramfar
      Nov 11 at 6:44













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Try this:



    def functionOne(textFile):
    textFileVar = open(textFile, 'r')

    def splitKeyword(argument):
    keywordList =
    for line in argument:
    keywordList.append(line.strip().split(','))
    return keywordList

    output = splitKeyword(textFileVar)
    print(output[1][0])
    return output

    results = functionOne("text1.txt")
    print(results)


    look at return keywordList in splitKeyword function. it returns the value(keywordList). but in other scopes you can not access that variable, so you need to store that in something.






    share|improve this answer












    Try this:



    def functionOne(textFile):
    textFileVar = open(textFile, 'r')

    def splitKeyword(argument):
    keywordList =
    for line in argument:
    keywordList.append(line.strip().split(','))
    return keywordList

    output = splitKeyword(textFileVar)
    print(output[1][0])
    return output

    results = functionOne("text1.txt")
    print(results)


    look at return keywordList in splitKeyword function. it returns the value(keywordList). but in other scopes you can not access that variable, so you need to store that in something.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 11 at 6:13









    mehrdad-pedramfar

    3,85211233




    3,85211233












    • is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
      – CosmicCat
      Nov 11 at 6:23










    • I think it is because of another line of your code. but remove it and check if you want.
      – mehrdad-pedramfar
      Nov 11 at 6:30










    • when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
      – CosmicCat
      Nov 11 at 6:37










    • that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
      – mehrdad-pedramfar
      Nov 11 at 6:44


















    • is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
      – CosmicCat
      Nov 11 at 6:23










    • I think it is because of another line of your code. but remove it and check if you want.
      – mehrdad-pedramfar
      Nov 11 at 6:30










    • when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
      – CosmicCat
      Nov 11 at 6:37










    • that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
      – mehrdad-pedramfar
      Nov 11 at 6:44
















    is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
    – CosmicCat
    Nov 11 at 6:23




    is the 'return output' needed? when I remove it, I get the console output I'm looking for but the word 'None' appears right under it
    – CosmicCat
    Nov 11 at 6:23












    I think it is because of another line of your code. but remove it and check if you want.
    – mehrdad-pedramfar
    Nov 11 at 6:30




    I think it is because of another line of your code. but remove it and check if you want.
    – mehrdad-pedramfar
    Nov 11 at 6:30












    when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
    – CosmicCat
    Nov 11 at 6:37




    when I keep return output, I don't get none, but when I remove it, i get none, will this cause any issues?
    – CosmicCat
    Nov 11 at 6:37












    that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
    – mehrdad-pedramfar
    Nov 11 at 6:44




    that is because of results = functionOne("text1.txt"). return output will return the value so results won't be empty(None), if you remove it, your function does not have any outputs so results would be None.
    – mehrdad-pedramfar
    Nov 11 at 6:44












    up vote
    0
    down vote













    Your keywordList is local to the function splitKeyword(), not to the function functionOne(). That's why you're getting a NameError.






    share|improve this answer





















    • Is there a workaround?
      – CosmicCat
      Nov 11 at 5:49










    • @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
      – Arjofocolovi
      Nov 11 at 5:51















    up vote
    0
    down vote













    Your keywordList is local to the function splitKeyword(), not to the function functionOne(). That's why you're getting a NameError.






    share|improve this answer





















    • Is there a workaround?
      – CosmicCat
      Nov 11 at 5:49










    • @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
      – Arjofocolovi
      Nov 11 at 5:51













    up vote
    0
    down vote










    up vote
    0
    down vote









    Your keywordList is local to the function splitKeyword(), not to the function functionOne(). That's why you're getting a NameError.






    share|improve this answer












    Your keywordList is local to the function splitKeyword(), not to the function functionOne(). That's why you're getting a NameError.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 11 at 5:45









    Arjofocolovi

    82




    82












    • Is there a workaround?
      – CosmicCat
      Nov 11 at 5:49










    • @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
      – Arjofocolovi
      Nov 11 at 5:51


















    • Is there a workaround?
      – CosmicCat
      Nov 11 at 5:49










    • @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
      – Arjofocolovi
      Nov 11 at 5:51
















    Is there a workaround?
    – CosmicCat
    Nov 11 at 5:49




    Is there a workaround?
    – CosmicCat
    Nov 11 at 5:49












    @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
    – Arjofocolovi
    Nov 11 at 5:51




    @CosmicCat Well your function splitKeyword() returns something, so you might as well use this return right? Take what it returns in a variable and then use this variable to print it.
    – Arjofocolovi
    Nov 11 at 5:51










    up vote
    0
    down vote













    keywordlist is a local variable to the function splitKeyword which return it so you can directly use this function and reduce the code.



    def functionOne(textFile):
    textFileVar = open(textFile, 'r')
    def splitKeyword(argument):
    keywordList =
    for line in argument:
    keywordList.append(line.strip().split(','))
    return keywordList

    print(splitKeyword(textFileVar))

    results = functionOne("text1.txt")
    print(results)





    share|improve this answer

























      up vote
      0
      down vote













      keywordlist is a local variable to the function splitKeyword which return it so you can directly use this function and reduce the code.



      def functionOne(textFile):
      textFileVar = open(textFile, 'r')
      def splitKeyword(argument):
      keywordList =
      for line in argument:
      keywordList.append(line.strip().split(','))
      return keywordList

      print(splitKeyword(textFileVar))

      results = functionOne("text1.txt")
      print(results)





      share|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        keywordlist is a local variable to the function splitKeyword which return it so you can directly use this function and reduce the code.



        def functionOne(textFile):
        textFileVar = open(textFile, 'r')
        def splitKeyword(argument):
        keywordList =
        for line in argument:
        keywordList.append(line.strip().split(','))
        return keywordList

        print(splitKeyword(textFileVar))

        results = functionOne("text1.txt")
        print(results)





        share|improve this answer












        keywordlist is a local variable to the function splitKeyword which return it so you can directly use this function and reduce the code.



        def functionOne(textFile):
        textFileVar = open(textFile, 'r')
        def splitKeyword(argument):
        keywordList =
        for line in argument:
        keywordList.append(line.strip().split(','))
        return keywordList

        print(splitKeyword(textFileVar))

        results = functionOne("text1.txt")
        print(results)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 11 at 5:52









        Moussa

        788




        788






























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