NameError occurring after calling nested function











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












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











share|improve this question




share|improve this question










asked Nov 11 at 5:38









CosmicCat

505




505








  • 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





















    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














     

    draft saved


    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53246143%2fnameerror-occurring-after-calling-nested-function%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    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
    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

    3,83211233




    3,83211233












    • 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






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53246143%2fnameerror-occurring-after-calling-nested-function%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Full-time equivalent

            さくらももこ

            13 indicted, 8 arrested in Calif. drug cartel investigation