What should be passed to a function which requires a filename when I have the file contents?









up vote
1
down vote

favorite
1












I have a function (not mine) which accepts a filename as input, then does things with the file content.



On the other hand, I receive the data which is the "file content" above from somewhere, not as a file.



To illustrate the case, consider the code below.




  • myfunc() is something I do not control

  • the expectation of the authors of myfunc() is that the data to process will be in a file ...

  • ... but I have it as a string

# a function (which I do not control) which uses a filename as input 
def myfunc(filename):
with open(filename) as f:
print(f.read())

# case 1: I have a real file
# creation of the file for the sake of the example.
# The authors of myfunc() expect that the the file is just there.
with open("afile.txt", "w") as f:
f.write("hello")

# normal usage of myfunc()
myfunc("afile.txt")

# case 2: I have the contents of the file and need to use myfunc()
# data below is received from somewhere, as a string
data = "world"

# I would like to pass it to myfunc
# ...


How to pass <something> as the argument of the function so that <something> holds the content of data but behaves as a file name?



The obvious solution is to dump data into a file and pass the filename to myfunc(). I would like to avoid this intermediate step to keep the whole thing in memory.










share|improve this question























  • What operating system are you on? Are OS-specific solutions acceptable? (E.g. you can pass /dev/stdin on some systems)
    – alexis
    Nov 9 at 13:54










  • Take a look in this question
    – Hemerson Tacon
    Nov 9 at 13:54






  • 1




    @HemersonTacon: thanks. That would be almost it, but I do not see how to access the filename of the created file. File-like objects are mentioned, which I read as the file handler (in other words f from f = open('afile.txt') and not afile.txt)
    – WoJ
    Nov 9 at 14:03






  • 1




    @HemersonTacon in the end it still writes a file, which wouldn't be that different for OP to do manually (which they know they can do, just want to avoid).
    – Idlehands
    Nov 9 at 14:07






  • 1




    @HemersonTacon: but then I would write the file to disk, which I explicitly wanted to avoid (see the last line of my question)
    – WoJ
    Nov 9 at 14:07














up vote
1
down vote

favorite
1












I have a function (not mine) which accepts a filename as input, then does things with the file content.



On the other hand, I receive the data which is the "file content" above from somewhere, not as a file.



To illustrate the case, consider the code below.




  • myfunc() is something I do not control

  • the expectation of the authors of myfunc() is that the data to process will be in a file ...

  • ... but I have it as a string

# a function (which I do not control) which uses a filename as input 
def myfunc(filename):
with open(filename) as f:
print(f.read())

# case 1: I have a real file
# creation of the file for the sake of the example.
# The authors of myfunc() expect that the the file is just there.
with open("afile.txt", "w") as f:
f.write("hello")

# normal usage of myfunc()
myfunc("afile.txt")

# case 2: I have the contents of the file and need to use myfunc()
# data below is received from somewhere, as a string
data = "world"

# I would like to pass it to myfunc
# ...


How to pass <something> as the argument of the function so that <something> holds the content of data but behaves as a file name?



The obvious solution is to dump data into a file and pass the filename to myfunc(). I would like to avoid this intermediate step to keep the whole thing in memory.










share|improve this question























  • What operating system are you on? Are OS-specific solutions acceptable? (E.g. you can pass /dev/stdin on some systems)
    – alexis
    Nov 9 at 13:54










  • Take a look in this question
    – Hemerson Tacon
    Nov 9 at 13:54






  • 1




    @HemersonTacon: thanks. That would be almost it, but I do not see how to access the filename of the created file. File-like objects are mentioned, which I read as the file handler (in other words f from f = open('afile.txt') and not afile.txt)
    – WoJ
    Nov 9 at 14:03






  • 1




    @HemersonTacon in the end it still writes a file, which wouldn't be that different for OP to do manually (which they know they can do, just want to avoid).
    – Idlehands
    Nov 9 at 14:07






  • 1




    @HemersonTacon: but then I would write the file to disk, which I explicitly wanted to avoid (see the last line of my question)
    – WoJ
    Nov 9 at 14:07












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I have a function (not mine) which accepts a filename as input, then does things with the file content.



On the other hand, I receive the data which is the "file content" above from somewhere, not as a file.



To illustrate the case, consider the code below.




  • myfunc() is something I do not control

  • the expectation of the authors of myfunc() is that the data to process will be in a file ...

  • ... but I have it as a string

# a function (which I do not control) which uses a filename as input 
def myfunc(filename):
with open(filename) as f:
print(f.read())

# case 1: I have a real file
# creation of the file for the sake of the example.
# The authors of myfunc() expect that the the file is just there.
with open("afile.txt", "w") as f:
f.write("hello")

# normal usage of myfunc()
myfunc("afile.txt")

# case 2: I have the contents of the file and need to use myfunc()
# data below is received from somewhere, as a string
data = "world"

# I would like to pass it to myfunc
# ...


How to pass <something> as the argument of the function so that <something> holds the content of data but behaves as a file name?



The obvious solution is to dump data into a file and pass the filename to myfunc(). I would like to avoid this intermediate step to keep the whole thing in memory.










share|improve this question















I have a function (not mine) which accepts a filename as input, then does things with the file content.



On the other hand, I receive the data which is the "file content" above from somewhere, not as a file.



To illustrate the case, consider the code below.




  • myfunc() is something I do not control

  • the expectation of the authors of myfunc() is that the data to process will be in a file ...

  • ... but I have it as a string

# a function (which I do not control) which uses a filename as input 
def myfunc(filename):
with open(filename) as f:
print(f.read())

# case 1: I have a real file
# creation of the file for the sake of the example.
# The authors of myfunc() expect that the the file is just there.
with open("afile.txt", "w") as f:
f.write("hello")

# normal usage of myfunc()
myfunc("afile.txt")

# case 2: I have the contents of the file and need to use myfunc()
# data below is received from somewhere, as a string
data = "world"

# I would like to pass it to myfunc
# ...


How to pass <something> as the argument of the function so that <something> holds the content of data but behaves as a file name?



The obvious solution is to dump data into a file and pass the filename to myfunc(). I would like to avoid this intermediate step to keep the whole thing in memory.







python python-3.x file






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 9 at 14:03

























asked Nov 9 at 13:48









WoJ

6,181754123




6,181754123











  • What operating system are you on? Are OS-specific solutions acceptable? (E.g. you can pass /dev/stdin on some systems)
    – alexis
    Nov 9 at 13:54










  • Take a look in this question
    – Hemerson Tacon
    Nov 9 at 13:54






  • 1




    @HemersonTacon: thanks. That would be almost it, but I do not see how to access the filename of the created file. File-like objects are mentioned, which I read as the file handler (in other words f from f = open('afile.txt') and not afile.txt)
    – WoJ
    Nov 9 at 14:03






  • 1




    @HemersonTacon in the end it still writes a file, which wouldn't be that different for OP to do manually (which they know they can do, just want to avoid).
    – Idlehands
    Nov 9 at 14:07






  • 1




    @HemersonTacon: but then I would write the file to disk, which I explicitly wanted to avoid (see the last line of my question)
    – WoJ
    Nov 9 at 14:07
















  • What operating system are you on? Are OS-specific solutions acceptable? (E.g. you can pass /dev/stdin on some systems)
    – alexis
    Nov 9 at 13:54










  • Take a look in this question
    – Hemerson Tacon
    Nov 9 at 13:54






  • 1




    @HemersonTacon: thanks. That would be almost it, but I do not see how to access the filename of the created file. File-like objects are mentioned, which I read as the file handler (in other words f from f = open('afile.txt') and not afile.txt)
    – WoJ
    Nov 9 at 14:03






  • 1




    @HemersonTacon in the end it still writes a file, which wouldn't be that different for OP to do manually (which they know they can do, just want to avoid).
    – Idlehands
    Nov 9 at 14:07






  • 1




    @HemersonTacon: but then I would write the file to disk, which I explicitly wanted to avoid (see the last line of my question)
    – WoJ
    Nov 9 at 14:07















What operating system are you on? Are OS-specific solutions acceptable? (E.g. you can pass /dev/stdin on some systems)
– alexis
Nov 9 at 13:54




What operating system are you on? Are OS-specific solutions acceptable? (E.g. you can pass /dev/stdin on some systems)
– alexis
Nov 9 at 13:54












Take a look in this question
– Hemerson Tacon
Nov 9 at 13:54




Take a look in this question
– Hemerson Tacon
Nov 9 at 13:54




1




1




@HemersonTacon: thanks. That would be almost it, but I do not see how to access the filename of the created file. File-like objects are mentioned, which I read as the file handler (in other words f from f = open('afile.txt') and not afile.txt)
– WoJ
Nov 9 at 14:03




@HemersonTacon: thanks. That would be almost it, but I do not see how to access the filename of the created file. File-like objects are mentioned, which I read as the file handler (in other words f from f = open('afile.txt') and not afile.txt)
– WoJ
Nov 9 at 14:03




1




1




@HemersonTacon in the end it still writes a file, which wouldn't be that different for OP to do manually (which they know they can do, just want to avoid).
– Idlehands
Nov 9 at 14:07




@HemersonTacon in the end it still writes a file, which wouldn't be that different for OP to do manually (which they know they can do, just want to avoid).
– Idlehands
Nov 9 at 14:07




1




1




@HemersonTacon: but then I would write the file to disk, which I explicitly wanted to avoid (see the last line of my question)
– WoJ
Nov 9 at 14:07




@HemersonTacon: but then I would write the file to disk, which I explicitly wanted to avoid (see the last line of my question)
– WoJ
Nov 9 at 14:07












2 Answers
2






active

oldest

votes

















up vote
1
down vote













This is a bit convoluted, so bear with me... but if you absolutely want to avoid writing to disk you could shadow the builtin open() function with your own...



from io import StringIO
from contextlib import contextmanager

@contextmanager
def myfunc_open(*args, **kwargs):
resource = StringIO(args[0])
try:
yield resource
finally:
resource.close()


We'll use StringIO since it has the IO functions you'd expect within the with open() context. Depending on the context you might want to use BytesIO instead.



Then, before you call upon myfunc simply back up open() and replace with your own:



def myfunc(filepath):
with open(filepath, 'r') as file:
print(file.read())
try:
myfunc('This is doomed to fail')
except:
print("Normal myfunc won't work")

_backup_open = open # back up before you shadow it!
open = myfunc_open
myfunc('This will however be read')

open = _backup_open # Perfectly balanced, as all things should be.



Result:



# Normal myfunc won't work
# This will however be read


Note: If myfunc is imported from another module, it'll require a little bit of work around:



import myfunc_module

myfunc_module.open = myfunc_open
myfunc_module.myfunc('This works too')

myfunc_module.open = _backup_open


Note you do have to import the whole module instead of from module import func so you can shadow the open within its namespace. Sample here.



This may be the evilest stroke of genius I have so far in my hobbyist journey...






share|improve this answer





























    up vote
    -2
    down vote













    if you send data as an argument in myfunc() then it will give FileNotFoundError. if this file doesn't exist then that must mean it is the file content that you are trying to send as the argument. so, if use exception handling then this may do the trick



    try :
    myfunc(data)
    except FileNotFoundError:
    print(data)





    share|improve this answer
















    • 1




      I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
      – WoJ
      Nov 9 at 14:31










    • you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
      – Nehal Birla
      Nov 9 at 14:39










    • what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
      – WoJ
      Nov 9 at 14:49










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    This is a bit convoluted, so bear with me... but if you absolutely want to avoid writing to disk you could shadow the builtin open() function with your own...



    from io import StringIO
    from contextlib import contextmanager

    @contextmanager
    def myfunc_open(*args, **kwargs):
    resource = StringIO(args[0])
    try:
    yield resource
    finally:
    resource.close()


    We'll use StringIO since it has the IO functions you'd expect within the with open() context. Depending on the context you might want to use BytesIO instead.



    Then, before you call upon myfunc simply back up open() and replace with your own:



    def myfunc(filepath):
    with open(filepath, 'r') as file:
    print(file.read())
    try:
    myfunc('This is doomed to fail')
    except:
    print("Normal myfunc won't work")

    _backup_open = open # back up before you shadow it!
    open = myfunc_open
    myfunc('This will however be read')

    open = _backup_open # Perfectly balanced, as all things should be.



    Result:



    # Normal myfunc won't work
    # This will however be read


    Note: If myfunc is imported from another module, it'll require a little bit of work around:



    import myfunc_module

    myfunc_module.open = myfunc_open
    myfunc_module.myfunc('This works too')

    myfunc_module.open = _backup_open


    Note you do have to import the whole module instead of from module import func so you can shadow the open within its namespace. Sample here.



    This may be the evilest stroke of genius I have so far in my hobbyist journey...






    share|improve this answer


























      up vote
      1
      down vote













      This is a bit convoluted, so bear with me... but if you absolutely want to avoid writing to disk you could shadow the builtin open() function with your own...



      from io import StringIO
      from contextlib import contextmanager

      @contextmanager
      def myfunc_open(*args, **kwargs):
      resource = StringIO(args[0])
      try:
      yield resource
      finally:
      resource.close()


      We'll use StringIO since it has the IO functions you'd expect within the with open() context. Depending on the context you might want to use BytesIO instead.



      Then, before you call upon myfunc simply back up open() and replace with your own:



      def myfunc(filepath):
      with open(filepath, 'r') as file:
      print(file.read())
      try:
      myfunc('This is doomed to fail')
      except:
      print("Normal myfunc won't work")

      _backup_open = open # back up before you shadow it!
      open = myfunc_open
      myfunc('This will however be read')

      open = _backup_open # Perfectly balanced, as all things should be.



      Result:



      # Normal myfunc won't work
      # This will however be read


      Note: If myfunc is imported from another module, it'll require a little bit of work around:



      import myfunc_module

      myfunc_module.open = myfunc_open
      myfunc_module.myfunc('This works too')

      myfunc_module.open = _backup_open


      Note you do have to import the whole module instead of from module import func so you can shadow the open within its namespace. Sample here.



      This may be the evilest stroke of genius I have so far in my hobbyist journey...






      share|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        This is a bit convoluted, so bear with me... but if you absolutely want to avoid writing to disk you could shadow the builtin open() function with your own...



        from io import StringIO
        from contextlib import contextmanager

        @contextmanager
        def myfunc_open(*args, **kwargs):
        resource = StringIO(args[0])
        try:
        yield resource
        finally:
        resource.close()


        We'll use StringIO since it has the IO functions you'd expect within the with open() context. Depending on the context you might want to use BytesIO instead.



        Then, before you call upon myfunc simply back up open() and replace with your own:



        def myfunc(filepath):
        with open(filepath, 'r') as file:
        print(file.read())
        try:
        myfunc('This is doomed to fail')
        except:
        print("Normal myfunc won't work")

        _backup_open = open # back up before you shadow it!
        open = myfunc_open
        myfunc('This will however be read')

        open = _backup_open # Perfectly balanced, as all things should be.



        Result:



        # Normal myfunc won't work
        # This will however be read


        Note: If myfunc is imported from another module, it'll require a little bit of work around:



        import myfunc_module

        myfunc_module.open = myfunc_open
        myfunc_module.myfunc('This works too')

        myfunc_module.open = _backup_open


        Note you do have to import the whole module instead of from module import func so you can shadow the open within its namespace. Sample here.



        This may be the evilest stroke of genius I have so far in my hobbyist journey...






        share|improve this answer














        This is a bit convoluted, so bear with me... but if you absolutely want to avoid writing to disk you could shadow the builtin open() function with your own...



        from io import StringIO
        from contextlib import contextmanager

        @contextmanager
        def myfunc_open(*args, **kwargs):
        resource = StringIO(args[0])
        try:
        yield resource
        finally:
        resource.close()


        We'll use StringIO since it has the IO functions you'd expect within the with open() context. Depending on the context you might want to use BytesIO instead.



        Then, before you call upon myfunc simply back up open() and replace with your own:



        def myfunc(filepath):
        with open(filepath, 'r') as file:
        print(file.read())
        try:
        myfunc('This is doomed to fail')
        except:
        print("Normal myfunc won't work")

        _backup_open = open # back up before you shadow it!
        open = myfunc_open
        myfunc('This will however be read')

        open = _backup_open # Perfectly balanced, as all things should be.



        Result:



        # Normal myfunc won't work
        # This will however be read


        Note: If myfunc is imported from another module, it'll require a little bit of work around:



        import myfunc_module

        myfunc_module.open = myfunc_open
        myfunc_module.myfunc('This works too')

        myfunc_module.open = _backup_open


        Note you do have to import the whole module instead of from module import func so you can shadow the open within its namespace. Sample here.



        This may be the evilest stroke of genius I have so far in my hobbyist journey...







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 9 at 15:19

























        answered Nov 9 at 15:07









        Idlehands

        3,1101316




        3,1101316






















            up vote
            -2
            down vote













            if you send data as an argument in myfunc() then it will give FileNotFoundError. if this file doesn't exist then that must mean it is the file content that you are trying to send as the argument. so, if use exception handling then this may do the trick



            try :
            myfunc(data)
            except FileNotFoundError:
            print(data)





            share|improve this answer
















            • 1




              I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
              – WoJ
              Nov 9 at 14:31










            • you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
              – Nehal Birla
              Nov 9 at 14:39










            • what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
              – WoJ
              Nov 9 at 14:49














            up vote
            -2
            down vote













            if you send data as an argument in myfunc() then it will give FileNotFoundError. if this file doesn't exist then that must mean it is the file content that you are trying to send as the argument. so, if use exception handling then this may do the trick



            try :
            myfunc(data)
            except FileNotFoundError:
            print(data)





            share|improve this answer
















            • 1




              I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
              – WoJ
              Nov 9 at 14:31










            • you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
              – Nehal Birla
              Nov 9 at 14:39










            • what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
              – WoJ
              Nov 9 at 14:49












            up vote
            -2
            down vote










            up vote
            -2
            down vote









            if you send data as an argument in myfunc() then it will give FileNotFoundError. if this file doesn't exist then that must mean it is the file content that you are trying to send as the argument. so, if use exception handling then this may do the trick



            try :
            myfunc(data)
            except FileNotFoundError:
            print(data)





            share|improve this answer












            if you send data as an argument in myfunc() then it will give FileNotFoundError. if this file doesn't exist then that must mean it is the file content that you are trying to send as the argument. so, if use exception handling then this may do the trick



            try :
            myfunc(data)
            except FileNotFoundError:
            print(data)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 9 at 14:30









            Nehal Birla

            148




            148







            • 1




              I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
              – WoJ
              Nov 9 at 14:31










            • you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
              – Nehal Birla
              Nov 9 at 14:39










            • what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
              – WoJ
              Nov 9 at 14:49












            • 1




              I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
              – WoJ
              Nov 9 at 14:31










            • you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
              – Nehal Birla
              Nov 9 at 14:39










            • what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
              – WoJ
              Nov 9 at 14:49







            1




            1




            I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
            – WoJ
            Nov 9 at 14:31




            I think you misunderstood my question. I know that what is expected is a filename and that what I have is a string (the content of the expected file).
            – WoJ
            Nov 9 at 14:31












            you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
            – Nehal Birla
            Nov 9 at 14:39




            you want to print the file content right? if it is inside the file then you can easily print it using myfunc() but if you already have the file content as a string then this will work
            – Nehal Birla
            Nov 9 at 14:39












            what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
            – WoJ
            Nov 9 at 14:49




            what myfunc() does in the code in the question is just an example. The idea is that I have to use a function which accepts a filename (and does things with the contents of the file, does not matter what - in the example it prints them) but what I have is the contents of the file and not the file itself (and I do not want to dump it into a file).
            – WoJ
            Nov 9 at 14:49

















             

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