python2 how to handle file input as argument with subprocess.call - python-2.7

Question

I need to use a file as argument to the command but subprocess.call returns following error
>>> with open('test.txt') as F:
... subprocess.call(['cat', F])
...
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "/usr/lib64/python2.7/subprocess.py", line 524, in call
return Popen(*popenargs, **kwargs).wait()
File "/usr/lib64/python2.7/subprocess.py", line 711, in __init__
errread, errwrite)
File "/usr/lib64/python2.7/subprocess.py", line 1327, in _execute_child
raise child_exception
TypeError: execv() arg 2 must contain only strings
However when file is not stored as variable it works fine
>>> subprocess.call(['cat', 'test.txt'])
fruit=banana
fruit=mango
fruit=apple
0

Answer 1

Code with open('test.txt') as F opens the file and assigns its pointor to variable F.
If I understand correctly you wanted file's content. Since subprocess.call required a command like you execute from terminal/console(In your case cat test.txt). Here in the subprocess.call you are passing File Descriptor sometime called File Pointer instead of file itself.
Therefore from the code if you really want a file content, you can simply use F.read() or F.readlines(). For Example
with open('test.txt') as F:
print F.readlines()
Also note that subprocess.call expect arguments as string. In the first case its the file object i.e <open file 'test.txt', mode 'r' at 0x7fa64a6f5a50> that leads to a traceback whereas in the second case its string i.e test.txt that's why works without any issue.