IO error when reading csv files in python

Question

I have the following code:

for file in os.listdir('/home/sainik/Final/'+str(folderno)):
        if file.endswith('.csv'):
            print file
            with open(file,'rb') as csvfile:
                spamreader = csv.reader(csvfile)
                for row in spamreader:
                    print row

when running the code, I am getting the following error:

Traceback (most recent call last):
  File "/home/sainik/Final/Programs/sainik.py", line 28, in <module>
    with open(file,'rb') as csvfile:
IOError: [Errno 2] No such file or directory: '4.csv'

Kindly help.

lapinkoira · Accepted Answer · 2017-09-18 13:36:13Z

1

You are trying to open the file from the path you are running the script.

You should try to open the full path

with open('/home/sainik/Final/' + file)

answered Sep 18, 2017 at 13:36

lapinkoira

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Comments

Mitiku · Accepted Answer · 2017-09-22 06:46:10Z

1

You are passing only the filename for open function. You should pass path to the open function. Two possible ways to pass the path of file to open function, either relative path or full path.

try:

with open( os.path.join('/home/sainik/Final/',str(folderno),file),'rb') as csvfile:

edited Sep 22, 2017 at 6:46

answered Sep 18, 2017 at 13:37

Mitiku

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Comments

zipa · Accepted Answer · 2017-09-18 13:37:13Z

0

Your script is looking at it's own directory for file 4.csv. Try it like this:

for file in os.listdir('/home/sainik/Final/'+str(folderno)):
        if file.endswith('.csv'):
            print file
            with open(/home/sainik/Final/'+str(folderno)+'\/'+file,'rb') as csvfile:
                spamreader = csv.reader(csvfile)
                for row in spamreader:
                    print row

answered Sep 18, 2017 at 13:37

zipa

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