Cannot open file in python

Question

I'm having trouble with opening files, python (2.7) tells me that there is no such file, but there is.

        csvData = None
        csvHeader = None
        os.chdir("../result/files/")
        for fileName in os.listdir("."):
            if fileName.endswith(".csv"):
                print fileName
                with open("../result/files/"+fileName, 'rb') as csvFile:
                    readerCsv = reader(csvFile, delimiter = ';')
                    csvHeader = readerCsv.next()
                    _unused = readerCsv.next()
                    self.data = list.append(list(readerCsv))

this code is suposed to read 5 csv files and store their content in data (except for the first line)

At first chdir tells me he found Stats20120903.csv then I ask the program to open said file but I get an error :

IOError: [Errno 2] No such file or directory: '../files/Stats20120903.csv'

why?

Thanks

madjar · Accepted Answer · 2013-03-04 12:09:05Z

1

os.chdir does not tell you what is in the directory, it changes the current working path. All relative file path you give are relative to the working dir.

You should either

remove the os.chdir("../result/files/") and call os.listdir("../result/files/") or
not add "../result/files/" to the filename you want to open.

answered Mar 4, 2013 at 12:09

madjar

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1 Comment

sliders_alpha Over a year ago

removing os.chdir("../result/files/") and replacing "." by "../result/files/" worked, thanks

YXD · Accepted Answer · 2013-03-04 12:06:55Z

1

Do

with open(fileName, 'rb') as csvFile:

instead of looking relative to the directory you've changed to.

answered Mar 4, 2013 at 12:06

YXD

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Comments

djc · Accepted Answer · 2013-03-04 12:07:59Z

0

You could try using os.path.abspath() to find out what actual path Python is trying to use.

answered Mar 4, 2013 at 12:07

djc

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