I'm trying to do a couple things here with the script below (it is incomplete). The first thing is to loop through some subdirectories. I was able to do that successfully. The second thing was to open a specific file (it is the same name in each subdirectory) and find the minimum and maximum value in each column EXCEPT the first.
Right now I'm stuck on finding the max value in a single column because the files I'm reading have two rows which I want to ignore. Unfortunately, I'm getting the following error when attempting to run the code:
Traceback (most recent call last):
File "test_script.py", line 22, in <module>
with open(file) as f:
IOError: [Errno 2] No such file or directory: 'tc.out'
Here is the current state of my code:
import scipy as sp
import os
rootdir = 'mydir'; #mydir has been changed from the actual directory path
data = []
for root, dirs, files in os.walk(rootdir):
for file in files:
if file == "tc.out":
with open(file) as f:
for line in itertools.islice(f,3,None):
for line in file:
fields = line.split()
rowdata = map(float, fields)
data.extend(rowdata)
print 'Maximum: ', max(data)
To open a file you need to specify full path. You need to change the line
with open(file) as f:
to
with open(os.path.join(root, file)) as f:
When you write open(file), Python is trying to find the the file tc.out in the directory where you started the interpreter from. You should use the full path to that file in open:
with open(os.path.join(root, file)) as f:
Let me illustrate with an example:
I have a file named 'somefile.txt' in the directory /tmp/sto/deep/ (this is a Unix system, so I use forward slashes). And then I have this simple script which resides in the directory /tmp:
oliver#armstrong:/tmp$ cat myscript.py
import os
rootdir = '/tmp'
for root, dirs, files in os.walk(rootdir):
for fname in files:
if fname == 'somefile.txt':
with open(os.path.join(root, fname)) as f:
print('Filename: %s' % fname)
print('directory: %s' % root)
print(f.read())
When I execute this script from the /tmp directory, you'll see that fname is just the filename, the path leading to it is ommitted. That's why you need to join it with the first returned argument from os.walk.
oliver#armstrong:/tmp$ python myscript.py
Filename: somefile.txt
directory: /tmp/sto/deep
contents
Related
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file
I have a directory with further sub directories each having files with a specific extension. I am able to get the names of all the files using glob function:
for name in glob.glob('*/*[a-b]*'):
print(os.path.basename(name))
that prints the name of files I have in all the sub directories:
PF44_aa
PF52_aa
PF95_aa
PF38_aa
PF45_aa
PF63_aa
PF68_aa
PF39_aa
However, if I pass these file names as arguments to open the files and read the contents:
for name in glob.glob('*/*[a-b]*'):
filename=os.path.basename(name)
with open('%s' %filename) as fn:
content = fn.readlines()
I get the following error:
File "<ipython-input-194-147f38fc2684>", line 1, in <module>
with open('%s' %filename) as fn:
FileNotFoundError: [Errno 2] No such file or directory: 'PF44_aa'
I also tried giving the filename directly as input instead of %s:
for name in glob.glob('*/*[a-b]*'):
filename=os.path.basename(name)
with open(filename) as fn:
content = fn.readlines()
But still got the same error:
File "<ipython-input-193-fb125b5aa813>", line 1, in <module>
with open(filename) as fn:
FileNotFoundError: [Errno 2] No such file or directory: 'PF44_aa'
What am I doing wrong and how can I fix this?
You have to use complete path of the file to open it, you can't use just filename unless if its on the same directory as your python file. So you have to do little change in your script to make it work.
for name in glob.glob('*/*[a-b]*'):
with open(name) as fn:
content = fn.readlines()
filename is replaced by name.
here, "name" is complete path to your file.
Alternative method...
Start by first importing:
import shutil
import os
Then assign the directory to a list:
file_list = []
file_list = os.listdir('C:/filepath')
Now distinguish between files:
files = []
files = [x for x in file_list if "_aa" in x]
Now you can open and read the files in the files list using your method.
however do:
filepath + filename
with open(filepath + filename) as fn:
content = fn.readlines()
Currently you're just trying to open the file with its name, you need to include the full file path...
outcome:
"C:/filepath/PF44_aa"
I was trying to iterate over the files in a directory like this:
import os
path = r'E:/somedir'
for filename in os.listdir(path):
f = open(filename, 'r')
... # process the file
But Python was throwing FileNotFoundError even though the file exists:
Traceback (most recent call last):
File "E:/ADMTM/TestT.py", line 6, in <module>
f = open(filename, 'r')
FileNotFoundError: [Errno 2] No such file or directory: 'foo.txt'
So what is wrong here?
It is because os.listdir does not return the full path to the file, only the filename part; that is 'foo.txt', when open would want 'E:/somedir/foo.txt' because the file does not exist in the current directory.
Use os.path.join to prepend the directory to your filename:
path = r'E:/somedir'
for filename in os.listdir(path):
with open(os.path.join(path, filename)) as f:
... # process the file
(Also, you are not closing the file; the with block will take care of it automatically).
os.listdir(directory) returns a list of file names in directory. So unless directory is your current working directory, you need to join those file names with the actual directory to get a proper absolute path:
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
f = open(filepath,'r')
raw = f.read()
# ...
Here's an alternative solution using pathlib.Path.iterdir, which yields the full paths instead, removing the need to join paths:
from pathlib import Path
path = Path(r'E:/somedir')
for filename in path.iterdir():
with filename.open() as f:
... # process the file