I want to log into a ftp server (not a public url) and download a csv file which is located in a zip file and then save this to a particular directory:
#log in OK
# this is the zip file I want to download
fpath = strDate + ".zip"
#set where to save file
ExtDir = "A:\\LOCAL\\DIREC\\TORY\\"""
ExtDir = ExtDir + strdate + "\\"
ExtFile = ExtDir + "Download.zip"
#download files
#use zipfile.ZipFile as alternative method to open(ExtFile, 'w')
with zipfile.ZipFile(ExtFile,'w') as outzip:
ftp.retrbinary('RETR %s' % fpath , outzip.write)
outzip.close
I get this error
File "C:\Program Files (x86)\Python 2.7\lib\ftplib.py", line 419, in retrbinary callback(data)
File "C:\Program Files (x86)\Python 2.7\lib\zipfile.py", line 1123, in write st = os.stat(filename)
TypeError: stat() argument 1 must be encoded string without null bytes, not str
Fixed using:
ftp.retrlines('RETR %s' % fpath ,lambda s, w=outzip.write: w(s+"\n"))
Related
I want to open and read all csv file in a specific folder.
I'm on OS X El Capitan version 10.11.6, and I'm using Python 2.7.10.
I have the following function in phyton file:
def open_csv_files(dir):
for root,dirs,files in os.walk(dir):
for file in files:
if file.endswith(".csv"):
f= open(file)
print "FILE OPEN, AND DO SOMETHING... "
f.close
return
I call open_csv_file(./dati/esempi)
This procedure return
IOError: [Errno 2] No such file or directory: 'sensorfile_1.csv'
I try to call the procedure with absolute path /Users/Claudia/Desktop/Thesis/dati/esempi/ but I have the same error.
Moreover I define another procedure that print all filename in folder, this procedure print correctly all filenames in folder.
Thanks for the help.
You need to build absolute path to file based on values of root (base dir) and file name.
import os
def open_csv_files(directory):
for root, dirs, files in os.walk(directory):
for file_name in files:
if file_name.endswith(".csv"):
full_file_path = os.path.join(root, file_name)
with open(full_file_path) as fh:
print "Do something with", full_file_path
Anyone successfully tried to convert many ARFF files to CSV files from windows Command line.
I tried to use weka.core.converters.CSVSaver but it works for a single file only.
Can it be done for multiple files?
I found a way to solve this conversion by using R as shown in the following Script:
#### Set the default directory to the folder that contains all ARFF files
temp = list.files(pattern="*.arff")
library(foreign)
for (i in 1:length(temp)) assign(temp[i], read.arff(temp[i]))
for(i in 1:length(temp))
{
mydata=read.arff(temp[i])
t=temp[i]
x=paste(t,".csv")
write.csv(mydata,x,row.names=FALSE)
mydata=0
}
On a windows command line, type powershell
Change to the directory where your *.arff files reside in
Enter this command
dir *.arff | Split-Path -Leaf| ForEach-Object {Invoke-Expression "C:\Program Files\Weka-3-6\weka.jar;." weka.core.converters.CSVSaver -i $_ -o $_.csv"}
This assumes that your filenames do not contain any blanks, and all arff files reside in a single directory, and you want to convert them all. It will create a new csv file from each arff file. myfile.arff will be exported/converted to myfile.arff.csv
I write a simple python script in github: arff2csv.py.
paste my code.
"""trans multi-label *.arff file to *.csv file."""
import re
def trans_arff2csv(file_in, file_out):
"""trans *.arff file to *.csv file."""
columns = []
data = []
with open(file_in, 'r') as f:
data_flag = 0
for line in f:
if line[:2] == '#a':
# find indices
indices = [i for i, x in enumerate(line) if x == ' ']
columns.append(re.sub(r'^[\'\"]|[\'\"]$|\\+', '', line[indices[0] + 1:indices[-1]]))
elif line[:2] == '#d':
data_flag = 1
elif data_flag == 1:
data.append(line)
content = ','.join(columns) + '\n' + ''.join(data)
# save to file
with open(file_out, 'w') as f:
f.write(content)
if __name__ == '__main__':
from multi_label.arff2csv import trans_arff2csv
# setting arff file path
file_attr_in = r'D:\Downloads\birds\birds-test.arff'
# setting output csv file path
file_csv_out = r"D:\Downloads\birds\birds-test.csv"
# trans
trans_arff2csv(file_attr_in, file_csv_out)
I am having a bit of trouble in obtaining a file path so that I can open and execute my data from the specified (text) file. Below is the code I have written so far:
def pickfile():
options={}
options['defaultextension'] = '.txt'
options['filetypes'] = [('all files','.*'), ('text files', '.*txt')]
options['initialfile'] = 'sample.txt'
options['initialdir'] = 'C:\Users\profcs\Desktop'
filename=open(tkFileDialog.askopenfilename(**options))
if filename:
print(filename)
return
with open(filename, 'rb') as f:
reader = csv.reader(f)
try:
for row in reader:
print row
except csv.Error as e:
sys.exit('file %s, line %d: %s' % (filename, reader.line_num,e))
but1 = Button(widget1, text='Pick Your File', command=pickfile)
but1.pack(side=BOTTOM, padx=10, pady=1, anchor=SE)
but1.config(relief=RAISED, bd=2)
When I display a filename, I now get the path in this form:
================ RESTART: C:\Users\profcs\Desktop\BD TEST.py ================
<open file u'C:/Users/profcs/Desktop/sample.txt', mode 'r' at 0x01EFF128>
How can I filter this path and only get 'C:/Users/profcs/Desktop/sample.txt' so that I can open my file?
Thanks in advance.
filename.name gives you the path from filename object.
I hope this helps :
filename = open(tkFileDialog.askopenfilename(**options))
print (filename.name)
'C:/Users/profcs/Desktop/sample.txt'
In your case filename is an object which represents an open file.
I'm trying to zip files present in a directory and give a specific name (destination folder) to it. I want to pass the source and destination folders as input to the program.
But when ever I'm passeing the source file path it's giving me and error. I guess I'll face the same problem with the destination file path.
d:\SARFARAZ\Python>python zip.py
Enter source directry:D:\Sarfaraz\Python\Project_Euler
Traceback (most recent call last):
File "zip.py", line 17, in <module>
SrcPath = input("Enter source directry:")
File "<string>", line 1
D:\Sarfaraz\Python\Project_Euler
^
SyntaxError: invalid syntax
The code I've written is as follow:
import os
import zipfile
def zip(src, dst):
zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
abs_src = os.path.abspath(src)
for dirname, subdirs, files in os.walk(src):
for filename in files:
absname = os.path.abspath(os.path.join(dirname, filename))
arcname = absname[len(abs_src) + 1:]
print 'zipping %s as %s' % (os.path.join(dirname, filename),arcname)
zf.write(absname, arcname)
zf.close()
#zip("D:\\Sarfaraz\\Python\\Project_Euler", "C:\\Users\\md_sarfaraz\\Desktop")
SrcPath = input("Enter source directry:")
SrcPath = ("#'"+ str(SrcPath) +"'")
print SrcPath # checking source path
DestPath = input("Enter destination directry:")
DestPath = ("#'"+str(DestPath) +"'")
print DestPath
zip(SrcPath, DestPath)
i have made some changes to your code as follows:
import os
import zipfile
def zip(src, dst):
zf = zipfile.ZipFile(dst, "w", zipfile.ZIP_DEFLATED)
abs_src = os.path.abspath(src)
for dirname, subdirs, files in os.walk(src):
for filename in files:
absname = os.path.abspath(os.path.join(dirname, filename))
arcname = absname[len(abs_src) + 1:]
print 'zipping %s as %s' % (os.path.join(dirname, filename),arcname)
zf.write(absname, arcname)
zf.close()
# Changed from input() to raw_input()
SrcPath = raw_input("Enter source directory: ")
print SrcPath # checking source path
# done the same and also added the option of specifying the name of Zipped file.
DestZipFileName = raw_input("Enter destination Zip File Name: ") + ".zip" # i.e. test.zip
DestPathName = raw_input("Enter destination directory: ")
# Here added "\\" to make sure the zipped file will be placed in the specified directory.
# i.e. C:\\Users\\md_sarfaraz\\Desktop\\
# i.e. double \\ to escape the backlash character.
DestPath = DestPathName + "\\" + DestZipFileName
print DestPath # Checking Destination Zip File name & Path
zip(SrcPath, DestPath)
Good Luck!
I'm trying to download multiple files from a website.
The url resembles this: foo.com/foo-1.pdf.
Since I want those files to be stored in a directory of my choice,
I have written the following code:
import os
from urllib import urlretrieve
ext = ".pdf"
for i in range(1,37):
print "fetching file " + str(i)
url = "http://foo.com/Lec-" + str(i) + ext
myPath = "/dir/"
filename = "Lec-"+str(i)+ext
fullfilename = os.path.join(myPath, filename)
x = urlretrieve(url, fullfilename)
EDIT : Complete error message.
Traceback (most recent call last):
File "scraper.py", line 10, in <module>
x = urlretrieve(url, fullfilename)
File "/usr/lib/python2.7/urllib.py", line 94, in urlretrieve
return _urlopener.retrieve(url, filename, reporthook, data)
File "/usr/lib/python2.7/urllib.py", line 244, in retrieve
tfp = open(filename, 'wb')
IOError: [Errno 2] No such file or directory: /dir/Lec-1.pdf'
I'd be grateful if someone could point out where I have gone wrong.
Thanks in advance!
As for me your code works (Python3.9). So make sure your script has access to the directory you've specified. Also, it looks like you are trying to open a file which does not exist. So make sure you've downloaded the file before opening it:
fullfilename = os.path.abspath("d:/DownloadedFiles/Lec-1.pdf")
print(fullfilename)
if os.path.exists(fullfilename): # open file only if it exists
with open(fullfilename, 'rb') as file:
content = file.read() # read file's content
print(content[:150]) # print only the first 150 characters
The output would be as follows:
C:/Users/Administrator/PycharmProjects/Tests/dtest.py
d:\DownloadedFiles\Lec-1.pdf
b'%PDF-1.6\r%\xe2\xe3\xcf\xd3\r\n2346 0 obj <</Linearized 1/L 1916277/O 2349/E 70472/N 160/T 1869308/H [ 536 3620]>>\rendobj\r \r\nxref\r\n2346 12\r\n0000000016 00000 n\r'
Process finished with exit code 0