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Alternative of urllib.urlretrieve in Python 3.5

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1

I am currently doing a course on machine learning in UDACITY . In there they have written some code in python 2.7 but as i am currently using python 3.5 , i am getting some error . This is the code

import urllib
url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
urllib.urlretrieve(url, filename="../enron_mail_20150507.tgz")
print ("download complete!") 

I tried urllib.request .

 import urllib
 url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
 urllib.request(url, filename="../enron_mail_20150507.tgz")
 print ("download complete!")

But still gives me error .

urllib.request(url, filename="../enron_mail_20150507.tgz")
TypeError: 'module' object is not callable

I am using PyCharm as my IDE .

python python-3.x urllib2

share|improve this question

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if requests is available, I'd use that
  
  – Wayne Werner
  Jul 13 '16 at 17:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Someone just answerd it and worked for me . instead of urllib.urlretrive one would have to use urllib.request.urlretrieve
  
  – CodeHead
  Jul 13 '16 at 17:59
  

  
  
  
  

add a comment | 

1

I am currently doing a course on machine learning in UDACITY . In there they have written some code in python 2.7 but as i am currently using python 3.5 , i am getting some error . This is the code

import urllib
url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
urllib.urlretrieve(url, filename="../enron_mail_20150507.tgz")
print ("download complete!") 

I tried urllib.request .

 import urllib
 url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
 urllib.request(url, filename="../enron_mail_20150507.tgz")
 print ("download complete!")

But still gives me error .

urllib.request(url, filename="../enron_mail_20150507.tgz")
TypeError: 'module' object is not callable

I am using PyCharm as my IDE .

python python-3.x urllib2

share|improve this question

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if requests is available, I'd use that
  
  – Wayne Werner
  Jul 13 '16 at 17:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Someone just answerd it and worked for me . instead of urllib.urlretrive one would have to use urllib.request.urlretrieve
  
  – CodeHead
  Jul 13 '16 at 17:59
  

  
  
  
  

add a comment | 

I am currently doing a course on machine learning in UDACITY . In there they have written some code in python 2.7 but as i am currently using python 3.5 , i am getting some error . This is the code

import urllib
url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
urllib.urlretrieve(url, filename="../enron_mail_20150507.tgz")
print ("download complete!") 

I tried urllib.request .

 import urllib
 url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
 urllib.request(url, filename="../enron_mail_20150507.tgz")
 print ("download complete!")

But still gives me error .

urllib.request(url, filename="../enron_mail_20150507.tgz")
TypeError: 'module' object is not callable

I am using PyCharm as my IDE .

python python-3.x urllib2

share|improve this question

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

import urllib
url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
urllib.urlretrieve(url, filename="../enron_mail_20150507.tgz")
print ("download complete!") 

 import urllib
 url = "https://www.cs.cmu.edu/~./enron/enron_mail_20150507.tgz"
 urllib.request(url, filename="../enron_mail_20150507.tgz")
 print ("download complete!")

urllib.request(url, filename="../enron_mail_20150507.tgz")
TypeError: 'module' object is not callable

share|improve this question

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

share|improve this question

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

asked Jul 13 '16 at 17:48

CodeHeadCodeHead

57110

2 Answers
2

active

oldest

votes

7

You'd use urllib.request.urlretrieve. Note that this function "may become deprecated at some point in the future", so you might be better off using the less likely to be deprecated interface:

# Adapted from the source:
# https://hg.python.org/cpython/file/3.5/Lib/urllib/request.py#l170
with open(filename, 'wb') as out_file:
 with contextlib.closing(urllib.request.urlopen(url)) as fp:
 block_size = 1024 * 8
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)

For small enough files, you could just read and write the whole thing and drop the loop entirely.

share|improve this answer

edited Jul 13 '16 at 18:00

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks a lot , It worked fine .
  
  – CodeHead
  Jul 13 '16 at 18:04
  

  
  
  
  

add a comment | 

0

I know this question has long been answered but I'll contribute for any future viewer.

The proposed solution is good but the main issue if that it can generate empty files if you are using invalid urls.

As a workaround to this problem here is how I adapted the code:

def getfile(url,filename,timeout=45):
 with contextlib.closing(urlopen(url,timeout=timeout)) as fp:
 block_size = 1024 * 8
 block = fp.read(block_size)
 if block:
 with open(filename,'wb') as out_file:
 out_file.write(block)
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)
 else:
 raise Exception ('nonexisting file or connection error')

I hope this help.

share|improve this answer

answered Mar 24 at 11:52

Al rlAl rl

195

add a comment | 

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7

You'd use urllib.request.urlretrieve. Note that this function "may become deprecated at some point in the future", so you might be better off using the less likely to be deprecated interface:

# Adapted from the source:
# https://hg.python.org/cpython/file/3.5/Lib/urllib/request.py#l170
with open(filename, 'wb') as out_file:
 with contextlib.closing(urllib.request.urlopen(url)) as fp:
 block_size = 1024 * 8
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)

For small enough files, you could just read and write the whole thing and drop the loop entirely.

share|improve this answer

edited Jul 13 '16 at 18:00

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks a lot , It worked fine .
  
  – CodeHead
  Jul 13 '16 at 18:04
  

  
  
  
  

add a comment | 

7

You'd use urllib.request.urlretrieve. Note that this function "may become deprecated at some point in the future", so you might be better off using the less likely to be deprecated interface:

# Adapted from the source:
# https://hg.python.org/cpython/file/3.5/Lib/urllib/request.py#l170
with open(filename, 'wb') as out_file:
 with contextlib.closing(urllib.request.urlopen(url)) as fp:
 block_size = 1024 * 8
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)

For small enough files, you could just read and write the whole thing and drop the loop entirely.

share|improve this answer

edited Jul 13 '16 at 18:00

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks a lot , It worked fine .
  
  – CodeHead
  Jul 13 '16 at 18:04
  

  
  
  
  

add a comment | 

You'd use urllib.request.urlretrieve. Note that this function "may become deprecated at some point in the future", so you might be better off using the less likely to be deprecated interface:

# Adapted from the source:
# https://hg.python.org/cpython/file/3.5/Lib/urllib/request.py#l170
with open(filename, 'wb') as out_file:
 with contextlib.closing(urllib.request.urlopen(url)) as fp:
 block_size = 1024 * 8
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)

For small enough files, you could just read and write the whole thing and drop the loop entirely.

share|improve this answer

edited Jul 13 '16 at 18:00

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

# Adapted from the source:
# https://hg.python.org/cpython/file/3.5/Lib/urllib/request.py#l170
with open(filename, 'wb') as out_file:
 with contextlib.closing(urllib.request.urlopen(url)) as fp:
 block_size = 1024 * 8
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)

share|improve this answer

edited Jul 13 '16 at 18:00

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

answered Jul 13 '16 at 17:54

mgilsonmgilson

217k41429542

0

I know this question has long been answered but I'll contribute for any future viewer.

The proposed solution is good but the main issue if that it can generate empty files if you are using invalid urls.

As a workaround to this problem here is how I adapted the code:

def getfile(url,filename,timeout=45):
 with contextlib.closing(urlopen(url,timeout=timeout)) as fp:
 block_size = 1024 * 8
 block = fp.read(block_size)
 if block:
 with open(filename,'wb') as out_file:
 out_file.write(block)
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)
 else:
 raise Exception ('nonexisting file or connection error')

I hope this help.

share|improve this answer

answered Mar 24 at 11:52

Al rlAl rl

195

add a comment | 

0

I know this question has long been answered but I'll contribute for any future viewer.

The proposed solution is good but the main issue if that it can generate empty files if you are using invalid urls.

As a workaround to this problem here is how I adapted the code:

def getfile(url,filename,timeout=45):
 with contextlib.closing(urlopen(url,timeout=timeout)) as fp:
 block_size = 1024 * 8
 block = fp.read(block_size)
 if block:
 with open(filename,'wb') as out_file:
 out_file.write(block)
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)
 else:
 raise Exception ('nonexisting file or connection error')

I hope this help.

share|improve this answer

answered Mar 24 at 11:52

Al rlAl rl

195

add a comment | 

I know this question has long been answered but I'll contribute for any future viewer.

The proposed solution is good but the main issue if that it can generate empty files if you are using invalid urls.

As a workaround to this problem here is how I adapted the code:

def getfile(url,filename,timeout=45):
 with contextlib.closing(urlopen(url,timeout=timeout)) as fp:
 block_size = 1024 * 8
 block = fp.read(block_size)
 if block:
 with open(filename,'wb') as out_file:
 out_file.write(block)
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)
 else:
 raise Exception ('nonexisting file or connection error')

I hope this help.

share|improve this answer

answered Mar 24 at 11:52

Al rlAl rl

195

def getfile(url,filename,timeout=45):
 with contextlib.closing(urlopen(url,timeout=timeout)) as fp:
 block_size = 1024 * 8
 block = fp.read(block_size)
 if block:
 with open(filename,'wb') as out_file:
 out_file.write(block)
 while True:
 block = fp.read(block_size)
 if not block:
 break
 out_file.write(block)
 else:
 raise Exception ('nonexisting file or connection error')

share|improve this answer

answered Mar 24 at 11:52

Al rlAl rl

195

answered Mar 24 at 11:52

Al rlAl rl

195

answered Mar 24 at 11:52

Al rlAl rl

195