Styjun

Noob at soldering, can anyone explain why my circuit won't work?

Understanding basic photoresistor circuit

What does this quote in Small Gods refer to?

Will change of address affect direct deposit?

Can the sorting of a list be verified without comparing neighbors?

Thesis' "Future Work" section – is it acceptable to omit personal involvement in a mentioned project?

How to make the table in the figure in LaTeX?

What is the difference between "Plural" and "Mehrzahl"?

Pre-1993 comic in which Wolverine's claws were turned to rubber?

Was there ever any real use for a 6800-based Apple I?

How to make a language evolve quickly?

Can I do brevets (long distance rides) on my hybrid bike? If yes, how to start?

How can a Lich look like a human without magic?

How do I compare the result of "1d20+x, with advantage" to "1d20+y, without advantage", assuming x < y?

Should these notes be played as a chord or one after another?

Why use steam instead of just hot air?

histogram using edges

Cropping a message using array splits

What is the best way for a skeleton to impersonate human without using magic?

Why in a Ethernet LAN, a packet sniffer can obtain all packets sent over the LAN?

Control variables and other independent variables

Make all the squares explode

Why does a C.D.F need to be right-continuous?

Adding slope values to attribute table (QGIS 3)

Check if a file type is a media file?

How to find the mime type of a file in python?How do I check if a list is empty?How do I check whether a file exists without exceptions?What's the canonical way to check for type in Python?How do I check if a string is a number (float)?Check if a given key already exists in a dictionaryDetermine the type of an object?How do I list all files of a directory?How to read a file line-by-line into a list?Delete a file or folder.endswith missing file extention in a loop

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

3

2

I am trying to loop through a list of files, and return those files that are media files (images, video, gif, audio, etc.).

Seeing as there are a lot of media types, is there a library or perhaps better way to check this, than listing all types then checking a file against that list?

Here's what I'm doing so far:

import os
types = [".mp3", ".mpeg", ".gif", ".jpg", ".jpeg"]
files = ["test.mp3", "test.tmp", "filename.mpg", ".AutoConfig"]

media_files = []
for file in files:
 root, extention = os.path.splitext(file)
 print(extention)
 if extention in types:
 media_files.append(file)

print("Found media files are:")
print(media_files)

But note it didn't include filename.mpg, since I forgot to put .mpg in my types list. (Or, more likely, I didn't expect that list to include a .mpg file, so didn't think to list it out.)

python python-3.x

share|improve this question

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, you can you mimetype check. Here is a example: stackoverflow.com
  
  – Cpp Forever
  Mar 22 at 21:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you're running on UNIX/Linux, you can use file to determine media type.
  
  – tk421
  Mar 22 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @CppForever - I found that, and am studying that library, but am not sure how to check without something like - if mime.from_file("media.mp3") == "application/mp3" or ...:? I am missing understanding something I think...
  
  – BruceWayne
  Mar 22 at 21:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @CppForever so do I just heck generally "is the file a mime type" without having to check exactly what kind?
  
  – BruceWayne
  Mar 22 at 21:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  After you get mime type for example audio/mp3 you can split by / character and get the first part and check if it is audio or video or image
  
  – Cpp Forever
  Mar 22 at 21:43
  

  
  
  
  

 | 
show 1 more comment

3

2

I am trying to loop through a list of files, and return those files that are media files (images, video, gif, audio, etc.).

Seeing as there are a lot of media types, is there a library or perhaps better way to check this, than listing all types then checking a file against that list?

Here's what I'm doing so far:

import os
types = [".mp3", ".mpeg", ".gif", ".jpg", ".jpeg"]
files = ["test.mp3", "test.tmp", "filename.mpg", ".AutoConfig"]

media_files = []
for file in files:
 root, extention = os.path.splitext(file)
 print(extention)
 if extention in types:
 media_files.append(file)

print("Found media files are:")
print(media_files)

But note it didn't include filename.mpg, since I forgot to put .mpg in my types list. (Or, more likely, I didn't expect that list to include a .mpg file, so didn't think to list it out.)

python python-3.x

share|improve this question

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, you can you mimetype check. Here is a example: stackoverflow.com
  
  – Cpp Forever
  Mar 22 at 21:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you're running on UNIX/Linux, you can use file to determine media type.
  
  – tk421
  Mar 22 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @CppForever - I found that, and am studying that library, but am not sure how to check without something like - if mime.from_file("media.mp3") == "application/mp3" or ...:? I am missing understanding something I think...
  
  – BruceWayne
  Mar 22 at 21:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @CppForever so do I just heck generally "is the file a mime type" without having to check exactly what kind?
  
  – BruceWayne
  Mar 22 at 21:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  After you get mime type for example audio/mp3 you can split by / character and get the first part and check if it is audio or video or image
  
  – Cpp Forever
  Mar 22 at 21:43
  

  
  
  
  

 | 
show 1 more comment

I am trying to loop through a list of files, and return those files that are media files (images, video, gif, audio, etc.).

Seeing as there are a lot of media types, is there a library or perhaps better way to check this, than listing all types then checking a file against that list?

Here's what I'm doing so far:

import os
types = [".mp3", ".mpeg", ".gif", ".jpg", ".jpeg"]
files = ["test.mp3", "test.tmp", "filename.mpg", ".AutoConfig"]

media_files = []
for file in files:
 root, extention = os.path.splitext(file)
 print(extention)
 if extention in types:
 media_files.append(file)

print("Found media files are:")
print(media_files)

But note it didn't include filename.mpg, since I forgot to put .mpg in my types list. (Or, more likely, I didn't expect that list to include a .mpg file, so didn't think to list it out.)

python python-3.x

share|improve this question

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

import os
types = [".mp3", ".mpeg", ".gif", ".jpg", ".jpeg"]
files = ["test.mp3", "test.tmp", "filename.mpg", ".AutoConfig"]

media_files = []
for file in files:
 root, extention = os.path.splitext(file)
 print(extention)
 if extention in types:
 media_files.append(file)

print("Found media files are:")
print(media_files)

share|improve this question

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

share|improve this question

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

asked Mar 22 at 21:22

BruceWayneBruceWayne

18.2k113464

2 Answers
2

active

oldest

votes

2

For this purpose you need to get internet media type for file, split it by / character and check if it starts with audio,video,image.

Here is a sample code:

import mimetypes
mimetypes.init()

mimestart = mimetypes.guess_type("test.mp3")[0]

if mimestart != None:
 mimestart = mimestart.split('/')[0]

 if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: This method assume the file type by its extension and don't open the actual file, it is based only on the file extension

share|improve this answer

edited Mar 23 at 10:44

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You should check for None (i.e. unknown type) when using guess_type. However, you should note that this method will only check the extension, so it cannot detect the file's actual type.
  
  – ekhumoro
  Mar 22 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  While mimetypes does exactly what the OP asks, maybe also point to pypi.org/project/python-libmagic which inspects file contents, not just the filename. libmagic is the library behind the Unix file command.
  
  – tripleee
  Mar 23 at 10:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thanks so much for this! I really appreciate both answers too, I like this since I know the filetypes and can check without opening the file. Thanks so much! :D
  
  – BruceWayne
  Mar 23 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For anyone - Here's a list of common Web MIME types. And here's a longer list (but I note it doesn't include mp3 for some reason).
  
  – BruceWayne
  Mar 23 at 21:52
  

  
  
  
  

add a comment | 

1

There is another method that is based not on the file extension but on the file contents using the media type library pypi.org/project/python-libmagic:

Here is the sample code for this library:

import magic

magic = magic.Magic()
mimestart = magic.from_file("test.mp3").split('/')[0]

if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: for using this code sample you need to install python-libmagic trough pip.

share|improve this answer

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I assume this method, where it actually checks the file contents itself, is most often used when you can't trust the extension, for whatever reason? Thanks for this!
  
  – BruceWayne
  Mar 23 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For example in linux the executables don't have extension but have a signature.
  
  – Cpp Forever
  Mar 23 at 18:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  ohhh okay! I am running Windows but using raspberry pi so that will likely come in handy. Thanks again!!
  
  – BruceWayne
  Mar 23 at 18:54
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55307951%2fcheck-if-a-file-type-is-a-media-file%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

For this purpose you need to get internet media type for file, split it by / character and check if it starts with audio,video,image.

Here is a sample code:

import mimetypes
mimetypes.init()

mimestart = mimetypes.guess_type("test.mp3")[0]

if mimestart != None:
 mimestart = mimestart.split('/')[0]

 if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: This method assume the file type by its extension and don't open the actual file, it is based only on the file extension

share|improve this answer

edited Mar 23 at 10:44

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You should check for None (i.e. unknown type) when using guess_type. However, you should note that this method will only check the extension, so it cannot detect the file's actual type.
  
  – ekhumoro
  Mar 22 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  While mimetypes does exactly what the OP asks, maybe also point to pypi.org/project/python-libmagic which inspects file contents, not just the filename. libmagic is the library behind the Unix file command.
  
  – tripleee
  Mar 23 at 10:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thanks so much for this! I really appreciate both answers too, I like this since I know the filetypes and can check without opening the file. Thanks so much! :D
  
  – BruceWayne
  Mar 23 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For anyone - Here's a list of common Web MIME types. And here's a longer list (but I note it doesn't include mp3 for some reason).
  
  – BruceWayne
  Mar 23 at 21:52
  

  
  
  
  

add a comment | 

2

For this purpose you need to get internet media type for file, split it by / character and check if it starts with audio,video,image.

Here is a sample code:

import mimetypes
mimetypes.init()

mimestart = mimetypes.guess_type("test.mp3")[0]

if mimestart != None:
 mimestart = mimestart.split('/')[0]

 if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: This method assume the file type by its extension and don't open the actual file, it is based only on the file extension

share|improve this answer

edited Mar 23 at 10:44

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You should check for None (i.e. unknown type) when using guess_type. However, you should note that this method will only check the extension, so it cannot detect the file's actual type.
  
  – ekhumoro
  Mar 22 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  While mimetypes does exactly what the OP asks, maybe also point to pypi.org/project/python-libmagic which inspects file contents, not just the filename. libmagic is the library behind the Unix file command.
  
  – tripleee
  Mar 23 at 10:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thanks so much for this! I really appreciate both answers too, I like this since I know the filetypes and can check without opening the file. Thanks so much! :D
  
  – BruceWayne
  Mar 23 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For anyone - Here's a list of common Web MIME types. And here's a longer list (but I note it doesn't include mp3 for some reason).
  
  – BruceWayne
  Mar 23 at 21:52
  

  
  
  
  

add a comment | 

For this purpose you need to get internet media type for file, split it by / character and check if it starts with audio,video,image.

Here is a sample code:

import mimetypes
mimetypes.init()

mimestart = mimetypes.guess_type("test.mp3")[0]

if mimestart != None:
 mimestart = mimestart.split('/')[0]

 if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: This method assume the file type by its extension and don't open the actual file, it is based only on the file extension

share|improve this answer

edited Mar 23 at 10:44

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

import mimetypes
mimetypes.init()

mimestart = mimetypes.guess_type("test.mp3")[0]

if mimestart != None:
 mimestart = mimestart.split('/')[0]

 if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

share|improve this answer

edited Mar 23 at 10:44

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

answered Mar 22 at 21:52

Cpp ForeverCpp Forever

1525

1

There is another method that is based not on the file extension but on the file contents using the media type library pypi.org/project/python-libmagic:

Here is the sample code for this library:

import magic

magic = magic.Magic()
mimestart = magic.from_file("test.mp3").split('/')[0]

if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: for using this code sample you need to install python-libmagic trough pip.

share|improve this answer

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I assume this method, where it actually checks the file contents itself, is most often used when you can't trust the extension, for whatever reason? Thanks for this!
  
  – BruceWayne
  Mar 23 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For example in linux the executables don't have extension but have a signature.
  
  – Cpp Forever
  Mar 23 at 18:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  ohhh okay! I am running Windows but using raspberry pi so that will likely come in handy. Thanks again!!
  
  – BruceWayne
  Mar 23 at 18:54
  

  
  
  
  

add a comment | 

1

There is another method that is based not on the file extension but on the file contents using the media type library pypi.org/project/python-libmagic:

Here is the sample code for this library:

import magic

magic = magic.Magic()
mimestart = magic.from_file("test.mp3").split('/')[0]

if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: for using this code sample you need to install python-libmagic trough pip.

share|improve this answer

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I assume this method, where it actually checks the file contents itself, is most often used when you can't trust the extension, for whatever reason? Thanks for this!
  
  – BruceWayne
  Mar 23 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For example in linux the executables don't have extension but have a signature.
  
  – Cpp Forever
  Mar 23 at 18:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  ohhh okay! I am running Windows but using raspberry pi so that will likely come in handy. Thanks again!!
  
  – BruceWayne
  Mar 23 at 18:54
  

  
  
  
  

add a comment | 

There is another method that is based not on the file extension but on the file contents using the media type library pypi.org/project/python-libmagic:

Here is the sample code for this library:

import magic

magic = magic.Magic()
mimestart = magic.from_file("test.mp3").split('/')[0]

if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

NOTE: for using this code sample you need to install python-libmagic trough pip.

share|improve this answer

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525

import magic

magic = magic.Magic()
mimestart = magic.from_file("test.mp3").split('/')[0]

if mimestart == 'audio' or mimestart == 'video' or mimestart == 'image':
 print("media types")

share|improve this answer

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525

answered Mar 23 at 11:15

Cpp ForeverCpp Forever

1525