By default, when running Flask application using the built-in server (Flask.run
), it monitors its Python files and automatically reloads the app if its code changes:
* Detected change in "/home/xion/hello-world/app.py", reloading
* Restarting with reloader
Unfortunately, this seems to work for *.py files only, and I don"t seem to find any way to extend this functionality to other files. Most notably, it would be extremely useful to have Flask restart the app when a template changes. I"ve lost count on how many times I was fiddling with markup in templates and getting confused by not seeing any changes, only to find out that the app was still using the old version of Jinja template.
So, is there a way to have Flask monitor files in templates directory, or does it require diving into the framework"s source?
Edit: I"m using Ubuntu 10.10. Haven"t tried that on any other platforms really.
After further inquiry, I have discovered that changes in templates indeed are updated in real time, without reloading the app itself. However, this seems to apply only to those templates that are passed to flask.render_template
.
But it so happens that in my app, I have quite a lot of reusable, parametrized components which I use in Jinja templates. They are implemented as {% macro %}
s, reside in dedicated "modules" and are {% import %}
ed into actual pages. All nice and DRY... except that those imported templates are apparently never checked for modifications, as they don"t pass through render_template
at all.
(Curiously, this doesn"t happen for templates invoked through {% extends %}
. As for {% include %}
, I have no idea as I don"t really use them.)
So to wrap up, the roots of this phenomenon seems to lie somewhere between Jinja and Flask or Werkzeug. I guess it may warrant a trip to bug tracker for either of those projects :) Meanwhile, I"ve accepted the jd."s answer because that"s the solution I actually used - and it works like a charm.
Reload Flask app when template file changes _files: Questions
How do I list all files of a directory?
5 answers
How can I list all files of a directory in Python and add them to a list
?
Answer #1
os.listdir()
will get you everything that"s in a directory - files and directories.
If you want just files, you could either filter this down using os.path
:
from os import listdir
from os.path import isfile, join
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]
or you could use os.walk()
which will yield two lists for each directory it visits - splitting into files and dirs for you. If you only want the top directory you can break the first time it yields
from os import walk
f = []
for (dirpath, dirnames, filenames) in walk(mypath):
f.extend(filenames)
break
or, shorter:
from os import walk
filenames = next(walk(mypath), (None, None, []))[2] # [] if no file
Answer #2
I prefer using the glob
module, as it does pattern matching and expansion.
import glob
print(glob.glob("/home/adam/*"))
It does pattern matching intuitively
import glob
# All files ending with .txt
print(glob.glob("/home/adam/*.txt"))
# All files ending with .txt with depth of 2 folder
print(glob.glob("/home/adam/*/*.txt"))
It will return a list with the queried files:
["/home/adam/file1.txt", "/home/adam/file2.txt", .... ]
Answer #3
os.listdir()
- list in the current directory
With listdir in os module you get the files and the folders in the current dir
import os
arr = os.listdir()
print(arr)
>>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
Looking in a directory
arr = os.listdir("c:\files")
glob
from glob
with glob you can specify a type of file to list like this
import glob
txtfiles = []
for file in glob.glob("*.txt"):
txtfiles.append(file)
glob
in a list comprehension
mylist = [f for f in glob.glob("*.txt")]
get the full path of only files in the current directory
import os
from os import listdir
from os.path import isfile, join
cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if
os.path.isfile(os.path.join(cwd, f))]
print(onlyfiles)
["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]
Getting the full path name with
os.path.abspath
You get the full path in return
import os
files_path = [os.path.abspath(x) for x in os.listdir()]
print(files_path)
["F:\documentiapplications.txt", "F:\documenticollections.txt"]
Walk: going through sub directories
os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.
import os
# Getting the current work directory (cwd)
thisdir = os.getcwd()
# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
for file in f:
if file.endswith(".docx"):
print(os.path.join(r, file))
os.listdir()
: get files in the current directory (Python 2)
In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.
import os
arr = os.listdir(".")
print(arr)
>>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
To go up in the directory tree
# Method 1
x = os.listdir("..")
# Method 2
x= os.listdir("/")
Get files:
os.listdir()
in a particular directory (Python 2 and 3)
import os
arr = os.listdir("F:\python")
print(arr)
>>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
Get files of a particular subdirectory with
os.listdir()
import os
x = os.listdir("./content")
os.walk(".")
- current directory
import os
arr = next(os.walk("."))[2]
print(arr)
>>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]
next(os.walk("."))
andos.path.join("dir", "file")
import os
arr = []
for d,r,f in next(os.walk("F:\_python")):
for file in f:
arr.append(os.path.join(r,file))
for f in arr:
print(files)
>>> F:\_python\dict_class.py
>>> F:\_python\programmi.txt
next(os.walk("F:\")
- get the full path - list comprehension
[os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
>>> ["F:\_python\dict_class.py", "F:\_python\programmi.txt"]
os.walk
- get full path - all files in sub dirs**
x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]
print(x)
>>> ["F:\_python\dict.py", "F:\_python\progr.txt", "F:\_python\readl.py"]
os.listdir()
- get only txt files
arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
print(arr_txt)
>>> ["work.txt", "3ebooks.txt"]
Using
glob
to get the full path of the files
If I should need the absolute path of the files:
from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\*.txt")]
for f in x:
print(f)
>>> F:acquistionline.txt
>>> F:acquisti_2018.txt
>>> F:ootstrap_jquery_ecc.txt
Using
os.path.isfile
to avoid directories in the list
import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)
>>> ["a simple game.py", "data.txt", "decorator.py"]
Using
pathlib
from Python 3.4
import pathlib
flist = []
for p in pathlib.Path(".").iterdir():
if p.is_file():
print(p)
flist.append(p)
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speak_gui2.py
>>> thumb.PNG
With list comprehension
:
flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]
Alternatively, use pathlib.Path()
instead of pathlib.Path(".")
Use glob method in pathlib.Path()
import pathlib
py = pathlib.Path().glob("*.py")
for file in py:
print(file)
>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py
Get all and only files with os.walk
import os
x = [i[2] for i in os.walk(".")]
y=[]
for t in x:
for f in t:
y.append(f)
print(y)
>>> ["append_to_list.py", "data.txt", "data1.txt", "data2.txt", "data_180617", "os_walk.py", "READ2.py", "read_data.py", "somma_defaltdic.py", "substitute_words.py", "sum_data.py", "data.txt", "data1.txt", "data_180617"]
Get only files with next and walk in a directory
import os
x = next(os.walk("F://python"))[2]
print(x)
>>> ["calculator.bat","calculator.py"]
Get only directories with next and walk in a directory
import os
next(os.walk("F://python"))[1] # for the current dir use (".")
>>> ["python3","others"]
Get all the subdir names with
walk
for r,d,f in os.walk("F:\_python"):
for dirs in d:
print(dirs)
>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints
os.scandir()
from Python 3.5 and greater
import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)
>>> ["calculator.bat","calculator.py"]
# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.
import os
with os.scandir() as i:
for entry in i:
if entry.is_file():
print(entry.name)
>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG
Examples:
Ex. 1: How many files are there in the subdirectories?
In this example, we look for the number of files that are included in all the directory and its subdirectories.
import os
def count(dir, counter=0):
"returns number of files in dir and subdirs"
for pack in os.walk(dir):
for f in pack[2]:
counter += 1
return dir + " : " + str(counter) + "files"
print(count("F:\python"))
>>> "F:\python" : 12057 files"
Ex.2: How to copy all files from a directory to another?
A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.
import os
import shutil
from path import path
destination = "F:\file_copied"
# os.makedirs(destination)
def copyfile(dir, filetype="pptx", counter=0):
"Searches for pptx (or other - pptx is the default) files and copies them"
for pack in os.walk(dir):
for f in pack[2]:
if f.endswith(filetype):
fullpath = pack[0] + "\" + f
print(fullpath)
shutil.copy(fullpath, destination)
counter += 1
if counter > 0:
print("-" * 30)
print(" ==> Found in: `" + dir + "` : " + str(counter) + " files
")
for dir in os.listdir():
"searches for folders that starts with `_`"
if dir[0] == "_":
# copyfile(dir, filetype="pdf")
copyfile(dir, filetype="txt")
>>> _compiti18Compito Contabilità 1conti.txt
>>> _compiti18Compito Contabilità 1modula4.txt
>>> _compiti18Compito Contabilità 1moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files
Ex. 3: How to get all the files in a txt file
In case you want to create a txt file with all the file names:
import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
for eachfile in os.listdir():
mylist += eachfile + "
"
file.write(mylist)
Example: txt with all the files of an hard drive
"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""
import os
# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding="utf-8") as testo:
for root, dirs, files in os.walk("D:\"):
for file in files:
listafile.append(file)
percorso.append(root + "\" + file)
testo.write(file + "
")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
for file in listafile:
testo_ordinato.write(file + "
")
with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
for file in percorso:
file_percorso.write(file + "
")
os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")
All the file of C: in one text file
This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.
import os
with open("file.txt", "w", encoding="utf-8") as filewrite:
for r, d, f in os.walk("C:\"):
for file in f:
filewrite.write(f"{r + file}
")
How to write a file with all paths in a folder of a type
With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.
import os
def searchfiles(extension=".ttf", folder="H:\"):
"Create a txt file with all the file of a type"
with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
for r, d, f in os.walk(folder):
for file in f:
if file.endswith(extension):
filewrite.write(f"{r + file}
")
# looking for png file (fonts) in the hard disk H:
searchfiles(".png", "H:\")
>>> H:4bs_18Dolphins5.png
>>> H:4bs_18Dolphins6.png
>>> H:4bs_18Dolphins7.png
>>> H:5_18marketing htmlassetsimageslogo2.png
>>> H:7z001.png
>>> H:7z002.png
(New) Find all files and open them with tkinter GUI
I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list.
import tkinter as tk
import os
def searchfiles(extension=".txt", folder="H:\"):
"insert all files in the listbox"
for r, d, f in os.walk(folder):
for file in f:
if file.endswith(extension):
lb.insert(0, r + "\" + file)
def open_file():
os.startfile(lb.get(lb.curselection()[0]))
root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()
Reload Flask app when template file changes find: Questions
Finding the index of an item in a list
5 answers
Given a list ["foo", "bar", "baz"]
and an item in the list "bar"
, how do I get its index (1
) in Python?
Answer #1
>>> ["foo", "bar", "baz"].index("bar")
1
Reference: Data Structures > More on Lists
Caveats follow
Note that while this is perhaps the cleanest way to answer the question as asked, index
is a rather weak component of the list
API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index
follow. It is probably worth initially taking a look at the documentation for it:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a
ValueError
if there is no such item.The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
Linear time-complexity in list length
An index
call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index
a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000)
is roughly five orders of magnitude faster than straight l.index(999_999)
, because the former only has to search 10 entries, while the latter searches a million:
>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514
Only returns the index of the first match to its argument
A call to index
searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.
>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
Most places where I once would have used index
, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index
, take a look at these excellent Python features.
Throws if element not present in list
A call to index
results in a ValueError
if the item"s not present.
>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
If the item might not be present in the list, you should either
- Check for it first with
item in my_list
(clean, readable approach), or - Wrap the
index
call in atry/except
block which catchesValueError
(probably faster, at least when the list to search is long, and the item is usually present.)
Answer #2
One thing that is really helpful in learning Python is to use the interactive help function:
>>> help(["foo", "bar", "baz"])
Help on list object:
class list(object)
...
|
| index(...)
| L.index(value, [start, [stop]]) -> integer -- return first index of value
|
which will often lead you to the method you are looking for.
Answer #3
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate()
:
for i, j in enumerate(["foo", "bar", "baz"]):
if j == "bar":
print(i)
The index()
function only returns the first occurrence, while enumerate()
returns all occurrences.
As a list comprehension:
[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
Here"s also another small solution with itertools.count()
(which is pretty much the same approach as enumerate):
from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
This is more efficient for larger lists than using enumerate()
:
$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop