I am a newbie in Python, and I only have been programming in the language for the third day. I convert from Matlab because I am graduating from university and I just cannot afford the heavy cost of the software.
I am trying to add several filenames to a list, which fails. The following is my code:
import os
dirname = r"D:\MyFiles"
temp = os.listdir(dirname)
fl = []
for fn in temp:
if fn.startswith('L0116'):
fl = fl.append(fn)
What I don't understand is, the variable fl when initiated with [] returns as the list type, but then I fail to append the list with the above loop. I recheck my code using interactive prompt and I found out that the type changes to str and fails to append more.
I am stuck at this and I tried to google for about an hour without any clear clue to how I should do this.
Is there a better way? Or which part did I do it wrong?
Thank you so much for your help!
The append() method mutates the list in place. You should just append fn like this:
fl.append(fn)
don't try to assign the result of fl.append(fn) (which is None, by the way)
Also, don't use r"foo" strings for filenames. They are designed for use with regular expressions and using them for filenames or other kinds of strings will bite you. For example, you cannot end such a string with a backslash.
append works inplace, so no need to assign it to the list itself (fl = fl.append(fn)) just do fl.append(fn)
List comprehension:
import os
dirname = r"D:\MyFiles"
fl = [fn for fn in os.listdir(dirname) if fn.startswith('L0116')]
List append in python does not return a type, instead, you simply call the method like so:
import os
dirname = r"D:\MyFiles"
temp = os.listdir(dirname)
fl = []
for fn in temp:
if fn.startswith('L0116'):
fl.append(fn)
I haven't tested this code, but it should work. Your logic was resetting the list every time.
Related
so i'm currently learning about mail merging and was issued a challenge on it. The idea is to open a names file, read the name on the current line and then replace it in the letter and save that letter as a new item.
I figured a good idea to do this would be a for loop.
Open file > for loop > append names to list > loop the list and replace ect.
Except when I try to actually append the names to the list, i get this:
['Aang\nZuko\nAppa\nKatara\nSokka\nMomo\nUncle Iroh\nToph']
The code I am using is:
invited_names = []
with open ("./Input/Names/invited_names.txt") as names:
invited_names.append(names.read())
for item in invited_names:
new_names = [str.strip("\n") for str in invited_names]
print(new_names)
Have tried to replace the \n and now .strip but I have not been able to remove the \n. Any ideas?
EDIT: not sure if it helps but the .txt file for the names looks like this:
Aang
Zuko
Appa
Katara
Sokka
Momo
Uncle Iroh
Toph
As you can see, read() only returns a giant string of what you have in your invited_names.txt file. But instead, you can use readlines() which returns a list which contains strings of every line (Thanks to codeflush.dev for the comment). Then use extend() method to add this list to another list invited_names.
Again, you are using for loop and list comprehension at the same time. As a result, you are running the same list comprehension code for many times. So, you can cut off any of them. But I prefer you should keep the list comprehension because it is efficient.
Try this code:
invited_names = []
with open ("./Input/Names/invited_names.txt") as names:
invited_names.extend(names.readlines()) # <--
new_names = [str.strip("\n") for str in invited_names]
print(new_names)
I am trying to create a string separated by comma from the below given list
['D:\\abc\\pqr\\123\\aaa.xlsx', 'D:\\abc\\pqr\\123\\bbb.xlsx', 'D:\\abc\\pqr\\123\\ccc.xlsx']
New string should contain only the filename like below which is separated by comma
'aaa.xlsx,bbb.xlsx,ccc.xlsx'
I have achieved this using the below code
n = []
for p in input_list:
l = p.split('\\')
l = l[len(l)-1]
n.append(l)
a = ','.join(n)
print(a)
But instead of using multiple lines of code i would like to achieve this in single line using a list comprehension or regular expression.
Thanks in advance...
Simply do a
main_list = ['D:\\abc\\pqr\\123\\aaa.xlsx', 'D:\\abc\\pqr\\123\\bbb.xlsx', 'D:\\abc\\pqr\\123\\ccc.xlsx']
print([x.split("\\")[-1] for x in main_list])
OUTPUT:
['aaa.xlsx', 'bbb.xlsx', 'ccc.xlsx']
In case u want to get the string of this simply do a
print(",".join([x.split("\\")[-1] for x in main_list]))
OUTPUT:
aaa.xlsx,bbb.xlsx,ccc.xlsx
Another way to do the same is:
print(",".join(map(lambda x : x.split("\\")[-1],main_list)))
OUTPUT:
aaa.xlsx,bbb.xlsx,ccc.xlsx
Do see that os.path.basename is OS-dependent and may create problems on cross-platform scripts.
Using os.path.basename with str.join
Ex:
import os
data = ['D:\\abc\\pqr\\123\\aaa.xlsx', 'D:\\abc\\pqr\\123\\bbb.xlsx', 'D:\\abc\\pqr\\123\\ccc.xlsx']
print(",".join(os.path.basename(i) for i in data))
Output:
aaa.xlsx,bbb.xlsx,ccc.xlsx
I'm trying to make a python script that renames files randomly from a list and I used numbers.remove(place) on it but it keeps choosing values that are supposed to have been removed.
I used to just use random.randint but now I have moved to choosing from a list then removing the chosen value from the list but it seems to keep choosing chosen values.
'''python
from os import chdir, listdir, rename
from random import choice
def main():
chdir('C:\\Users\\user\\Desktop\\Folders\\Music')
for f in listdir():
if f.endswith('.mp4'):
numbers = [str(x) for x in range(0, 100)]
had = []
print(f'numbers = {numbers}')
place = choice(numbers)
print(f'place = {place}')
numbers.remove(place)
print(f'numbers = {numbers}')
while place in had:
input('Place has been had.')
place = choice(numbers)
had.append(place)
name = place + '.mp4'
print(f'name = {name}')
print(f'\n\nRenaming {f} to {name}.\n\n')
try:
rename(f, name)
except FileExistsError:
pass
if __name__ == '__main__':
main()
'''
It should randomly number the files without choosing the same value for a file twice but it does that and I have no idea why.
When you call listdir() the first time, that's the same list that you're iterating over the entire time. Yes, you're changing the contents of the directory, but python doesn't really care about that because you only asked for the contents of the directory at a specific point in time - before you began modifying it.
I would do this in two separate steps:
# get the current list of files in the directory
dirlist = os.listdir()
# choose a new name for each file
to_rename = zip(
dirlist,
[f'{num}.mp4' for num in random.sample(range(100), len(dirlist))]
)
# actually rename each file
for oldname, newname in to_rename:
try:
os.rename(oldname, newname)
except FileExistsError:
pass
This method is more concise than the one you're using. First, I use random.sample() on the iterable range(100) to generate non-overlapping numbers from that range (without having to do the extra step of using had like you're doing now). I generate exactly as many as I need, and then use the built-in zip() function to bind together the original filenames with these new numbers.
Then, I do the rename() operations all at once.
I have a script using argparse to gather a list of user defined directories. On the command line they may or may not specify a trailing "/" symbol. I'd like to do something up front so that all variables have the trailing "/" so I can reliably do:
# What I want:
with open(args.a + filename, "w") as fileout:
#do stuff
print('whatever', file=fileout)
rather than having to include an extra "/" in the name like this:
# What I have:
with open(args.a + "/" + filename, "w") as fileout:
#do stuff
print('whatever', file=fileout)
I also know that dir/ect/ory and dir//ect//ory are nearly equivalent save some fringe cases which are not applicable, but putting + "/" + all over the place seems wrong/wasteful.
In attempting to make a small function to run on all relevant variable I'm only seeing the desired outcome when I explicitly call the function on the variable not on a list containing the elements.
def trailingSlash(x):
if x.endswith("/"):
return x
else:
return x + "/"
a = 'ok/'
b = 'notok'
c = 'alsonotok'
for _ in [a, b, c]:
_ = trailingSlash(_)
print(a,b,c) #gives ok/ notok alsonotok
c = trailingSlash(c)
print(c) #gives alsonotok/
I understand why changing a list as you are iterating over it is generally bad, and understand that in the for loop the iterator is not actually pointing to a, b, or c. I also know if I wanted the values in a new list i could do something like [trailingSlash(x) for x [a,b,c]] but I need to maintain the a,b,c handle. in I know that I can also solve this by specifically calling x = trailingSlash(x) on every individual variable, but seems like there should be a better way. Any solutions I'm missing?
You can use os.path.join() to ignore the whole issue. It behaves no matter whether there are slashes at the end or not, and is platform-independent as a bonus (that is, it uses \\ instead of / when running on Windows, for example):
import os
...
os.path.join("dir/", "ect", "ory")
# "dir/ect/ory" on Unix, "dir\\ect\\ory" on Windows
In your case you'd want to do
with open(os.path.join(args.a, filename), "w") as fileout:
...
this question about matlab:
i'm running a loop and each iteration a new set of data is produced, and I want it to be saved in a new file each time. I also overwrite old files by changing the name. Looks like this:
name_each_iter = strrep(some_source,'.string.mat','string_new.(j).mat')
and what I#m struggling here is the iteration so that I obtain files:
...string_new.1.mat
...string_new.2.mat
etc.
I was trying with various combination of () [] {} as well as 'string_new.'j'.mat' (which gave syntax error)
How can it be done?
Strings are just vectors of characters. So if you want to iteratively create filenames here's an example of how you would do it:
for j = 1:10,
filename = ['string_new.' num2str(j) '.mat'];
disp(filename)
end
The above code will create the following output:
string_new.1.mat
string_new.2.mat
string_new.3.mat
string_new.4.mat
string_new.5.mat
string_new.6.mat
string_new.7.mat
string_new.8.mat
string_new.9.mat
string_new.10.mat
You could also generate all file names in advance using NUM2STR:
>> filenames = cellstr(num2str((1:10)','string_new.%02d.mat'))
filenames =
'string_new.01.mat'
'string_new.02.mat'
'string_new.03.mat'
'string_new.04.mat'
'string_new.05.mat'
'string_new.06.mat'
'string_new.07.mat'
'string_new.08.mat'
'string_new.09.mat'
'string_new.10.mat'
Now access the cell array contents as filenames{i} in each iteration
sprintf is very useful for this:
for ii=5:12
filename = sprintf('data_%02d.mat',ii)
end
this assigns the following strings to filename:
data_05.mat
data_06.mat
data_07.mat
data_08.mat
data_09.mat
data_10.mat
data_11.mat
data_12.mat
notice the zero padding. sprintf in general is useful if you want parameterized formatted strings.
For creating a name based of an already existing file, you can use regexp to detect the '_new.(number).mat' and change the string depending on what regexp finds:
original_filename = 'data.string.mat';
im = regexp(original_filename,'_new.\d+.mat')
if isempty(im) % original file, no _new.(j) detected
newname = [original_filename(1:end-4) '_new.1.mat'];
else
num = str2double(original_filename(im(end)+5:end-4));
newname = sprintf('%s_new.%d.mat',original_filename(1:im(end)-1),num+1);
end
This does exactly that, and produces:
data.string_new.1.mat
data.string_new.2.mat
data.string_new.3.mat
...
data.string_new.9.mat
data.string_new.10.mat
data.string_new.11.mat
when iterating the above function, starting with 'data.string.mat'