I'm trying to increment the video file names every time they get into my folder. I tried the + and the join() method but I can't seem to figure it out. I tried integers without quotation marks but the join method wont let me use an integer so I tried with quotation marks but now it won't increment
Here is my code
VideoNumber += "99"
folderLocation = ("C:/Users/someone/Documents", VideoNumber, ".mp4")
x = "/".join(folderLocation)
print(x)
You can format integers into a string using an f-string or the format() method on strings.
video_number += 99
video_path = f"C:/Users/someone/Documents/{video_number}.mp4"
print(video_path)
Just as an example of how to make your original code work, you could keep your number as an integer and then convert it to a string using str() (though note this has a bug because you will have an extra / between the number and .mp4).
VideoNumber += 99
folderLocation = ("C:/Users/someone/Documents", str(VideoNumber), ".mp4")
x = "/".join(folderLocation)
print(x)
You can cast the integer into string, so your code will be like this
folderLocation = ("C:/Users/someone/Documents", str(VideoNumber), ".mp4")
Related
Problem solved! was newfilename[0,3] instead of newfilename[0: 3]
I know this question has been asked before and I have look around on all the answers and the types of problems people have been having related to this error message, but was unable to find anyone with the same type of problem.
I am sowing the whole method just in case. So here is my problem;
When I am trying to get is a substring of "newfilename" using newfilename[int, int] and the compiler keeps thinking I don't have an integer there when I do, at least from my checking I do.
What I'm doing with this code: I am cutting of the end of a filename such as 'foo.txt' to get 'foo' that is saved as newfilename. Then I am adding the number (converted to a string) to the end of it to get 'foo 1' and after that adding back the '.txt' to get the final result of 'foo 1.txt'. The problem occurs when I try to get the substring out and delete the last four characters of the filename to get just 'foo'. After that, I do another check to see if there is a file like that still in the folder and if so I do another set of cutting and pasting to add 1 to the previous file. To be honest, I have not tested of the while loop will work I just thought it should work technically, but my code does not reach that far because of this error lol.
My error:
File "C:/Users/Reaper/IdeaProjects/Curch Rec Managment/Setup.py", line 243, in moveFiles
print(newfilename[0, 3])
TypeError: string indices must be integers
NOTE this error is from when I tried to hard code the numbers it to see if it would work
Here is the current error with the hard code commented out:
newfilename = newfilename[0, int(newfilename.__len__() - 4)] + " 1.m4a"
TypeError: string indices must be integers
What I have tried: I have tried hard coding the numbers is by literally typing in newfilename[0, 7] and still got the same error. I have tried doing this in a separate python file and it seems to work there fine. Also, what is really confusing me is that it works in another part of my program just fine as shown here:
nyear = str(input("Enter new Year: "))
if nyear[0:2] != "20" or nyear.__len__() > 4:
print("Sorry incorrect year. Please try again")
So I have been at it for a while now trying to figure out what in the world is going on and can't get there. Decided I would sleep on it but would post the question just in case. If someone could point out what may be wrong that would be awesome! Or tell me the compilers are just being stupid, well I guess that will do as well.
My function code
def moveFiles(pathList, source, filenameList):
# moves files to new location
# counter keeps track of file name position in list
cnter = 0
for x in pathList:
filename = filenameList[cnter]
#print(x + "\\" + filename)
# new filename
if filename.find("PR") == 0:
newfilename = filename[3:filename.__len__()]
else:
newfilename = filename[2:filename.__len__()]
# checking if file exists and adding numbers to the end if it does
if os.path.isfile(x + "\\" + newfilename):
print("File Name exists!!")
# adding a 1 to the end
print(newfilename)
# PROBLEM ON NEXT TWO LINES, also prob. on any line with the following calls
print(newfilename[0, 3])
newfilename = newfilename[0, int(newfilename.__len__() - 4)] + " 1.m4a"
print("Adding 1:", newfilename)
# once again check if the file exists and adding 1 to the last number
while os.path.isfile(x + "\\" + newfilename):
# me testing if maybe i just can't have math operations withing the substring call
print("File exists again!!")
num = newfilename.__len__() - 6
num2 = newfilename.__len__() - 4
num3 = int(newfilename[num, num2])
num = newfilename.__len__() - 5
newfilename = newfilename[0, num] + str(num3 + 1)
print("Adding 1:", newfilename)
# moving file and deleting prefix
if not os.path.isdir(x):
os.makedirs(x)
os.rename(source + "\\" + filename, x + "\\" + newfilename)
cnter += 1
I think you need this:
print(newfilename[0:3])
I have hex variable that I want to print as hex
data = '\x99\x02'
print (data)
Result is: ™
I want to the python to print 0x9902
Thank you for your help
Please check this one.
data = r'\x99\x02'
a, b = [ x for x in data.split(r'\x') if x]
d = int(a+b, base=16)
print('%#x'%d)
You have to convert every char to its number - ord(char) - and convert every number to hex value - '{:02x}'.format() - and concatenate these values to string. And add string '0x'.
data = '\x99\x02'
print('0x' + ''.join('{:02x}'.format(ord(char)) for char in data))
EDIT: The same but first string is converted to bytes using encode('raw_unicode_escape')
data = '\x99\x02'
print('0x' + ''.join('{:02x}'.format(code) for code in data.encode('raw_unicode_escape')))
and if you have already bytes then you don't have to encode()
data = b'\x99\x02'
print('0x' + ''.join('{:02x}'.format(code) for code in data))
BTW: Similar way you can convert to binary using {:08b}
data = '\x99\x02'
print(''.join('{:08b}'.format(code) for code in data.encode('raw_unicode_escape')))
I wrote a function to see encrypt a given text file. The below code is a small portion of the function.
#pad it before encrypting it
elif len(chunk) % 16 != 0:
chunk += ' ' * (16 - len(chunk) % 16)
#write encrypted data into output file
out_file.write(encryptor.encrypt(chunk))
Whenever I try to use the function I get an error that points to the last line saying
"TyprError: can't concat str to bytes". I'm not sure what I need to do in order to fix this error. I've tried a few things and they end leading me into more similar errors. Any guidance would be greatly appreciated.
The encryptor is below.
encryptor = PKCS1_OAEP.new(pub_key)
Your encryption method encryptor.encrypt() very likely accepts bytes as argument, not str. It also returns bytes very likely. So I suggest you to use the encode/decode methods as follows (example of utf-8 coding):
out_file.write(encryptor.encrypt(chunk.encode('utf-8')).decode('utf-8'))
You are trying to mix and match incompatible data types. Here is an example that will throw the same error:
str1 = bytes(123)
str2 = str(123)
str1 + str2
Go through your example and find where you are trying to concat a bytes value to a str value and make them match types.
I am reading a large amount of data from an excel spreadsheet in which I read (and reformat and rewrite) from the spreadsheet using the following general structure:
book = open_workbook('file.xls')
sheettwo = book.sheet_by_index(1)
out = open('output.file', 'w')
for i in range(sheettwo.nrows):
z = i + 1
toprint = """formatting of the data im writing. important stuff is to the right -> """ + str(sheettwo.cell(z,y).value) + """ more formatting! """ + str(sheettwo.cell(z,x).value.encode('utf-8')) + """ and done"""
out.write(toprint)
out.write("\n")
where x and y are arbitrary cells in this case, with x being less arbitrary and containing utf-8 characters
So far I have only been using the .encode('utf-8') in cells where I know there will be errors otherwise or foresee an error without using utf-8.
My question is basically this: is there a disadvantage to using .encode('utf-8') on all of the cells even if it is unnecessary? Efficiency is not an issue. the main issue is that it works even if there is a utf-8 character in a place there shouldn't be. If no errors would occur if I just lump the ".encode('utf-8')" onto every cell read, I will probably end up doing that.
The XLRD Documentation states it clearly: "From Excel 97 onwards, text in Excel spreadsheets has been stored as Unicode.". Since you are likely reading in files newer than 97, they are containing Unicode codepoints anyway. It is therefore necessary that keep the content of these cells as Unicode within Python and do not convert them to ASCII (which you do in with the str() function). Use this code below:
book = open_workbook('file.xls')
sheettwo = book.sheet_by_index(1)
#Make sure your writing Unicode encoded in UTF-8
out = open('output.file', 'w')
for i in range(sheettwo.nrows):
z = i + 1
toprint = u"formatting of the data im writing. important stuff is to the right -> " + unicode(sheettwo.cell(z,y).value) + u" more formatting! " + unicode(sheettwo.cell(z,x).value) + u" and done\n"
out.write(toprint.encode('UTF-8'))
This answer is really a few mild comments on the accepted answer, but they need better formatting than the SO comment facility provides.
(1) Avoiding the SO horizontal scrollbar enhances the chance that people will read your code. Try wrapping your lines, for example:
toprint = u"".join([
u"formatting of the data im writing. "
u"important stuff is to the right -> ",
unicode(sheettwo.cell(z,y).value),
u" more formatting! ",
unicode(sheettwo.cell(z,x).value),
u" and done\n"
])
out.write(toprint.encode('UTF-8'))
(2) Presumably you are using unicode() to convert floats and ints to unicode; it does nothing for values that are already unicode. Be aware that unicode(), like str(), gives you only 12 digits of precision for floats:
>>> unicode(123456.78901234567)
u'123456.789012'
If that is a bother, you might like to try something like this:
>>> def full_precision(x):
>>> ... return unicode(repr(x) if isinstance(x, float) else x)
>>> ...
>>> full_precision(u'\u0400')
u'\u0400'
>>> full_precision(1234)
u'1234'
>>> full_precision(123456.78901234567)
u'123456.78901234567'
(3) xlrd builds Cell objects on the fly when demanded.
sheettwo.cell(z,y).value # slower
sheettwo.cell_value(z,y) # faster
I have done this operation millions of times, just using the + operator! I have no idea why it is not working this time, it is overwriting the first part of the string with the new one! I have a list of strings and just want to concatenate them in one single string! If I run the program from Eclipse it works, from the command-line it doesn't!
The list is:
["UNH+1+XYZ:08:2:1A+%CONVID%'&\r", "ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&\r", "DUM'&\r"]
I want to discard the first and the last elements, the code is:
ediMsg = ""
count = 1
print "extract_the_info, lineList ",lineList
print "extract_the_info, len(lineList) ",len(lineList)
while (count < (len(lineList)-1)):
temp = ""
# ediMsg = ediMsg+str(lineList[count])
# print "Count "+str(count)+" ediMsg ",ediMsg
print "line value : ",lineList[count]
temp = lineList[count]
ediMsg += " "+temp
print "ediMsg : ",ediMsg
count += 1
print "count ",count
Look at the output:
extract_the_info, lineList ["UNH+1+XYZ:08:2:1A+%CONVID%'&\r", "ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&\r", "DUM'&\r"]
extract_the_info, len(lineList) 8
line value : ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&
ediMsg : ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&
count 2
line value : DUM'&
DUM'& : ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&
count 3
Why is it doing so!?
While the two answers are correct (use " ".join()), your problem (besides very ugly python code) is this:
Your strings end in "\r", which is a carriage return. Everything is fine, but when you print to the console, "\r" will make printing continue from the start of the same line, hence overwrite what was written on that line so far.
You should use the following and forget about this nightmare:
''.join(list_of_strings)
The problem is not with the concatenation of the strings (although that could use some cleaning up), but in your printing. The \r in your string has a special meaning and will overwrite previously printed strings.
Use repr(), as such:
...
print "line value : ", repr(lineList[count])
temp = lineList[count]
ediMsg += " "+temp
print "ediMsg : ", repr(ediMsg)
...
to print out your result, that will make sure any special characters doesn't mess up the output.
'\r' is the carriage return character. When you're printing out a string, a '\r' will cause the next characters to go at the start of the line.
Change this:
print "ediMsg : ",ediMsg
to:
print "ediMsg : ",repr(ediMsg)
and you will see the embedded \r values.
And while your code works, please change it to the one-liner:
ediMsg = ' '.join(lineList[1:-1])
Your problem is printing, and it is not string manipulation. Try using '\n' as last char instead of '\r' in each string in:
lineList = [
"UNH+1+TCCARQ:08:2:1A+%CONVID%'&\r",
"ORG+1A+77499505:PARAF0103+++A+FR:EUR++11730788+1A'&\r",
"DUM'&\r",
"FPT+CC::::::::N'&\r",
"CCD+CA:5132839000000027:0450'&\r",
"CPY+++AF'&\r",
"MON+712:1.00:EUR'&\r",
"UNT+8+1'\r"
]
I just gave it a quick look. It seems your problem arises when you are printing the text. I haven't done such things for a long time, but probably you only get the last line when you print. If you check the actual variable, I'm sure you'll find that the value is correct.
By last line, I'm talking about the \r you got in the text strings.