I'm trying read multiple text files, doing word segmentation (use jieba) and then save the results to CSV files respectively. It shows
AttributeError: '_io.TextIOWrapper' object has no attribute 'decode'
Thanks for anyone's help.
The python code is:
import jieba
import csv
import glob
list_of_files = glob.glob('C:/Users/user/Desktop/speech./*.txt')
for file_name in list_of_files:
FI = open(file_name, 'r')
FO = open(file_name, 'w')
seglist = jieba.cut(FI, cut_all=False)
w = csv.writer(FO)
w.writerows(seglist)
FI.close()
FO.close()
It seems that you need to send bytes to cut and not a file object
try this code instead:
list_of_files = glob.glob('C:/Users/user/Desktop/speech./*.txt')
for file_name in list_of_files:
with open(file_name, 'rb') as f:
text = f.read()
seglist = jieba.cut(text, cut_all=False)
with open(file_name, 'w') as f:
w = csv.writer(f)
w.writerows(seglist)
From what I read from source code as example and jieba.cut definition it seems jieba.cut needs string as parameter.
But you are giving an instance of file.
seglist = jieba.cut(FI.read(), cut_all=False)
Fixed the issue from what I saw. (FI.read() is the fix).
Btw, do not call variables like FI / FO, this is valid name for constants or maybe classes, but not variable.
Explicit is better than implicit: prefere something like: file_output & file_input.
Related
My tkinker Python script is not working, it says the error ""tuple" object has no attribute "read"", how could I fix this?
My code is here: https://pastebin.com/kLnmutcg
My theory why it says this error is in this code:
def save():
filename = filedialog.askopenfilename(filetypes=(("txt files","*.txt"),("All files","*.*")))
file = open=(filename, "wt")
def open():
filename = filedialog.askopenfilename(filetypes=(("txt files","*.txt"),("All files","*.*")))
file = open=(filename, "rt")
read = file.read()
text_box.insert(tk.END, read)
I'm trying to make a notepad clone of sorts.
The correct syntax is:
file = open(filename, "wt")
By using file = open=(filename, "wt") you create two variables: file and open, that both contain a tuple (filename, "wt")
Also do not use open as variable/function name, this is a python builtin. You can find a list of python builtins in the documentation.
I'm working with TSV file (here below my_file) and trying to write it down to another temp file with a random ID (here below my_temp_file)and this is what I wrote:
def temp_generator():
while True:
my_string = 'tmp' + str(random.randint(1,1000000000))
if not os.path.exists(my_string):
return my_string
randomID = temp_generator()
my_temp_file = open('mytemp_'+randomID + '.tsv', 'w')
with open(my_file, 'r+') as mf:
for line in mf:
my_temp_file.write(line)
my_temp_file.close()
mf.close()
The output is something like:
mytemp_1283189.tsv
Now I'd like to work with my_temp_file.tsv in order to modify its content and rename it but if I try to open it with:
with open (my_temp_file.tsv, 'r') as mtf:
data = mtf.read()
print(data)
This is what I obtain:
TypeError: expected str, bytes or os.PathLike object, not _io.TextIOWrapper
What can I do?
Issue
The pattern handle = open(path)
is opening a file at path from path and returns the handle assigned to handle. You can use handle to .write, .read, or .close. But you can not open it again or use it as input to open - which expects a Path-like object, e.g. a filename.
Fixed
def temp_generator():
while True:
my_string = 'tmp' + str(random.randint(1,1000000000))
if not os.path.exists(my_string):
return my_string
randomID = temp_generator()
# copy from input (my_file) to output, a random temp file (my_temp_file)
my_temp_file = 'mytemp_' + randomID + '.tsv'
with open(my_temp_file, 'w') as mtf:
with open(my_file, 'r+') as mf: # my_file is supposed to be a Path-like object
for line in mf:
mtf.write(line)
# since with..open used, no close needed (auto-close!)
# modify output (content and rename the file)
# remember: my_temp_file is holding a Path or filename
with open(my_temp_file, 'r') as mtf: # open the file again
data = mtf.read()
print(data)
See also:
[Solved] Python TypeError: expected str, bytes or os.PathLike object, not _io.TextIOWrapper
Python documentation about TextIOWrapper: io — Core tools for working with streams
Your error is here, assuming you are actually using open(my_temp_file
my_temp_file = open('mytemp_'+randomID + '.tsv', 'w')
with open(my_file, 'r+') as mf:
You've already opened the file, so you shouldn't open it again using the file handle as the parameter. You should prefer only using the with way of opening files, too
For example
my_temp_file = 'mytemp_'+randomID + '.tsv'
with open(my_temp_file, 'r+') as mf:
Even then, if you're going to eventually rename the file, just make it the name you want from the start
I am trying to merge all text files in a folder. I have this part working, but when I try to append the file name before the contents of each text file, I'm getting a error that reads: TypeError: a bytes-like object is required, not 'str'
The code below must be pretty close, but something is definitely off. Any thoughts what could be wrong?
import glob
folder = 'C:\\my_path\\'
read_files = glob.glob(folder + "*.txt")
with open(folder + "final_result.txt", "wb") as outfile:
for f in read_files:
with open(f, "rb") as infile:
outfile.write(f)
outfile.write(infile.read())
outfile.close
outfile.write(f) seems to be your problem because you opened the file with in binary mode with 'wb'. You can convert to bytes using encode You'll likely not want to close outfile in your last line either (although you aren't calling the function anyway). So something like this might work for you:
import glob
folder = 'C:\\my_path\\'
read_files = glob.glob(folder + "*.txt")
with open(folder + "final_result.txt", "wb") as outfile:
for f in read_files:
with open(f, "rb") as infile:
outfile.write(f.encode('utf-8'))
outfile.write(infile.read())
I am using a script to strip exif data from uploaded JPGs in Python, before writing them to disk. I'm using Flask, and the file is brought in through requests
file = request.files['file']
strip the exif data, and then save it
f = open(file)
image = f.read()
f.close()
outputimage = stripExif(image)
f = ('output.jpg', 'w')
f.write(outputimage)
f.close()
f.save(os.path.join(app.config['IMAGE_FOLDER'], filename))
Open isn't working because it only takes a string as an argument, and if I try to just set f=file, it throws an error about tuple objects not having a write attribute. How can I pass the current file into this function before it is read?
file is a FileStorage, described in http://werkzeug.pocoo.org/docs/datastructures/#werkzeug.datastructures.FileStorage
As the doc says, stream represents the stream of data for this file, usually under the form of a pointer to a temporary file, and most function are proxied.
You probably can do something like:
file = request.files['file']
image = file.read()
outputimage = stripExif(image)
f = open(os.path.join(app.config['IMAGE_FOLDER'], 'output.jpg'), 'w')
f.write(outputimage)
f.close()
Try the io package, which has a BufferedReader(), ala:
import io
f = io.BufferedReader(request.files['file'])
...
file = request.files['file']
image = stripExif(file.read())
file.close()
filename = 'whatever' # maybe you want to use request.files['file'].filename
dest_path = os.path.join(app.config['IMAGE_FOLDER'], filename)
with open(dest_path, 'wb') as f:
f.write(image)
I have the following code:
import re
#open the xml file for reading:
file = open('path/test.xml','r+')
#convert to string:
data = file.read()
file.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>",data))
file.close()
where I'd like to replace the old content that's in the file with the new content. However, when I execute my code, the file "test.xml" is appended, i.e. I have the old content follwed by the new "replaced" content. What can I do in order to delete the old stuff and only keep the new?
You need seek to the beginning of the file before writing and then use file.truncate() if you want to do inplace replace:
import re
myfile = "path/test.xml"
with open(myfile, "r+") as f:
data = f.read()
f.seek(0)
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>", r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", data))
f.truncate()
The other way is to read the file then open it again with open(myfile, 'w'):
with open(myfile, "r") as f:
data = f.read()
with open(myfile, "w") as f:
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>", r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", data))
Neither truncate nor open(..., 'w') will change the inode number of the file (I tested twice, once with Ubuntu 12.04 NFS and once with ext4).
By the way, this is not really related to Python. The interpreter calls the corresponding low level API. The method truncate() works the same in the C programming language: See http://man7.org/linux/man-pages/man2/truncate.2.html
file='path/test.xml'
with open(file, 'w') as filetowrite:
filetowrite.write('new content')
Open the file in 'w' mode, you will be able to replace its current text save the file with new contents.
Using truncate(), the solution could be
import re
#open the xml file for reading:
with open('path/test.xml','r+') as f:
#convert to string:
data = f.read()
f.seek(0)
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>",data))
f.truncate()
import os#must import this library
if os.path.exists('TwitterDB.csv'):
os.remove('TwitterDB.csv') #this deletes the file
else:
print("The file does not exist")#add this to prevent errors
I had a similar problem, and instead of overwriting my existing file using the different 'modes', I just deleted the file before using it again, so that it would be as if I was appending to a new file on each run of my code.
See from How to Replace String in File works in a simple way and is an answer that works with replace
fin = open("data.txt", "rt")
fout = open("out.txt", "wt")
for line in fin:
fout.write(line.replace('pyton', 'python'))
fin.close()
fout.close()
in my case the following code did the trick
with open("output.json", "w+") as outfile: #using w+ mode to create file if it not exists. and overwrite the existing content
json.dump(result_plot, outfile)
Using python3 pathlib library:
import re
from pathlib import Path
import shutil
shutil.copy2("/tmp/test.xml", "/tmp/test.xml.bak") # create backup
filepath = Path("/tmp/test.xml")
content = filepath.read_text()
filepath.write_text(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", content))
Similar method using different approach to backups:
from pathlib import Path
filepath = Path("/tmp/test.xml")
filepath.rename(filepath.with_suffix('.bak')) # different approach to backups
content = filepath.read_text()
filepath.write_text(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", content))