I'm not sure how to formulate this question, but I'm looking for a magic function that makes this code
for x in magicfunc("H̶e̕l̛l͠o͟ ̨w̡o̷r̀l҉ḑ!͜"):
print(x)
Behave like this:
H̶
e̕
l̛
l͠
o͟
̨
w̡
o̷
r̀
l҉
ḑ
!͜
Basically, is there a built in unicode function or method that takes a string and outputs an array per glyph with all their respective unicode decorators and diacritical marks and such? The same way that a text editor moves the cursor over to the next letter instead of iterating all of the combining characters.
If not, I'll write the function myself, no help needed. Just wondering if it already exists.
You can use unicodedata.combining to find out if a character is combining:
def combine(s: str) -> Iterable[str]:
buf = None
for x in s:
if unicodedata.combining(x) != 0:
# combining character
buf += x
else:
if buf is not None:
yield buf
buf = x
if buf is not None:
yield buf
Result:
>>> for x in combine("H̶e̕l̛l͠o͟ ̨w̡o̷r̀l҉ḑ!͜"):
... print(x)
...
H̶
e̕
l̛
l͠
o͟
̨
w̡
o̷
r̀
l
ḑ
!͜
Issue is that COMBINING CYRILLIC MILLIONS SIGN is not recognized as combining, not sure why. You could also test if COMBINING is in the unicodedata.name(x) for the character, that should solve it.
The 3rd party regex module can search by glyph:
>>> import regex
>>> s="H̶e̕l̛l͠o͟ ̨w̡o̷r̀l҉ḑ!͜"
>>> for x in regex.findall(r'\X',s):
... print(x)
...
H̶
e̕
l̛
l͠
o͟
̨
w̡
o̷
r̀
l҉
ḑ
!͜
This question already has answers here:
Convert "\x" escaped string into readable string in python
(4 answers)
Closed 7 months ago.
I have this str (coming from a file I can't fix):
In [131]: s
Out[131]: '\\xce\\xb8Oph'
This is close to the repr of a string encoded in utf8:
In [132]: repr('θOph'.encode('utf8'))
Out[132]: "b'\\xce\\xb8Oph'"
I need the original encoded string. I can do it with
In [133]: eval("b'{}'".format(s)).decode('utf8')
Out[133]: 'θOph'
But I would be ... sad? if there were no simpler option to get it. Is there a better way?
Your solution is OK, the only thing is that eval is dangerous when used with arbitrary inputs. The safe alternative is to use ast.literal_eval:
>>> s = '\\xce\\xb8Oph'
>>> from ast import literal_eval
>>> literal_eval("b'{}'".format(s)).decode('utf8')
'\u03b8Oph'
With eval you are subject to:
>>> eval("b'{}'".format("1' and print('rm -rf /') or b'u r owned")).decode('utf8')
rm -rf /
'u r owned'
Since ast.literal_eval is the opposite of repr for literals, I guess it is what you are looking for.
[updade]
If you have a file with escaped unicode, you may want to open it with the unicode_escape encoding as suggested in the answer by Ginger++. I will keep my answer because the question was "how to convert repr into encoded string", not "how to decode file with escaped unicode".
Just open your file with unicode_escape encoding, like:
with open('name', encoding="unicode_escape") as f:
pass # your code here
Original answer:
>>> '\\xce\\xb8Oph'.encode('utf-8').decode('unicode_escape')
'θOph'
You can get rid of that encoding to UTF-8, if you read your file in binary mode instead of text mode:
>>> b'\\xce\\xb8Oph'.decode('unicode_escape')
'θOph'
Unfortunately, this is really problematic. It's \ killing you softly here.
I can only think of:
s = '\\xce\\xb8Oph\\r\\nMore test\\t\\xc5\\xa1'
n = ""
x = 0
while x!=len(s):
if s[x]=="\\":
sx = s[x+1:x+4]
marker = sx[0:1]
if marker=="x": n += chr(int(sx[1:], 16)); x += 4
elif marker in ("'", '"', "\\", "n", "r", "v", "t", "0"):
# Pull this dict out of a loop to speed things up
n += {"'": "'", '"': '"', "\\": "\\", "n": "\n", "r": "\r", "t": "\t", "v": "\v", "0": "\0"}[marker]
x += 2
else: n += s[x]; x += 1
else: n += s[x]; x += 1
print repr(n), repr(s)
print repr(n.decode("UTF-8"))
There might be some other trick to pull this off, but at the moment this is all I got.
To make a teeny improvement on GingerPlusPlus's answer:
import tempfile
with tempfile.TemporaryFile(mode='rb+') as f:
f.write(r'\xce\xb8Oph'.encode())
f.flush()
f.seek(0)
print(f.read().decode('unicode_escape').encode('latin1').decode())
If you open the file in binary mode (i.e. rb, since you're reading, I added + since I was also writing to the file) you can skip the first encode call. It's still awkward, because you have to bounce through the decode/encode hop, but you at least do get to avoid that first encoding call.
I open my file like so :
f = open("filename.ext", "rb") # ensure binary reading with b
My first line of data looks like this (when using f.readline()):
'\x04\x00\x00\x00\x12\x00\x00\x00\x04\x00\x00\x00\xb4\x00\x00\x00\x01\x00\x00\x00\x08\x00\x00\x00\x00\x00\x00\x00\x18\x00\x00\x00\x01\x00\x00\x00\x02\x00\x00\x00\x03\x00\x00\x00\x04\x00\x00\x00\x05\x00\x00\x00\x06\x00\x00\x00:\x00\x00\x00;\x00\x00\x00<\x00\x00\x007\x00\x00\x008\x00\x00\x009\x00\x00\x00\x07\x00\x00\x00\x08\x00\x00\x00\t\x00\x00\x00\n'
Thing is, I want to read this data byte by byte (f.read(4)). While debugging, I realized that when it gets to the end of the first line, it still takes in the newline character \n and it is used as the first byte of the following int I read. I don't want to simply use .splitlines()because some data could have an n inside and I don't want to corrupt it. I'm using Python 2.7.10, by the way. I also read that opening a binary file with the b parameter "takes care" of the new line/end of line characters; why is not the case with me?
This is what happens in the console as the file's position is right before the newline character:
>>> d = f.read(4)
>>> d
'\n\x00\x00\x00'
>>> s = struct.unpack("i", d)
>>> s
(10,)
(Followed from discussion with OP in chat)
Seems like the file is in binary format and the newlines are just mis-interpreted values. This can happen when writing 10 to the file for example.
This doesn't mean that newline was intended, and it is probably not. You can just ignore it being printed as \n and just use it as data.
You should just be able to replace the bytes that indicate it is a newline.
>>> d = f.read(4).replace(b'\x0d\x0a', b'') #\r\n should be bytes b'\x0d\x0a'
>>> diff = 4 - len(d)
>>> while diff > 0: # You can probably make this more sophisticated
... d += f.read(diff).replace(b'\x0d\x0a', b'') #\r\n should be bytes b'\x0d\x0a'
... diff = 4 - len(d)
>>>
>>> s = struct.unpack("i", d)
This should give you an idea of how it will work. This approach could mess with your data's byte alignment.
If you really are seeing "\n" in your print of d then try .replace(b"\n", b"")
When reading a file (UTF-8 Unicode text, csv) with Python on Linux, either with:
csv.reader()
file()
values of some columns get a zero as their first characeter (there are no zeroues in input), other get a few zeroes, which are not seen when viewing file with Geany or any other editor. For example:
Input
10016;9167DE1;Tom;Sawyer ;Street 22;2610;Wil;;378983561;tom#hotmail.com;1979-08-10 00:00:00.000;0;1;Wil;081208608;NULL;2;IZMH726;2010-08-30 15:02:55.777;2013-06-24 08:17:22.763;0;1;1;1;NULL
Output
10016;9167DE1;Tom;Sawyer ;Street 22;2610;Wil;;0378983561;tom#hotmail.com;1979-08-10 00:00:00.000;0;1;Wil;081208608;NULL;2;IZMH726;2010-08-30 15:02:55.777;2013-06-24 08:17:22.763;0;1;1;1;NULL
See 378983561 > 0378983561
Reading with:
f = file('/home/foo/data.csv', 'r')
data = f.read()
split_data = data.splitlines()
lines = list(line.split(';') for line in split_data)
print data[51220][8]
>>> '0378983561' #should have been '478983561' (reads like this in Geany etc.)
Same result with csv.reader().
Help me solve the mystery, what could be the cause of this? Could it be related to encoding/decoding?
The data you're getting is a string.
print data[51220][8]
>>> '0478983561'
If you want to use this as an integer, you should parse it.
print int(data[51220][8])
>>> 478983561
If you want this as a string, you should convert it back to a string.
print repr(int(data[51220][8]))
>>> '478983561'
csv.reader treats all columns as strings. Conversion to the appropriate type is up to you as in:
print int(data[51220][8])
I'm trying to convert characters in one list into characters in another list at the same index in Japanese (zenkaku to hangaku moji, for those interested), and I can't get the comparison to work. I am decoding into utf-8 before I compare (decoding into ascii broke the program), but the comparison doesn't ever return true. Does anyone know what I'm doing wrong? Here's the code (indents are a little wacky due to SO's editor):
#!C:\Python27\python.exe
# coding=utf-8
import os
import shutil
import sys
zk = [
'。',
'、',
'「',
'」',
'（',
'）',
'！',
'？',
'・',
'／',
'ア','イ','ウ','エ','オ',
'カ','キ','ク','ケ','コ',
'サ','シ','ス','セ','ソ',
'ザ','ジ','ズ','ゼ','ゾ',
'タ','チ','ツ','テ','ト',
'ダ','ヂ','ヅ','デ','ド',
'ラ','リ','ル','レ','ロ',
'マ','ミ','ム','メ','モ',
'ナ','ニ','ヌ','ネ','ノ',
'ハ','ヒ','フ','ヘ','ホ',
'バ','ビ','ブ','ベ','ボ',
'パ','ピ','プ','ペ','ポ',
'ヤ','ユ','ヨ','ヲ','ン','ッ'
]
hk = [
'｡',
'､',
'｢',
'｣',
'(',
')',
'!',
'?',
'･',
'/',
'ｱ','ｲ','ｳ','ｴ','ｵ',
'ｶ','ｷ','ｸ','ｹ','ｺ',
'ｻ','ｼ','ｽ','ｾ','ｿ',
'ｻﾞ','ｼﾞ','ｽﾞ','ｾﾞ','ｿﾞ',
'ﾀ','ﾁ','ﾂ','ﾃ','ﾄ',
'ﾀﾞ','ﾁﾞ','ﾂﾞ','ﾃﾞ','ﾄﾞ',
'ﾗ','ﾘ','ﾙ','ﾚ','ﾛ',
'ﾏ','ﾐ','ﾑ','ﾒ','ﾓ',
'ﾅ','ﾆ','ﾇ','ﾈ','ﾉ',
'ﾊ','ﾋ','ﾌ','ﾍ','ﾎ',
'ﾊﾞ','ﾋﾞ','ﾌﾞ','ﾍﾞ','ﾎﾞ',
'ﾊﾟ','ﾋﾟ','ﾌﾟ','ﾍﾟ','ﾎﾟ',
'ﾔ','ﾕ','ﾖ','ｦ','ﾝ','ｯ'
]
def main():
if len(sys.argv) > 1:
filename = sys.argv[1]
else:
print("Please specify a file to check.")
return
try:
f = open(filename, 'r')
except IOError as e:
print("Sorry! The file doesn't exist.")
return
filecontent = f.read()
f.close()
#y = zk[29]
#print y.decode('utf-8')
for f in filecontent:
for z in zk:
if f == z.decode('utf-8'):
print f
print filename
if __name__ == "__main__":
main()
Am I missing a step?
Several.
zk = [
u'。',
u'、',
u'「',
...
...
f = codecs.open(filename, 'r', encoding='utf-8')
...
I'll let you work out the rest now that the hard work's been done.
Make sure that zk and hk lists contain Unicode strings. Either use Unicode literals e.g., u'a' or decode them at runtime:
fromutf8 = lambda s: s.decode('utf-8') if not isinstance(s, unicode) else s
zk = map(fromutf8, zk)
hk = map(fromutf8, hk)
You could use unicode.translate() to convert characters in one list into characters in another list at the same index:
import codecs
translation_table = dict(zip(map(ord,zk), hk))
with codecs.open(sys.argv[1], encoding='utf-8') as f:
for line in f:
print line.translate(translation_table),
You need to convert everything to the same form, and the form is Unicode strings. Unicode strings have no encoding in the sense .encode() or .decode(). When having a non-unicode string, it is actually a stream of bytes that expresses the value in some encoding. When converting to Unicode, you have to .decode(). When storing Unicode string to a sequence of bytes, you have to .encode() the abstraction to concrete bytes.
This way, when loading Unicode strings from an UTF-8 encoded file, or you have to read it into the old strings (non Unicode, sequences of bytes) and then .decode('utf-8'), or you can use `codecs.open(..., encoding='utf-8') -- then you get Unicode strings automatically.
The form # coding=utf-8 is not the usual, but it is OK... if the editor (I mean the tool that you use to write the text) also thinks this way. Then the old strings are displayed by the editor correctly. In the case they should be .decode('utf-8')d to get Unicode. Old strings with ASCII characters only in the same source can also be converted to Unicode using the .decode('utf-8').
To summarize: you are de coding from bytes to Unicode, and you are en coding the Unicode strings into sequence of bytes. It seems from the question that you are doing the opposite.
The following is completely wrong:
for f in filecontent:
for z in zk:
if f == z.decode('utf-8'):
print f
because the filecontent is the result of f.read(). This way it is a sequence of bytes. The f in the loop is one byte. The z.decode('utf-8') returns one Unicode character. They cannot be compared. (By the way, the f is a kind of misleading name for a byte value.)