Trying to split string, but I have issues related to encoding - python-3.x

I want to split a string , in two base on the last space in the string.
Example:
full_text = "0.808 um"
value, unit = full_text.rsplit(" ")
This should have worked bu I get the error:
ValueError: not enough values to unpack (expected 2, got 1)
So I printed, on what happens on split:
['0.808\xa0um']
In my example the string is static, but in reality I receive them from a database, and I don't know when a space is space or not.
I want to maintain the encoding for characters (not space) received, but also want to split.

You would simply need to expect more and different kinds of whitespace to split by. In your case you're dealing with a no-break space. The regular expression \s would match it and a few other kinds of whitespace:
>>> import re
>>> re.split(r'\s', '0.808\xa0um')
['0.808', 'um']

Related

How to substitute a repeating character with the same number of a different character in regex python?

Assume there's a string
"An example striiiiiing with other words"
I need to replace the 'i's with '*'s like 'str******ng'. The number of '*' must be same as 'i'. This replacement should happen only if there are consecutive 'i' greater than or equal to 3. If the number of 'i' is less than 3 then there is a different rule for that. I can hard code it:
import re
text = "An example striiiiing with other words"
out_put = re.sub(re.compile(r'i{3}', re.I), r'*'*3, text)
print(out_put)
# An example str***iing with other words
But number of i could be any number greater than 3. How can we do that using regex?
The i{3} pattern only matches iii anywhere in the string. You need i{3,} to match three or more is. However, to make it all work, you need to pass your match into a callable used as a replacement argument to re.sub, where you can get the match text length and multiply correctly.
Also, it is advisable to declare the regex outside of re.sub, or just use a string pattern since patterns are cached.
Here is the code that fixes the issue:
import re
text = "An example striiiiing with other words"
rx = re.compile(r'i{3,}', re.I)
out_put = rx.sub(lambda x: r'*'*len(x.group()), text)
print(out_put)
# => An example str*****ng with other words

How to replace hex value in a string

While importing data from a flat file, I noticed some embedded hex-values in the string (<0x00>, <0x01>).
I want to replace them with specific characters, but am unable to do so. Removing them won't work either.
What it looks like in the exported flat file: https://i.imgur.com/7MQpoMH.png
Another example: https://i.imgur.com/3ZUSGIr.png
This is what I've tried:
(and mind, <0x01> represents a none-editable entity. It's not recognized here.)
import io
with io.open('1.txt', 'r+', encoding="utf-8") as p:
s=p.read()
# included in case it bears any significance
import re
import binascii
s = "Some string with hex: <0x01>"
s = s.encode('latin1').decode('utf-8')
# throws e.g.: >>> UnicodeDecodeError: 'utf-8' codec can't decode byte 0xfc in position 114: invalid start byte
s = re.sub(r'<0x01>', r'.', s)
s = re.sub(r'\\0x01', r'.', s)
s = re.sub(r'\\\\0x01', r'.', s)
s = s.replace('\0x01', '.')
s = s.replace('<0x01>', '.')
s = s.replace('0x01', '.')
or something along these lines in hopes to get a grasp of it while iterating through the whole string:
for x in s:
try:
base64.encodebytes(x)
base64.decodebytes(x)
s.strip(binascii.unhexlify(x))
s.decode('utf-8')
s.encode('latin1').decode('utf-8')
except:
pass
Nothing seems to get the job done.
I'd expect the characters to be replacable with the methods I've dug up, but they are not. What am I missing?
NB: I have to preserve umlauts (äöüÄÖÜ)
-- edit:
Could I introduce the hex-values in the first place when exporting? If so, is there a way to avoid that?
with io.open('out.txt', 'w', encoding="utf-8") as temp:
temp.write(s)
Judging from the images, these are actually control characters.
Your editor displays them in this greyed-out way showing you the value of the bytes using hex notation.
You don't have the characters "0x01" in your data, but really a single byte with the value 1, so unhexlify and friends won't help.
In Python, these characters can be produced in string literals with escape sequences using the notation \xHH, with two hexadecimal digits.
The fragment from the first image is probably equal to the following string:
"sich z\x01 B. irgendeine"
Your attempts to remove them were close.
s = s.replace('\x01', '.') should work.

How to match a part of string before a character into one variable and all after it into another

I have a problem with splitting string into two parts on special character.
For example:
12345#data
or
1234567#data
I have 5-7 characters in first part separated with "#" from second part, where are another data (characters,numbers, doesn't matter what)
I need to store two parts on each side of # in two variables:
x = 12345
y = data
without "#" character.
I was looking for some Lua string function like splitOn("#") or substring until character, but I haven't found that.
Use string.match and captures.
Try this:
s = "12345#data"
a,b = s:match("(.+)#(.+)")
print(a,b)
See this documentation:
First of all, although Lua does not have a split function is its standard library, it does have string.gmatch, which can be used instead of a split function in many cases. Unlike a split function, string.gmatch takes a pattern to match the non-delimiter text, instead of the delimiters themselves
It is easily achievable with the help of a negated character class with string.gmatch:
local example = "12345#data"
for i in string.gmatch(example, "[^#]+") do
print(i)
end
See IDEONE demo
The [^#]+ pattern matches one or more characters other than # (so, it "splits" a string with 1 character).

split string by char

scala has a standard way of splitting a string in StringOps.split
it's behaviour somewhat surprised me though.
To demonstrate, using the quick convenience function
def sp(str: String) = str.split('.').toList
the following expressions all evaluate to true
(sp("") == List("")) //expected
(sp(".") == List()) //I would have expected List("", "")
(sp("a.b") == List("a", "b")) //expected
(sp(".b") == List("", "b")) //expected
(sp("a.") == List("a")) //I would have expected List("a", "")
(sp("..") == List()) // I would have expected List("", "", "")
(sp(".a.") == List("", "a")) // I would have expected List("", "a", "")
so I expected that split would return an array with (the number a separator occurrences) + 1 elements, but that's clearly not the case.
It is almost the above, but remove all trailing empty strings, but that's not true for splitting the empty string.
I'm failing to identify the pattern here. What rules does StringOps.split follow?
For bonus points, is there a good way (without too much copying/string appending) to get the split I'm expecting?
For curious you can find the code here.https://github.com/scala/scala/blob/v2.12.0-M1/src/library/scala/collection/immutable/StringLike.scala
See the split function with the character as an argument(line 206).
I think, the general pattern going on over here is, all the trailing empty splits results are getting ignored.
Except for the first one, for which "if no separator char is found then just send the whole string" logic is getting applied.
I am trying to find if there is any design documentation around these.
Also, if you use string instead of char for separator it will fall back to java regex split. As mentioned by #LRLucena, if you provide the limit parameter with a value more than size, you will get your trailing empty results. see http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String,%20int)
You can use split with a regular expression. I´m not sure, but I guess that the second parameter is the largest size of the resulting array.
def sp(str: String) = str.split("\\.", str.length+1).toList
Seems to be consistent with these three rules:
1) Trailing empty substrings are dropped.
2) An empty substring is considered trailing before it is considered leading, if applicable.
3) First case, with no separators is an exception.
split follows the behaviour of http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
That is split "around" the separator character, with the following exceptions:
Regardless of anything else, splitting the empty string will always give Array("")
Any trailing empty substrings are removed
Surrogate characters only match if the matched character is not part of a surrogate pair.

Splitting a Comma-Separated String in Scala: Missing Trailing Empty Strings?

I have a data file in csv format.
I am trying to split each line using the basic split command line.split(',')
But when I get a string like this "2,,",
instead of returning an array as thus Array(2,"","")
I just get an Array: Array(2).
I am most definitely missing something basic, could someone help point out the correct way to split a comma separated string here?
This is inherited from Java. You can achieve behavior you want by using the split(String regex, int limit) overload:
"2,,".split(",", -1) // = Array(2, "", "")
Note the String instead of Char.
As explained by the Java Docs, the limit parameter is used as follows:
The limit parameter controls the number of times the pattern is
applied and therefore affects the length of the resulting array. If
the limit n is greater than zero then the pattern will be applied at
most n - 1 times, the array's length will be no greater than n, and
the array's last entry will contain all input beyond the last matched
delimiter. If n is non-positive then the pattern will be applied as
many times as possible and the array can have any length. If n is zero
then the pattern will be applied as many times as possible, the array
can have any length, and trailing empty strings will be discarded.
Source
Using split(separator: Char) will call the overload above, using a limit of zero.

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