So say I have a string with some underscores like hi_there.
Is there a way to auto-convert that string into "hi there"?
(the original string, by the way, is a variable name that I'm converting into a plot title).
Surprising that no-one has yet mentioned strrep:
>> strrep('string_with_underscores', '_', ' ')
ans =
string with underscores
which should be the official way to do a simple string replacements. For such a simple case, regexprep is overkill: yes, they are Swiss-knifes that can do everything possible, but they come with a long manual. String indexing shown by AndreasH only works for replacing single characters, it cannot do this:
>> s = 'string*-*with*-*funny*-*separators';
>> strrep(s, '*-*', ' ')
ans =
string with funny separators
>> s(s=='*-*') = ' '
Error using ==
Matrix dimensions must agree.
As a bonus, it also works for cell-arrays with strings:
>> strrep({'This_is_a','cell_array_with','strings_with','underscores'},'_',' ')
ans =
'This is a' 'cell array with' 'strings with' 'underscores'
Try this Matlab code for a string variable 's'
s(s=='_') = ' ';
If you ever have to do anything more complicated, say doing a replacement of multiple variable length strings,
s(s == '_') = ' ' will be a huge pain. If your replacement needs ever get more complicated consider using regexprep:
>> regexprep({'hi_there', 'hey_there'}, '_', ' ')
ans =
'hi there' 'hey there'
That being said, in your case #AndreasH.'s solution is the most appropriate and regexprep is overkill.
A more interesting question is why you are passing variables around as strings?
regexprep() may be what you're looking for and is a handy function in general.
regexprep('hi_there','_',' ')
Will take the first argument string, and replace instances of the second argument with the third. In this case it replaces all underscores with a space.
In Matlab strings are vectors, so performing simple string manipulations can be achieved using standard operators e.g. replacing _ with whitespace.
text = 'variable_name';
text(text=='_') = ' '; //replace all occurrences of underscore with whitespace
=> text = variable name
I know this was already answered, however, in my case I was looking for a way to correct plot titles so that I could include a filename (which could have underscores). So, I wanted to print them with the underscores NOT displaying with as subscripts. So, using this great info above, and rather than a space, I escaped the subscript in the substitution.
For example:
% Have the user select a file:
[infile inpath]=uigetfile('*.txt','Get some text file');
figure
% this is a problem for filenames with underscores
title(infile)
% this correctly displays filenames with underscores
title(strrep(infile,'_','\_'))
Related
I have a text file. I would like to remove all decimal points and their trailing numbers, unless text is preceding.
e.g 12.29,14.6,8967.334 should be replaced with 12,14,8967
e.g happypants2.3#email.com should not be modified.
My code is:
import re
txt1 = "9.9,8.8,22.2,88.7,morris1.43#email.com,chat22.3#email.com,123.6,6.54"
txt1 = re.sub(r',\d+[.]\d+', r'\d+',txt1)
print(txt1)
unless there is an easier way of completing this, how do I modify r'\d+' so it just returns the number without a decimal place?
You need to make use of groups in your regex. You put the digits before the '.' into parentheses, and then you can use '\1' to refer to them later:
txt1 = re.sub(r',(\d+)[.]\d+', r',\1',txt1)
Note that in your attempted replacement code you forgot to replace the comma, so your numbers would have been glommed together. This still isn't perfect though; the first number, since it doesn't begin with a comma, isn't processed.
Instead of checking for a comma, the better way is to check word boundaries, which can be done using \b. So the solution is:
import re
txt1 = "9.9,8.8,22.2,88.7,morris1.43#email.com,chat22.3#email.com,123.6,6.54"
txt1 = re.sub(r'\b(\d+)[.]\d+\b', r'\1',txt1)
print(txt1)
Considering these are the only two types of string that is present in your file, you can explicitly check for these conditions.
This may not be an efficient way, but what I have done is split the str and check if the string contains #email.com. If thats true, I am just appending to a new list. For your 1st condition to satisfy, we can convert the str to int which will eliminate the decimal points.
If you want everything back to a str variable, you can use .join().
Code:
txt1 = "9.9,8.8,22.2,88.7,morris1.43#email.com,chat22.3#email.com,123.6,6.54"
txt_list = []
for i in (txt1.split(',')):
if '#email.com' in i:
txt_list.append(i)
else:
txt_list.append(str(int(float(i))))
txt_new = ",".join(txt_list)
txt_new
Output:
'9,8,22,88,morris1.43#email.com,chat22.3#email.com,123,6'
Does anyone know a method that allows you to replace all the characters in a word with a single character?
If not, can anyone suggest a way to basically print _ (underscore) the number of times which is the length of the string itself without using any loops or ifs in the code?
mystring = '_'*len(mystring)
Of course, I'm guessing at the name of your string variable and the character that you want to use.
Or, if you just want to print it out, you can:
print('_'*len(mystring))
import re
str = "abcdefghi"
print(re.sub('[a-z]','_',str))
I am a beginner in programming. I have a string for example "test:1" and "test:2". And I want to remove ":1" and ":2" (including :). How can I do it using regular expression?
Hi andrew it's pretty easy. Think of a string as if it is an array of chars (letters) cause it actually IS. If the part of the string you want to delete is allways at the end of the string and allways the same length it goes like this:
var exampleString = 'test:1';
exampleString.length -= 2;
Thats it you just deleted the last two values(letters) of the string(charArray)
If you cant be shure it's allways at the end or the amount of chars to delete you'd to use the version of szymon
There are at least a few ways to do it with Groovy. If you want to stick to regular expression, you can apply expression ^([^:]+) (which means all characters from the beginning of the string until reaching :) to a StringGroovyMethods.find(regexp) method, e.g.
def str = "test:1".find(/^([^:]+)/)
assert str == 'test'
Alternatively you can use good old String.split(String delimiter) method:
def str = "test:1".split(':')[0]
assert str == 'test'
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).
Is there a multiline string literal syntax in Matlab or is it necessary to concatenate multiple lines?
I found the verbatim package, but it only works in an m-file or function and not interactively within editor cells.
EDIT: I am particularly after readbility and ease of modifying the literal in the code (imagine it contains indented blocks of different levels) - it is easy to make multiline strings, but I am looking for the most convenient sytax for doing that.
So far I have
t = {...
'abc'...
'def'};
t = cellfun(#(x) [x sprintf('\n')],t,'Unif',false);
t = horzcat(t{:});
which gives size(t) = 1 8, but is obviously a bit of a mess.
EDIT 2: Basically verbatim does what I want except it doesn't work in Editor cells, but maybe my best bet is to update it so it does. I think it should be possible to get current open file and cursor position from the java interface to the Editor. The problem would be if there were multiple verbatim calls in the same cell how would you distinguish between them.
I'd go for:
multiline = sprintf([ ...
'Line 1\n'...
'Line 2\n'...
]);
Matlab is an oddball in that escape processing in strings is a function of the printf family of functions instead of the string literal syntax. And no multiline literals. Oh well.
I've ended up doing two things. First, make CR() and LF() functions that just return processed \r and \n respectively, so you can use them as pseudo-literals in your code. I prefer doing this way rather than sending entire strings through sprintf(), because there might be other backslashes in there you didn't want processed as escape sequences (e.g. if some of your strings came from function arguments or input read from elsewhere).
function out = CR()
out = char(13); % # sprintf('\r')
function out = LF()
out = char(10); % # sprintf('\n');
Second, make a join(glue, strs) function that works like Perl's join or the cellfun/horzcat code in your example, but without the final trailing separator.
function out = join(glue, strs)
strs = strs(:)';
strs(2,:) = {glue};
strs = strs(:)';
strs(end) = [];
out = cat(2, strs{:});
And then use it with cell literals like you do.
str = join(LF, {
'abc'
'defghi'
'jklm'
});
You don't need the "..." ellipses in cell literals like this; omitting them does a vertical vector construction, and it's fine if the rows have different lengths of char strings because they're each getting stuck inside a cell. That alone should save you some typing.
Bit of an old thread but I got this
multiline = join([
"Line 1"
"Line 2"
], newline)
I think if makes things pretty easy but obviously it depends on what one is looking for :)