I wanted to know how to remove first character of a string in octave. I am manipulating the string in a loop and after every loop, I want to remove the first character of the remaining string.
Thanks in advance.
If it's just a one-line string then:
short_string = long_string(2:end)
But if you have a cell array of strings then either do it as above if you have a loop already, otherwise you can use this shorthand to do it in one line:
short_strings = cellfun(#(x)(x(2:end)), long_strings, 'uni', false)
Or else if you have a matrix of strings (i.e. all the same length), then you can vectorize it as:
short_strings = long_strings(:, 2:end)
Related
Basically, I have a string in Lua, which is a user output, but I want to trim the first character cause I don't really need it. How can I do that?
You cannot delete the first character of a string.
But you can copy a substring that starts from the second character using string.sub
string.sub(s, i [,j])
Returns the substring of s that starts at i and continues until j; i
and j can be negative. If j is absent, then it is assumed to be equal
to -1 (which is the same as the string length)....
So str = str:sub(2) will give you the substring of str that starts at character 2 which is what you want.
heres the question:
Remove First and Last Character
It's pretty straightforward. Your goal is to create a function that removes the first and last characters of a string. You're given one parameter, the original string. You don't have to worry with strings with less than two characters.
i have no idea how to do it because of the fact that it wants me to remove first and last letter and not the first and last word
Slicing is the best way to go about that.
test_string = 'This is a test!'
print (test_string [1:-1])
From your response that removing the first and last word would be done by list1.pop() and list1.pop(0) I am assuming that you are given a list of strings and are required to remove the first and last character of each string in the list.
If my assumption is correct you can do a combination of slicing and list comprehension.
list_of_str = ["words", "words", "words", "words"]
list_of_shortened_str = [word[1:-1] for word in list_of_str]
# list_of_shortened_str = ["ord", "ord", "ord", "ord"]
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'
I have a few lines of text like this:
abc
def
ghi
and I want to assign these multiple lines to a Matlab variable for further processing.
I am copying these from very large text file and want to process it in Matlab Instead of saving the text into a file and then reading line by line for processing.
I tried to handle the above text lines as single string but am getting an error whilst trying to assign to a variable:
x = 'abc
def
ghi'
Error:
x = 'abc
|
Error: String is not terminated properly.
Any suggestions which could help me understand and solve the issue will be highly appreciated.
I frequently do this, namely copy text from elsewhere which I want to hard-code into a MATLAB script (in my case it's generally SQL code I want to manipulate and call from MATLAB).
To achieve this I have a helper function in clipboard2cellstr.m defined as follows:
function clipboard2cellstr
str = clipboard('paste');
str = regexprep(str, '''', ''''''); % Double any single quotes
strs = regexp(str, '\s*\r?\n\r?', 'split');
cs = sprintf('{\n''%s''\n}', strjoin(strs, sprintf('''\n''')));
clipboard('copy', cs);
disp(cs)
disp('(Copied to Clipboard)')
end
I then copy the text using Ctrl-c (or however) and run clipboard2cellstr. This changes the contents of the clipboard to something I can paste into the MATLAB editor using Ctrl-v (or however).
For example, copying this line
and this line
and this one, and then running the function generates this:
{
'For example, copying this line'
'and this line'
'and this one, and then running the function generates this:'
}
which is valid MATLAB which can be pasted directly in.
Your error is because you ended the line when MATLAB was expecting a closing quote character. You must use array notation to have multi-line or multi-element arrays.
You can assign like this if you use array notation
x = ['abc'
'def'
'hij']
>> x = 3×3 char array
Note: with this method, your rows must have the same number of characters, as you are really dealing with a character array. You can think of a character array like a numeric matrix, hence why it must be "rectangular".
If you have MATLAB R2016b or newer, you can use the string data type. This uses double quotes "..." rather than single quotes '...', and can be multi-line. You must still use array notation:
x = ["abc"
"def"
"hijk"]
>> x = 3×1 string array
We can have different numbers of characters in each line, as this is simply a 3 element string array, not a character array.
Alternatively, use a cell array of character arrays (or strings)
x = {'abc'
'def'
'hijk'}
>> x = 3×1 cell array
Again, you can have character arrays or strings of different lengths within a cell array.
In all of the above examples, a newline is simply for readability and can be replaced by a semi-colon ; to denote the next line of the array.
The option you choose will depend on what you want to do with the text. If you're reading from a file, I would suggest the string array or the cell array, as they can deal with different length lines. For backwards compatibility, use a cell array. You may find cellfun relevant for operating on cell arrays. For native string operations, use a string array.
The string I am given is as follows:
scrap1 =
a le h
ke fd
zyq b
ner i
You'll notice there are 2 blank spaces indicating a space (ASCII 32) in each row. I need to find the mean ASCII value in each column without taking into account the spaces (32). So first I would convert to with double(scrap1) but then how do I find the mean without taking into account the spaces?
If it's only the ASCII 32 you want to omit:
d = double(scrap1);
result = mean(d(d~=32)); %// logical indexing to remove unwanted value, then mean
You can remove the intermediate spaces in the string with scrap1(scrap1 == ' ') = ''; This replaces any space in the input with an empty string. Then you can do the conversion to double and average the result. See here for other methods.
Probably, you can use regex to find the space and ignore it. "\s"
findSpace = regexp(scrap1, '\s', 'ignore')
% I am not sure about the ignore case, this what comes to my mind. but u can read more about regexp by typying doc regexp.