I have a list of filenames in a struct array, example:
4x1 struct array with fields:
name
date
bytes
isdir
datenum
where files.name
ans =
ts.01094000.crest.csv
ans =
ts.01100600.crest.csv
etc.
I have another list of numbers (say, 1094000). And I want to find the corresponding file name from the struct.
Please note, that 1094000 doesn't have preceding 0. Often there might be other numbers. So I want to search for '1094000' and find that name.
I know I can do it using Regex. But I have never used that before. And finding it difficult to write for numbers instead of text using strfind. Any suggestion or another method is welcome.
What I have tried:
regexp(files.name,'ts.(\d*)1094000.crest.csv','match');
I think the regular expression you'd want is more like
filenames = {'ts.01100600.crest.csv','ts.01094000.crest.csv'};
matches = regexp(filenames, ['ts\.0*' num2str(1094000) '\.crest\.csv']);
matches = ~cellfun('isempty', matches);
filenames(matches)
For a solution with strfind...
Pre-16b:
match = ~cellfun('isempty', strfind({files.name}, num2str(1094000)),'UniformOutput',true)
files(match)
16b+:
match = contains({files.name}, string(1094000))
files(match)
However, the strfind way might have issues if the number you are looking for exists in unexpected places such as looking for 10 in ["01000" "00101"].
If your filenames match the pattern ts.NUMBER.crest.csv, then in 16b+ you could do:
str = {files.name};
str = extractBetween(str,4,'.');
str = strip(str,'left','0');
matches = str == string(1094000);
files(matches)
Related
I have the following strings in a long string:
a=b=c=d;
a=b;
a=b=c=d=e=f;
I want to first search for above mentioned pattern (X=Y=...=Z) and then output like the following for each of the above mentioned strings:
a=d;
b=d;
c=d;
a=b;
a=f;
b=f;
c=f;
d=f;
e=f;
In general, I want all the variables to have an equal sign with the last variable on the extreme right of the string. Is there a way I can do it using regexprep in MATLAB. I am able to do it for a fixed length string, but for variable length, I have no idea how to achieve this. Any help is appreciated.
My attempt for the case of two equal signs is as follows:
funstr = regexprep(funstr, '([^;])+\s*=\s*+(\w+)+\s*=\s*([^;])+;', '$1 = $3; \n $2 = $3;\n');
Not a regexp but if you stick to Matlab you can make use of the cellfun function to avoid loop:
str = 'a=b=c=d=e=f;' ; %// input string
list = strsplit(str,'=') ;
strout = cellfun( #(a) [a,'=',list{end}] , list(1:end-1), 'uni', 0).' %'// Horchler simplification of the previous solution below
%// this does the same than above but more convoluted
%// strout = cellfun( #(a,b) cat(2,a,'=',b) , list(1:end-1) , repmat(list(end),1,length(list)-1) , 'uni',0 ).'
Will give you:
strout =
'a=f;'
'b=f;'
'c=f;'
'd=f;'
'e=f;'
Note: As Horchler rightly pointed out in comment, although the cellfun instruction allows to compact your code, it is just a disguised loop. Moreover, since it runs on cell, it is notoriously slow. You won't see the difference on such simple inputs, but keep this use when super performances are not a major concern.
Now if you like regex you must like black magic code. If all your strings are in a cell array from the start, there is a way to (over)abuse of the cellfun capabilities to obscure your code do it all in one line.
Consider:
strlist = {
'a=b=c=d;'
'a=b;'
'a=b=c=d=e=f;'
};
Then you can have all your substring with:
strout = cellfun( #(s)cellfun(#(a,b)cat(2,a,'=',b),s(1:end-1),repmat(s(end),1,length(s)-1),'uni',0).' , cellfun(#(s) strsplit(s,'=') , strlist , 'uni',0 ) ,'uni',0)
>> strout{:}
ans =
'a=d;'
'b=d;'
'c=d;'
ans =
'a=b;'
ans =
'a=f;'
'b=f;'
'c=f;'
'd=f;'
'e=f;'
This gives you a 3x1 cell array. One cell for each group of substring. If you want to concatenate them all then simply: strall = cat(2,strout{:});
I haven't had much experience w/ Matlab; but your problem can be solved by a simple string split function.
[parts, m] = strsplit( funstr, {' ', '='}, 'CollapseDelimiters', true )
Now, store the last part of parts; and iterate over parts until that:
len = length( parts )
for i = 1:len-1
print( strcat(parts(i), ' = ', parts(len)) )
end
I do not know what exactly is the print function in matlab. You can update that accordingly.
There isn't a single Regex that you can write that will cover all the cases. As posted on this answer:
https://stackoverflow.com/a/5019658/3393095
However, you have a few alternatives to achieve your final result:
You can get all the values in the line with regexp, pick the last value, then use a for loop iterating throughout the other values to generate the output. The regex to get the values would be this:
matchStr = regexp(str,'([^=;\s]*)','match')
If you want to use regexprep at any means, you should write a pattern generator and a replace expression generator, based on number of '=' in the input string, and pass these as parameters of your regexprep func.
You can forget about Regex and Split the input to generate the output looping throughout the values (similarly to alternative #1) .
At the moment I am saving a set of variables to a text file. I am doing following to check if my code works, but whenever I use a two-digit numbers such as 10 it would not print this number as the max number.
If my text file looked like this.
tom:5
tom:10
tom:1
It would output 5 as the max number.
name = input('name')
score = 4
if name == 'tom':
fo= open('tom.txt','a')
fo.write('Tom: ')
fo.write(str(score ))
fo.write("\n")
fo.close()
if name == 'wood':
fo= open('wood.txt','a')
fo.write('Wood: ')
fo.write(str(score ))
fo.write("\n")
fo.close()
tomL2 = []
woodL2 = []
fo = open('tom.txt','r')
tomL = fo.readlines()
tomLi = tomL2 + tomL
fo.close
tomLL=max(tomLi)
print(tomLL)
fo = open('wood.txt','r')
woodL = fo.readlines()
woodLi = woodL2 + woodL
fo.close
woodLL=max(woodLi)
print(woodLL)
You are comparing strings, not numbers. You need to convert them into numbers before using max. For example, you have:
tomL = fo.readlines()
This contains a list of strings:
['tom:5\n', 'tom:10\n', 'tom:1\n']
Strings are ordered lexicographically (much like how words would be ordered in an English dictionary). If you want to compare numbers, you need to turn them into numbers first:
tomL_scores = [int(s.split(':')[1]) for s in tomL]
The parsing is done in the following way:
….split(':') separates the string into parts using a colon as the delimiter:
'tom:5\n' becomes ['tom', '5\n']
…[1] chooses the second element from the list:
['tom', '5\n'] becomes '5\n'
int(…) converts a string into an integer:
'5\n' becomes 5
The list comprehension [… for s in tomL] applies this sequence of operations to every element of the list.
Note that int (or similarly float) are rather picky about what it accepts: it must be in the form of a valid numeric literal or it will be rejected with an error (although preceding and trailing whitespace is allowed). This is why you need ….split(':')[1] to massage the string into a form that it's willing to accept.
This will yield:
[5, 10, 1]
Now, you can apply max to obtain the largest score.
As a side-note, the statement
fo.close
will not close a file, since it doesn't actually call the function. To call the function you must enclose the arguments in parentheses, even if there are none:
fo.close()
I have a file that contains rows and columns of information like:
104857 Big Screen TV 567.95
573823 Blender 45.25
I need to parse this information into three separate items, a string containing the identification number on the left, a string containing the item name, and a double variable containing the price. The information is always found in the same columns, i.e. in the same order.
I am having trouble accomplishing this. Even when not reading from the file and just using a sample string, my attempt just outputs a jumbled mess:
string input = "104857 Big Screen TV 567.95";
string tempone = "";
string temptwo = input.substr(0,1);
tempone += temptwo;
for(int i=1 ; temptwo != " " && i < input.length() ; i++)
{
temptwo = input.substr(j,j);
tempone += temp2;
}
cout << tempone;
I've tried tweaking the above code for quite some time, but no luck, and I can't think of any other way to do it at the moment.
You can find the first space and the last space using std::find_first_of and std::find_last_of . You can use this to better split the string into 3 - first space comes after the first variable and the last space comes before the third variable, everything in between is the second variable.
How about following pseudocode:
string input = "104857 Big Screen TV 567.95";
string[] parsed_output = input.split(" "); // split input string with 'space' as delimiter
// parsed_output[0] = 104857
// parsed_output[1] = Big
// parsed_output[2] = Screen
// parsed_output[3] = TV
// parsed_output[4] = 567.95
int id = stringToInt(parsed_output[0]);
string product = concat(parsed_output[1], parsed_output[2], ... ,parsed_output[length-2]);
double price = stringToDouble(parsed_output[length-1]);
I hope, that's clear.
Well try breaking down the files components:
you know a number always comes first, and we also know a number has no white spaces.
The string following the number CAN have whitespaces, but won't contain any numbers(i would assume)
After this title, you're going to have more numbers(with no whitespaces)
from these components, you can deduce:
grabbing the first number is as simple as reading in using the filestream <<.
getting the string requires you to check until you reach a number, grabbing one character at a time and inserting that into a string. the last number is just like the first, using the filestream <<
This seems like homework so i'll let you put the rest together.
I would try a regular expression, something along these lines:
^([0-9]+)\s+(.+)\s+([0-9]+\.[0-9]+)$
I am not very good at regex syntax, but ([0-9]+) corresponds to a sequence of digits (this is the id), ([0-9]+\.[0-9]+) is the floating point number (price) and (.+) is the string that is separated from the two number by sequences of "space" characters: \s+.
The next step would be to check if you need this to work with prices like ".50" or "10".
I wanted to match the words in string with reverse order.
We wanted to put validation to prompt user, if name exists in reverse order.
For example:
If name column has the value, 'Viral,Tennis'
Now if user enters a new name with the value, 'Tennis,Viral'
Then how can we match reverse order of word using regex or some other way?
I am using C#.net for development.
You could take a look at the Regex.Split(String input, String regex) and do something like so:
String[] userEntry = Regex.Split(userString, "\\s+");
StringBuilder sb = new StringBuilder()
for (int i = userEntry.Length -1; i >= 0; i--)
{
sb.append(userEntry[i]).append(" ");
}
String result = sb.ToString();
//Do Validation
That would do the trick, however, you need to keep in mind that things will get a little bit messy if you do not want to change the order of special symbols such as the comma. You could easily remove those and do any validation without special symbols.
EDIT: It depends on what you mean by special symbols. The regex [^a-zA-z0-9]+ will match any character which is not a letter (upper or lower case) and which is also not a number. So you could easily do something like so:
string input = ...
string pattern = "[^a-zA-z0-9]+";
string replacement = "";
Regex rgx = new Regex(pattern);
string result = rgx.Replace(input, replacement);
The above should yield a string which is only made from letters and digits. White spaces will also be removed.
I have a list of several phrases in the following format
thisIsAnExampleSentance
hereIsAnotherExampleWithMoreWordsInIt
and I'm trying to end up with
This Is An Example Sentance
Here Is Another Example With More Words In It
Each phrase has the white space condensed and the first letter is forced to lowercase.
Can I use regex to add a space before each A-Z and have the first letter of the phrase be capitalized?
I thought of doing something like
([a-z]+)([A-Z])([a-z]+)([A-Z])([a-z]+) // etc
$1 $2$3 $4$5 // etc
but on 50 records of varying length, my idea is a poor solution. Is there a way to regex in a way that will be more dynamic? Thanks
A Java fragment I use looks like this (now revised):
result = source.replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
result = result.substring(0, 1).toUpperCase() + result.substring(1);
This, by the way, converts the string givenProductUPCSymbol into Given Product UPC Symbol - make sure this is fine with the way you use this type of thing
Finally, a single line version could be:
result = source.substring(0, 1).toUpperCase() + source(1).replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
Also, in an Example similar to one given in the question comments, the string hiMyNameIsBobAndIWantAPuppy will be changed to Hi My Name Is Bob And I Want A Puppy
For the space problem it's easy if your language supports zero-width-look-behind
var result = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "(?<=[a-z])([A-Z])", " $1");
or even if it doesn't support them
var result2 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "([a-z])([A-Z])", "$1 $2");
I'm using C#, but the regexes should be usable in any language that support the replace using the $1...$n .
But for the lower-to-upper case you can't do it directly in Regex. You can get the first character through a regex like: ^[a-z] but you can't convet it.
For example in C# you could do
var result4 = Regex.Replace(result, "^([a-z])", m =>
{
return m.ToString().ToUpperInvariant();
});
using a match evaluator to change the input string.
You could then even fuse the two together
var result4 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "^([a-z])|([a-z])([A-Z])", m =>
{
if (m.Groups[1].Success)
{
return m.ToString().ToUpperInvariant();
}
else
{
return m.Groups[2].ToString() + " " + m.Groups[3].ToString();
}
});
A Perl example with unicode character support:
s/\p{Lu}/ $&/g;
s/^./\U$&/;