how to swap 2 numbers using regexp in vim - regex

i have some hardcoded data that looks like this:
s1 = [ ...
[-225.053,-74.050,4.801]' ...
[-82.053,68.950,4.801]' ...
[-82.053,-74.050,4.801]' ...
[-82.053,-217.050,4.801]' ...
[60.947,-74.050,4.801]' ...
];
i've got a bunch of those entries. i realized that the first 2 entries of each vector were in the incorrect order, so rather than [a,b,c] it should be [b,a,c].
i want the code to be clear, so i'd like to swap the hardcoded values.
how do i swap the values in vim using regexp?

found a solution. the following will work assuming there are no spaces:
s/\(\-\?\d\+\.\d\+\),\(\-\?\d\+\.\d\+\),\(\-\?\d\+\.\d\+\)/\2,\1,\3/gc
that's 3 times this:
\(\-\?\d\+\.\d\+\)
which matches a number of the form [optional sign]a.b

Firstly, you should know how to swap two values.
:%s/\(123\),\(456\)/\2,\1/
Secondly, you should replace to the correct regular expression.
[-225.053,-74.050,4.801]
these number split by "," so you can use this
\[\(.*\),\(.*\),\(.*\)\]
So, finally, the regular expression is
:%s/\[\(.*\),\(.*\),\(.*\)\]/[\3,\1,\2]/

Related

Match return substring between two substrings using regexp

I have a list of records that are character vectors. Here's an example:
'1mil_0,1_1_1_lb200_ks_drivers_sorted.csv'
'1mil_0_1_lb100_ks_drivers_sorted.csv'
'1mil_1_1_lb2_100_100_ks_drivers_sorted.csv'
'1mil_1_1_lb100_ks_drivers_sorted.csv'
From these names I would like to extract whatever's between the two substrings 1mil_ and _ks_drivers_sorted.csv.
So in this case the output would be:
0,1_1_1_lb200
0_1_lb100
1_1_lb2_100_100
1_1_lb100
I'm using MATLAB so I thought to use regexp to do this, but I can't understand what kind of regular expression would be correct.
Or are there some other ways to do this without using regexp?
Let the data be:
x = {'1mil_0,1_1_1_lb200_ks_drivers_sorted.csv'
'1mil_0_1_lb100_ks_drivers_sorted.csv'
'1mil_1_1_lb2_100_100_ks_drivers_sorted.csv'
'1mil_1_1_lb100_ks_drivers_sorted.csv'};
You can use lookbehind and lookahead to find the two limiting substrings, and match everything in between:
result = cellfun(#(c) regexp(c, '(?<=1mil_).*(?=_ks_drivers_sorted\.csv)', 'match'), x);
Or, since the regular expression only produces one match, the following simpler alternative can be used (thanks #excaza for noticing):
result = regexp(x, '(?<=1mil_).*(?=_ks_drivers_sorted\.csv)', 'match', 'once');
In your example, either of the above gives
result =
4×1 cell array
'0,1_1_1_lb200'
'0_1_lb100'
'1_1_lb2_100_100'
'1_1_lb100'
For me the easy way to do this is just use espace or nothing to replace what you don't need in your string, and the rest is what you need.
If is a list, you can use a loop to do this.
Exemple to replace "1mil_" with "" and "_ks_drivers_sorted.csv" with ""
newChr = strrep(chr,'1mil_','')
newChr = strrep(chr,'_ks_drivers_sorted.csv','')

Split string and get last element

Let's say I have a column which has values like:
foo/bar
chunky/bacon/flavor
/baz/quz/qux/bax
I.e. a variable number of strings separated by /.
In another column I want to get the last element from each of these strings, after they have been split on /. So, that column would have:
bar
flavor
bax
I can't figure this out. I can split on / and get an array, and I can see the function INDEX to get a specific numbered indexed element from the array, but can't find a way to say "the last element" in this function.
Edit:
this one is simplier:
=REGEXEXTRACT(A1,"[^/]+$")
You could use this formula:
=REGEXEXTRACT(A1,"(?:.*/)(.*)$")
And also possible to use it as ArrayFormula:
=ARRAYFORMULA(REGEXEXTRACT(A1:A3,"(?:.*/)(.*)$"))
Here's some more info:
the RegExExtract function
Some good examples of syntax
my personal list of Regex Tricks
This formula will do the same:
=INDEX(SPLIT(A1,"/"),LEN(A1)-len(SUBSTITUTE(A1,"/","")))
But it takes A1 three times, which is not prefferable.
You could do this too
=index(SPLIT(A1, "/"), COLUMNS(SPLIT(A1, "/"))-1)
Also possible, perhaps best on a copy, with Find:
.+/
(Replace with blank) and Search using regular expressions ticked.
You can try use this!
You've got the array of String, so you can acess the last element by length
String message = "chunky/bacon/flavor";
String[] outSplited = message.split("/");
System.out.println(outSplited[outSplited.length -1]);

Matching an optional number in a Lua pattern

I'm parsing the output from the diff3 command and some lines look like this:
1:1,2c
2:0a
I am interested in the numbers in the middle. Its either a single number or a pair of numbers separated by commas. With regexes I can capture them both like this:
/^\d+:(\d+)(?:,(\d+))?[ac]$/
What is the simplest equivalent in Lua? I can't pass a direct translation of that regex to string.match because of the optional second number.
Using lua patterns, you could use the following:
^%d+:(%d+),?(%d*)[ac]$
Example:
local n,m = string.match("1:2,3c", "^%d+:(%d+),?(%d*)[ac]$")
print(n,m) --> 2 3
local n,m = string.match("2:0a", "^%d+:(%d+),?(%d*)[ac]$")
print(n,m) --> 0
You can achieve it using lua patterns too:
local num = str:match '^%d+:(%d+),%d+[ac]$' or str:match '^%d+:(%d+)[ac]$'

Simplest way to find out if at least one cell in a cell array matches a regular expression

I need to search a cell array and return a single boolean value indicating whether any cell matches a regular expression.
For example, suppose I want to find out if the cell array strs contains foo or -foo (case-insensitive). The regular expression I need to pass to regexpi is ^-?foo$.
Sample inputs:
strs={'a','b'} % result is 0
strs={'a','foo'} % result is 1
strs={'a','-FOO'} % result is 1
strs={'a','food'} % result is 0
I came up with the following solution based on How can I implement wildcard at ismember function of matlab? and Searching cell array with regex, but it seems like I should be able to simplify it:
~isempty(find(~cellfun('isempty', regexpi(strs, '^-?foo$'))))
The problem I have is that it looks rather cryptic for such a simple operation. Is there a simpler, more human-readable expression I can use to achieve the same result?
NOTE: The answer refers to the original regexp in the question: '-?foo'
You can avoid the find:
any(~cellfun('isempty', regexpi(strs, '-?foo')))
Another possibility: concatenate first all cells into a single string:
~isempty(regexpi([strs{:}], '-?foo'))
Note that you can remove the "-" sign in any of the above:
any(~cellfun('isempty', regexpi(strs, 'foo')))
~isempty(regexpi([strs{:}], 'foo'))
And that allows using strfind (with lower) instead of regexpi:
~isempty(strfind(lower([strs{:}]),'foo'))

Part of as string from a string using regular expressions

I have a string of 5 characters out of which the first two characters should be in some list and next three should be in some other list.
How could i validate them with regular expressions?
Example:
List for First two characters {VBNET, CSNET, HTML)}
List for next three characters {BEGINNER, EXPERT, MEDIUM}
My Strings are going to be: VBBEG, CSBEG, etc.
My regular expression should find that the input string first two characters could be either VB, CS, HT and the rest should also be like that.
Would the following expression work for you in a more general case (so that you don't have hardcoded values): (^..)(.*$)
- returns the first two letters in the first group, and the remaining letters in the second group.
something like this:
^(VB|CS|HT)(BEG|EXP|MED)$
This recipe works for me:
^(VB|CS|HT)(BEG|EXP|MED)$
I guess (VB|CS|HT)(BEG|EXP|MED) should do it.
If your strings are as well-defined as this, you don't even need regex - simple string slicing would work.
For example, in Python we might say:
mystring = "HTEXP"
prefix = mystring[0:2]
suffix = mystring[2:5]
if (prefix in ['HT','CS','VB']) AND (suffix in ['BEG','MED','EXP']):
pass # valid!
else:
pass # not valid. :(
Don't use regex where elementary string operations will do.