To extract the last sub string from strings

Hello everybody,

I would like to extract the last string just before 'EP'.

I tried first split the string with strsplit function. and then get the last right string using for-loop.

I assume that it can be done without using for-loop.

Please give me some help how to make the code without using for-loop.

load input.mat

temp = cellfun(@(x) strsplit(x, {' ','EP'}), input, 'UniformOutput', false); % split the string

for i=1:length(temp)

num(i,:) = str2double(temp{i}(end-1)); % fill the last right string in cells

end

Answer 1

[Dyuman Joshi]

You can use extractBetween with index of last space character and the common denominator at the end of the string ('EP')

load input.mat
num=cellfun(@(x) str2double(extractBetween(x,find(x==' ',1,'last')+1,'EP')), input)

num = 30×1

11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000

[Smithy]

Wow.. Thank you very mcuh it works really well. I really appreciate with it.

Answer 2

[Stephen23]

Avoid slow CELLFUN, STR2NUM, REGEXP, etc.

One SSCANF call is probably the most efficient approach by far, as well as being very simple:

S = load('input.mat');
C = S.input

C = 30×1 cell array
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}

V = sscanf([C{:}],'%*s %*s%fEP')

V = 30×1

11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000

[Dyuman Joshi]

Stephen, how do you know that cellfun, str2num, regexp etc are slower functions? And are there any resources where I can read more about this?

[Stephen23]

"how do you know that cellfun, str2num, regexp etc are slower functions?"

1. many years of reading this and other forums, learning from the combined knowledge of many users.
2. many years of writing unit tests for my own code (i.e. making modifications and comparing).
3. reading the documentation, to know what features functions have and how to use them.
4. knowledge about functions, e.g. STR2NUM calls EVAL inside (so is not optimised by the JIT engine), and CELLFUN by design must call a function handle repeatedly (slower than a loop).

Lets compare the answers given so far on this thread:

S = load('input.mat');
C = repmat(S.input,1e3,1); % bigger array -> easier to compare

timeit(@()funAtsushiUeno(C)) % REGEXP, CELLFUN, STR2NUM

ans = 0.5426

timeit(@()funDyumanJoshi(C)) % CELLFUN, EXTRACTBETWEEN, STR2DOUBLE

ans = 0.4321

timeit(@()funChunru(C)) % REGEXP, CELLFUN, STR2NUM

ans = 0.4083

timeit(@()funImageAnalyst(C)) % loop and indexing

ans = 0.6536

timeit(@()funKSSV(C)) % REGEXP, STRREP, STR2DOUBLE

ans = 0.3133

timeit(@()funS23(C)) % SSCANF

ans = 0.0377

So, my function is more than eight times faster than the next fastest function (from KSSV), as well as being the simplest. And I had a fair idea that would be the case, even before writing this test code.

"And are there any resources where I can read more about this?"

If you want to learn how to use MATLAB efficiently, my advice is to read this forum a lot. And when I write "a lot", I don't mean "just a little bit". And not just new threads: there are some really important topics that have been discussed in some old yet canonical threads on this forum.

https://www.mathworks.com/matlabcentral/answers/228557-experts-of-matlab-how-did-you-learn-any-advice-for-beginner-intermediate-users

V1 = funAtsushiUeno(C);
V2 = funDyumanJoshi(C);
V3 = funChunru(C);
V4 = funImageAnalyst(C);
V5 = funKSSV(C);
V6 = funS23(C);
isequal(V1(:), V2(:), V3(:), V4(:), V5(:), V6(:)) % checking the function outputs:

ans = logical
   1

function num = funAtsushiUeno(C)
num = regexp(C,'([\d+-e.]+)EP\s*$','tokens');
num = [num{:}];
num = [num{:}];
num = cellfun(@str2num, num);
end
function num = funDyumanJoshi(C) 
num=cellfun(@(x) str2double(extractBetween(x,find(x==' ',1,'last')+1,'EP')), C);
end
function y = funChunru(C)
s = regexp(C, '\s([\d\.]*)EP$', 'tokens');
y = cellfun(@(s) str2num(s{1}{1}), s);
end
function numbers = funImageAnalyst(data)
rows = numel(data);
numbers = zeros(rows, 1);
for row = 1 : rows
    words = strsplit(data{row});
    lastWord = words{end};
    numbers(row) = str2double(lastWord(1:end-2));
end
end
function s = funKSSV(C)
expression = '\d+\.?\d*EP';
s = regexp(C,expression,'match') ;
s = strrep([s{:}]','EP','') ;
s = str2double(s);
end
function V = funS23(C);
V = sscanf([C{:}],'%*s %*s%fEP');
end

Answer 3

[Chunru]

load(websave("input.mat", "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1114425/input.mat"))
input

input = 30×1 cell array
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}
    {'N88 220D/2 11EP'         }
    {'N88 220D/2 11EP'         }
    {'N88 260D/2 8EP'          }
    {'N88 260D/2 8EP'          }
    {'A35000D+N14 1260D 11.4EP'}
    {'A35000D+N14 1260D 11.4EP'}

s = regexp(input, '\s([\d\.]*)EP$', 'tokens');
y = cellfun(@(s) str2num(s{1}{1}), s)

y = 30×1

11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000

[Smithy]

It works so well. Thank you alot. I really appreciate with it.

Answer 4

[Atsushi Ueno]

This is a bit of a challenge because we need to deal with cell arrays with irregular number sizes.

I would use the regexp function.

load input.mat
num = regexp(input,'([\d+-e.]+)EP\s*$','tokens');
num = [num{:}];
num = [num{:}];
num = cellfun(@str2num, num)

num = 1×30

11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000 11.4000 11.4000 11.0000 11.0000 8.0000 8.0000 11.4000 11.4000

Answer 5

[Image Analyst]

Here's one way. Simple, straightforward, and intuitive. There may be more compact but cryptic methods though.

s = load('input.mat')
data = s.input; % DON'T call your variables input, which is a build in function!
rows = numel(data)
numbers = zeros(rows, 1);
for row = 1 : rows
    words = strsplit(data{row});
    lastWord = words{end};
    numbers(row) = str2double(lastWord(1:end-2));
end

[Smithy]

Thank you very much. I really really appreciate with it.

Answer 6

[KSSV]

expression = '\d+\.?\d*EP';
s = regexp(input,expression,'match') ;
s = strrep([s{:}]','EP','') ;
s = str2double(s)

[Smithy]

Thank you alot. It helps me alot. I really really appreciate with it.