Delete rows with characters in cell array

I need some basic help. I have a cell array:

• 1 TITLE 13122423
• 2 NAME Bob
• 3 PROVIDER James

and many more rows with text...

• 44 234 456 234 345
• 45 324 346 234 345
• 46 344 454 462 435

and many MANY (>4000) more with only numbers

• 4100 text
• 4101 text

and more text and mixed entries

Now what I want is to delete all the rows where the first column contain a character, and end up with only those rows containing numbers. Row 44 - 46 in this example.

I tried to use

rawdataTruncated(strncmp(rawdataTruncated(:, 1), 'A', 1), :) = [];

but then i need to go throught the whole alphabet, right?

[the cyclist]

Rather than describing what you have, it would be stupendously easier for us if you just defined a representative cell array with code, like

C = {'1 TITLE 13122423';
     '44 234 456 234 345';
     '4100 text'};

and then what you want the output to be.

So ... uh ... what would be output be from this array I just defined? It seems like maybe just the 2nd row.

Also, where you have written the numbers 1-3, 44-46, and 4100-4101, are those part of the cell array , or are you just indicating lines numbers here?

[JB]

That is true. I am completely new to matlab and programming , but thanks. In the array you describe I want to delete row 1 and 3. output={'44 234 456 234 345'}.

[John BG]

Hi JB

this is John BG <mailto:[e-mail] [e-mail]>

are the numbers positive integers only?

or may it be the case that the input data is going to have lines like

1.

negative figures

'45 -65 2345 43'

2. decimals

'234 0.345 -2.4'

3.

fractions

'12 50/3456'

4.

line with numbers and operations

'35 45+67-3456/4356*3.2 '

Answer 1

[Andrei Bobrov]

My ruble:

A={'1 TITLE 13122423';
'2 NAME Bob';
'10 PROVIDER James';
'44 234 456 234 345';
'48 344 454 462 435';
'4100 text';
'4101 text';
'4102 2more text';
'4103 0495- 3725'};

x = regexp(A,'^(\d+\s*)+$','match');
out = [x{:}]';

[Stephen23]

+1 nice and simple.

[Andrei Bobrov]

Thank you Stephen!

[John BG]

Andrei Bobrov

this is John BG <mailto:[e-mail] [e-mail]>

your answer doesn't catch a numbers only line with more than one space between 2 numbers, look:

A = {'1 TITLE 13122423';
     '2 NAME Bob';
     '10 PROVIDER James';
     '44 234 456 234 345';
     ' 48    344 454 462 435';
     '4100 text';
     '4101 text';
     '4102 2more text';
     '4103 0495 3725'}

x = regexp(A,'^(\d+\s*)+$','match');out = [x{:}]';
out = 
 2×1 cell array
    '44 234 456 234 345'
    '4103 0495 3725'

the line with numbers only starting with 48 is missing.

[Jan]

@John BG: Obviously the original poster JB is satisfied with this solution already. Then there is no reason to invent further exceptions as leading or multiple spaces, negative or fractional numbers, or even operators.

Steven Lord's comment was useful: Let the OP ask if other details are required and reduce cluttering comments.

[Andrei Bobrov]

Hm.

x = regexp(A,'^\s*(\d+\s*)+$','match');out = [x{:}]'

Answer 2

[the cyclist]

I feel like there is a way to do this without resorting to cellfun, and just using regexp, but I'm drawing a blank. However, here is one way.

C = {'TITLE 13122423';
   '234 456 234 345';
   'text'};

cellfun(@(x)not(isempty(x)),regexp(C,'^[0-9][0-9 ]+'))

regexp is checking that each cell starts with a numeric character, and has only spaces and numbers after that.

[Stephen23]

+1 Simpler:

>> idx = cellfun('isempty',regexp(C,'^[0-9][0-9 ]+'));
>> C(~idx)
ans =
    '234 456 234 345'

[John BG]

wrong:

A line with '-' renders this single line attempt, like the Cyclist's, useless, because it returns null indices to the whole cell:

A={'1 TITLE 13122423';
'2 NAME Bob';
'10 PROVIDER James';
'44 234 456 234 345';
'48 344 454 462 435';
'4100 text';
'4101 text';
'4102 2more text';
'4103 0495- 3725'}

    idx = cellfun('isempty',regexp(A,'^[0-9][0-9 ]+'))

    idx =

      9×1 logical array

       0
       0
       0
       0
       0
       0
       0
       0
       0

[Stephen23]

@John BG: luckily regular expressions are trivial to alter, as long as the requirements are clearly specified. Note that the original question does not show - in its example, and JB states "...end up with only those rows containing numbers", so we would have to ask JB exactly what characters are allowed in the output numbers: I certainly don't know what data JB is working with. Do you?

>> idx = cellfun('isempty',regexp(A,'^[0-9][-0-9 ]+$'));
>> A(~idx)
ans =
  '44 234 456 234 345'
  '48 344 454 462 435'
  '4103 0495- 3725'

Note that using regexp much more efficient (in terms of total coding, debugging, and running time) than trying to write your own string parser, as some beginners would try to do

[Guillaume]

The simple addition of a $ at the end of the regular expression is all that is needed to fix the problem.

Answer 3

[Guillaume]

As said, ignore attempts to build your own custom string parser, it's inefficient and a complete waste of time. Use a regular expression:

This one works, John BG can attempt to find flaws in it to his heart content:

A = {'1 TITLE 13122423';
     '2 NAME Bob';
     '10 PROVIDER James';
     '44 234 456 234 345';
     '48 344 454 462 435';
     '4100 text';
     '4101 text';
     '4102 2more text';
     '4103 0495- 3725'}

   isjustnumbers = ~cellfun('isempty', regexp(A, '^[0-9][0-9 ]+$'));
   filteredA = A(isjustnumbers)

[Stephen23]

@Guillaume: +1 I don't think a simple $ character has ever entertained me so much. Thank you!

[John BG]

yes, keep patching

[Stephen23]

@John BG: are you suggesting that you own code is perfect, and never needs to be fixed?

s1=s1+1; $ shift pointer while ' ' or number

???

[Jan]

+1. It works and is elegant.

[José-Luis]

+1. Simple and efficient.

Tongue in cheek: That moment when Matlab answers starts feeling like a battlefield...

Answer 4

[Jan]

If you want to remove all lines, which contain a non-digit and non-white-space:

index = cellfun(@(ac) all(isstrprop(ac, 'digit') | ...
                          isstrprop(ac, 'wspace')), C);
C = C(index)

A loop is faster than cellfun:

B = true(size(A));
for k = 1:numel(A)
  B(k) = ~all(isstrprop(A{k}, 'digit') | ...
              isstrprop(A{k}, 'wspace'));
end
A(B) = [];

[Alan Peters]

I love your profile picture, Jan. I think I share that expression half the time I'm trying to solve a problem with Matlab!

[Walter Roberson]

(Jan used to have a different profile picture. He also used to get a lot of email requests from people. One day I teased him that it was because he had such a handsome and elegant profile picture. Jan changed his picture to the funny face you see now. The number of email requests he got dropped considerably. ;-) )

Answer 5

[JB]

gnovice gave me this awesome code:

index = ~any(cellfun(@any, isstrprop(C, 'alpha')), 2); C = C(index, :)

Thanks guys and sorry for the confusion. I am completely new to matlab coding

[John BG]

Hi JB

this is John BG <mailto:[e-mail] [e-mail]>

gnovice single line takes characters like '@' as numbers, please have a look to my answer.

[Jan]

@JB: The suggested code removes all lines, which contain 'alpha' characters: "the alphabetic letter range". isstrprop does not consider '@' as a number.

Answer 6

[John BG]

Hi JB

thanks for pointing out the code supplied by gnovice.

I just tried it and, with gnovice characters like '@' are taken as numbers.

Let's say that there's s line like

'4103 0495@ 3725'

.

then

.

A={'1 TITLE 13122423';
'2 NAME Bob';
'10 PROVIDER James';
'44 234 456 234 345';
'48 344 454 462 435';
'4100 text';
'4101 text';
'4102 2more text';
'4103 0495@ 3725'}

index = ~any(cellfun(@any, isstrprop(A, 'alpha')), 2); C = A(index, :)

   =
    3×1 cell array
      '44 234 456 234 345'
      '48 344 454 462 435'
      '4103   0495@  3   725'

My answer does only take into account numbers, as requested

A={'1 TITLE 13122423';
'2 NAME Bob';
'10 PROVIDER James';
'44 234 456 234 345';
'48 344 454 462 435';
'4100 text';
'4101 text';
'4102 2more text';
'4103 0495@ 3725'}

B=[];   % log to record locations of lines to delete
L0='0123456789';
L1=' 0123456789';
for k=1:1:length(A)
  L=A{k,:}
  s1=1;while L(s1)~=' '       % start from left until find ' '
    s1=s1+1; 
  end 
  while L(s1)==' '     % in case more than one consecutive ' '
    s1=s1+1
  end
  while s1<length(L) && ~isempty(strfind(L1,L(s1)))
    s1=s1+1;      % shift pointer while ' ' or number    
  end

  if isempty(strfind(L0,L(s1)))    % check 
    B=[B k];
  end
    
end
A(B)=[]  
 =
  2×1 cell array
    '44 234 456 234 345'
    '48 344 454 462 435'

.

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[JB]

Hi John. Thanks a lot for pointing this out. When I use your coding I get an error in the loop:

s1=s1+1; $ shift pointer while ' ' or number

Error: The input character is not valid in MATLAB statements or expressions.

[Guillaume]

@JB, you'd be better off forgetting about this solution altogether, it's extremely long-winded, inefficient and unnecessarily complicated. It's basically a poorly formed attempt at building a string parser, when matlab already has built-in ones

Use the regexp answer provided by the cyclist. It'll be a lot more reliable.

[John BG]

Hi JB

this is John BG <mailto:[e-mail] [e-mail]>

where

s1=s1+1; $ shift pointer while ' ' or number

is

s1=s1+1; % shift pointer while ' ' or number

typing error, sorry.

look what happens when one if the lines contains the character '-'

A={'1 TITLE 13122423';
'2 NAME Bob';
'10 PROVIDER James';
'44 234 456 234 345';
'48 344 454 462 435';
'4100 text';
'4101 text';
'4102 2more text';
'4103 0495- 3725'}
cellfun(@(x)not(isempty(x)),regexp(A,'^[0-9][0-9 ]+'))

ans =

    9×1 logical array

     1
     1
     1
     1
     1
     1
     1
     1
     1

It returns the complete cell!

[John BG]

A={'1 TITLE 13122423';
'2 NAME Bob';
'10 PROVIDER James';
'44 234 456 234 345';
'48 344 454 462 435';
'4100 text';
'4101 text';
'4102 2more text';
'4103 0495- 3725'}

idx = cellfun('isempty',regexp(A,'^[0-9][0-9 ]+'))

idx =

    9×1 logical array

     0
     0
     0
     0
     0
     0
     0
     0
     0

[John BG]

Hi JB

this is John BG <mailto:[e-mail] [e-mail]>

the single line approach eliminates lines that contain numbers only but with a preceding ' ',

like 5th line ' 48 344 454 462 435'.

A = {'1 TITLE 13122423';
     '2 NAME Bob';
     '10 PROVIDER James';
     '44 234 456 234 345';
     ' 48 344 454 462 435';
     '4100 text';
     '4101 text';
     '4102 2more text';
     '4103 0495 3725'}
cellfun('isempty', regexp(A, '^[0-9][0-9 ]+$'))
ans =
  9×1 logical array
   0
   0
   0
   1
   0
   0
   0
   0
   1

.

the question reads

' .. and end up with only those rows containing numbers'

The code that I have supplied does not skip lines that as requested have numbers only, but that have spaces ahead of numbers.

A = {'1 TITLE 13122423';
     '2 NAME Bob';
     '10 PROVIDER James';
     '44 234 456 234 345';
     ' 48 344 454 462 435';
     '4100 text';
     '4101 text';
     '4102 2more text';
     '4103 0495 3725'}

L0='0123456789';
L1=' 0123456789';
for k=1:1:length(A)
  L=A{k,:}
  s1=1;while L(s1)~=' '       % start from left until find ' '
    s1=s1+1; 
  end 
  while L(s1)==' '     % in case more than one consecutive ' '
    s1=s1+1
  end
  while s1<length(L) && ~isempty(strfind(L1,L(s1)))
    s1=s1+1;      % shift pointer while ' ' or number    
  end
  if isempty(strfind(L0,L(s1)))    % check 
    B=[B k];
  end
    
end
A(B)=[] 

..
A =

    3×1 cell array
      '44 234 456 234 345'
      ' 48 344 454 462 435'
      '4103 0495 3725'

regards

John BG <mailto:[e-mail] [e-mail]>

[Stephen23]

In case anyone else cares:

>> idx = cellfun('isempty',regexp(A,'^[0-9 ]+$'));
>> A(~idx)
  '44 234 456 234 345'
  ' 48 344 454 462 435'
  '4103 0495 3725'

[Jan]

The handling leading spaces in your code is not efficient, but this problem does not occur in the original question. The two loops while L(s1)~=' ' and while L(s1)==' ' are fragile: they fail, if the string does not contain any space, or if only spaces are following an initial sequence of non-space characters. This loop version could be simplified:

B = true(size(A));
L = ' 0123456789';
for k = 1:numel(A)
  B(k) = ~all(ismember(A{k}, L));
end
A(B) = [];

But I prefer regexp and isstrprop methods.

[Steven Lord]

I recommend everyone except the original poster JB take a step back from this question for a little while. Remember that one of the tips for a helpful answer is "Be honest and considerate with all responses to all contributors." and a little breather might help make that easier.

JB, please add a comment to the original question indicating if you are satisfied with a combination of one or more of the responses you've received. If you aren't, clarify what aspect of the problem has not yet been solved. If you feel one of the answers was most useful in solving the problem, consider accepting it.

[John BG]

Steven Lord

this is John BG <mailto:[e-mail] [e-mail]>

Your intervention is highly appreciated.

Too many comments often clutter the answers, and it may be the case that the question originator, here JB, decides just to walk away from such discussions, sometimes diverging to speed and code compactness accessory considerations, while it's not yet know whether JB finds any useful code at all, so far.

Awaiting JB response