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1

I use MATLAB version R2015a.

I get a series answer from solving the optimal problem several times, and I want to get their sum and average them. However, some of them are NaN. How do I write code to ignore those NaN and sum the others which are not NaN?

matlab sum average nan

share|improve this question

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What have you tried? Have you heard of rmmissing or isnan?
  
  – Joseph Sible
  Mar 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have tried the for loop to sum them,or use the sum instruction directly
  
  – Shine Sun
  Mar 26 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Adding 2x2 matrix with NaNs
  
  – Cris Luengo
  Mar 26 at 3:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @CrisLuengo, that post is definitely related, though I feel this set of answers if more specific and updated now.
  
  – SecretAgentMan
  Mar 26 at 13:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @SecretAgentMan: I voted to close as duplicate because that question has the same answers as his one (nansum and sum+isnan), even though the questions are not identical. This qestion is more generic, the other one is rather specific, with the cell array complicating the question. If the vote ages away, which seems likely, at least there is a link between the questions now, it will help people find answers.
  
  – Cris Luengo
  Mar 26 at 13:19
  

  
  
  
  

add a comment | 

1

I use MATLAB version R2015a.

I get a series answer from solving the optimal problem several times, and I want to get their sum and average them. However, some of them are NaN. How do I write code to ignore those NaN and sum the others which are not NaN?

matlab sum average nan

share|improve this question

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What have you tried? Have you heard of rmmissing or isnan?
  
  – Joseph Sible
  Mar 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have tried the for loop to sum them,or use the sum instruction directly
  
  – Shine Sun
  Mar 26 at 2:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Adding 2x2 matrix with NaNs
  
  – Cris Luengo
  Mar 26 at 3:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @CrisLuengo, that post is definitely related, though I feel this set of answers if more specific and updated now.
  
  – SecretAgentMan
  Mar 26 at 13:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @SecretAgentMan: I voted to close as duplicate because that question has the same answers as his one (nansum and sum+isnan), even though the questions are not identical. This qestion is more generic, the other one is rather specific, with the cell array complicating the question. If the vote ages away, which seems likely, at least there is a link between the questions now, it will help people find answers.
  
  – Cris Luengo
  Mar 26 at 13:19
  

  
  
  
  

add a comment | 

I use MATLAB version R2015a.

I get a series answer from solving the optimal problem several times, and I want to get their sum and average them. However, some of them are NaN. How do I write code to ignore those NaN and sum the others which are not NaN?

matlab sum average nan

share|improve this question

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

share|improve this question

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

share|improve this question

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

edited Mar 26 at 19:37

Cris Luengo

26.7k66 gold badges2424 silver badges5858 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

asked Mar 26 at 1:44

Shine SunShine Sun

2688 bronze badges

3 Answers
3

active

oldest

votes

2

You can use inbuilt functions as suggested in the above answer. If you want to know the logic and use a loop..you can follow as shown below:

A = [NaN 1 2 NaN 3 4 7 -1 NaN] ; 
count = 0 ;
thesum = 0 ; 
for i = 1:length(A)
 if ~isnan(A(i))
 count = count+1 ;
 thesum = thesum+A(i) ;
 end
end

share|improve this answer

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Aside from this just being the same as sum(A(~isnan(A))), I don't understand what the point of the count variable is?
  
  – Wolfie
  Mar 26 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ohh yes forgot to add......count is used to find mean in case needed.
  
  – Siva Srinivas Kolukula
  Mar 26 at 12:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For future readers: note this is probably the most inefficient solution on this page. MATLAB is optimised for vector operations. Loops can be helpful to foster understanding, but it's good practise to heed the advice of others here suggesting sleeker solutions.
  
  – Wolfie
  Mar 26 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wolfie, thanks for keeping me in check. I guess I meant a way to get the count without looping is count = sum(~isnan(A)).
  
  – SecretAgentMan
  Mar 26 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wolfie I agree with you...my intention was to show with loop for learning things. :)
  
  – Siva Srinivas Kolukula
  Mar 27 at 6:58
  

  
  
  
  

add a comment | 

3

Option 1: toolbox free solution using sum and isnan from base MATLAB.

A = [1 2 3 4 5 6 7 8 9 NaN];
sum(A(~isnan(A))) % No toolbox required

Option 2: nansum (see this answer from OP)
Note: nansum requires the Statistics toolbox.

nansum(A) % Requires Statistics toolbox

Code tested using MATLAB R2018b.

---

Update from comments

Great suggestion from @Cris Luengo for those with more recent versions. Requires no toolbox.

sum(A,'omitnan') % No toolbox required

Another suggestion from @Ben Voigt for some applications. Also requires no toolbox.

sum(A(isfinite(A))) % No toolbox required

share|improve this answer

edited Mar 26 at 16:13

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In the more recent versions of MATLAB you can also do sum(A,'omitnan').
  
  – Cris Luengo
  Mar 26 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Better than ~isnan(A) might be isfinite(A), depending on how the other special values ought to be handled.
  
  – Ben Voigt
  Mar 26 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Great suggestions. Will update answer to include your notes w/ attribution.
  
  – SecretAgentMan
  Mar 26 at 13:05
  

  
  
  
  

add a comment | 

1

You can use the omitnan argument

A = [1 2 3 4 5 6 7 8 9 NaN];
s = sum( A, 'omitnan' )

Note, this is literally the same code as used by the nansum function from the Statistics toolbox, which was introduced before R2006a, so I would think compatibility is pretty good.

share|improve this answer

answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  When was it introduced? I don’t remember for sure, but I think it was after R2015a? (OP specifies this version)
  
  – Cris Luengo
  Mar 26 at 12:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just updated my answer based on @CrisLuengo 's comment then saw your answer here. My mistake. Should I delete that portion of my answer or just leave it @Wolfie?
  
  – SecretAgentMan
  Mar 26 at 13:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Cris I don't have access to that documentation, might be able to dig through change notes, or use doc sum in your version of choice to see whether it's present. As I say, it's literally the same code as nansum (see edit nansum), but I don't know if it always was.
  
  – Wolfie
  Mar 26 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It makes sense that they would rewrite nansum to use the new feature. It likely speeds things up. Core functions are more likely to be implemented in a compiled language, and toolbox functions are more likely to be implemented in M-files (though not always the case).
  
  – Cris Luengo
  Mar 26 at 13:30
  

  
  
  
  

add a comment | 

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2

You can use inbuilt functions as suggested in the above answer. If you want to know the logic and use a loop..you can follow as shown below:

A = [NaN 1 2 NaN 3 4 7 -1 NaN] ; 
count = 0 ;
thesum = 0 ; 
for i = 1:length(A)
 if ~isnan(A(i))
 count = count+1 ;
 thesum = thesum+A(i) ;
 end
end

share|improve this answer

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Aside from this just being the same as sum(A(~isnan(A))), I don't understand what the point of the count variable is?
  
  – Wolfie
  Mar 26 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ohh yes forgot to add......count is used to find mean in case needed.
  
  – Siva Srinivas Kolukula
  Mar 26 at 12:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For future readers: note this is probably the most inefficient solution on this page. MATLAB is optimised for vector operations. Loops can be helpful to foster understanding, but it's good practise to heed the advice of others here suggesting sleeker solutions.
  
  – Wolfie
  Mar 26 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wolfie, thanks for keeping me in check. I guess I meant a way to get the count without looping is count = sum(~isnan(A)).
  
  – SecretAgentMan
  Mar 26 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wolfie I agree with you...my intention was to show with loop for learning things. :)
  
  – Siva Srinivas Kolukula
  Mar 27 at 6:58
  

  
  
  
  

add a comment | 

2

You can use inbuilt functions as suggested in the above answer. If you want to know the logic and use a loop..you can follow as shown below:

A = [NaN 1 2 NaN 3 4 7 -1 NaN] ; 
count = 0 ;
thesum = 0 ; 
for i = 1:length(A)
 if ~isnan(A(i))
 count = count+1 ;
 thesum = thesum+A(i) ;
 end
end

share|improve this answer

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Aside from this just being the same as sum(A(~isnan(A))), I don't understand what the point of the count variable is?
  
  – Wolfie
  Mar 26 at 10:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ohh yes forgot to add......count is used to find mean in case needed.
  
  – Siva Srinivas Kolukula
  Mar 26 at 12:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For future readers: note this is probably the most inefficient solution on this page. MATLAB is optimised for vector operations. Loops can be helpful to foster understanding, but it's good practise to heed the advice of others here suggesting sleeker solutions.
  
  – Wolfie
  Mar 26 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wolfie, thanks for keeping me in check. I guess I meant a way to get the count without looping is count = sum(~isnan(A)).
  
  – SecretAgentMan
  Mar 26 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wolfie I agree with you...my intention was to show with loop for learning things. :)
  
  – Siva Srinivas Kolukula
  Mar 27 at 6:58
  

  
  
  
  

add a comment | 

You can use inbuilt functions as suggested in the above answer. If you want to know the logic and use a loop..you can follow as shown below:

A = [NaN 1 2 NaN 3 4 7 -1 NaN] ; 
count = 0 ;
thesum = 0 ; 
for i = 1:length(A)
 if ~isnan(A(i))
 count = count+1 ;
 thesum = thesum+A(i) ;
 end
end

share|improve this answer

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

A = [NaN 1 2 NaN 3 4 7 -1 NaN] ; 
count = 0 ;
thesum = 0 ; 
for i = 1:length(A)
 if ~isnan(A(i))
 count = count+1 ;
 thesum = thesum+A(i) ;
 end
end

share|improve this answer

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

answered Mar 26 at 5:16

Siva Srinivas KolukulaSiva Srinivas Kolukula

1,22611 gold badge77 silver badges1313 bronze badges

3

Option 1: toolbox free solution using sum and isnan from base MATLAB.

A = [1 2 3 4 5 6 7 8 9 NaN];
sum(A(~isnan(A))) % No toolbox required

Option 2: nansum (see this answer from OP)
Note: nansum requires the Statistics toolbox.

nansum(A) % Requires Statistics toolbox

Code tested using MATLAB R2018b.

---

Update from comments

Great suggestion from @Cris Luengo for those with more recent versions. Requires no toolbox.

sum(A,'omitnan') % No toolbox required

Another suggestion from @Ben Voigt for some applications. Also requires no toolbox.

sum(A(isfinite(A))) % No toolbox required

share|improve this answer

edited Mar 26 at 16:13

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In the more recent versions of MATLAB you can also do sum(A,'omitnan').
  
  – Cris Luengo
  Mar 26 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Better than ~isnan(A) might be isfinite(A), depending on how the other special values ought to be handled.
  
  – Ben Voigt
  Mar 26 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Great suggestions. Will update answer to include your notes w/ attribution.
  
  – SecretAgentMan
  Mar 26 at 13:05
  

  
  
  
  

add a comment | 

3

Option 1: toolbox free solution using sum and isnan from base MATLAB.

A = [1 2 3 4 5 6 7 8 9 NaN];
sum(A(~isnan(A))) % No toolbox required

Option 2: nansum (see this answer from OP)
Note: nansum requires the Statistics toolbox.

nansum(A) % Requires Statistics toolbox

Code tested using MATLAB R2018b.

---

Update from comments

Great suggestion from @Cris Luengo for those with more recent versions. Requires no toolbox.

sum(A,'omitnan') % No toolbox required

Another suggestion from @Ben Voigt for some applications. Also requires no toolbox.

sum(A(isfinite(A))) % No toolbox required

share|improve this answer

edited Mar 26 at 16:13

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In the more recent versions of MATLAB you can also do sum(A,'omitnan').
  
  – Cris Luengo
  Mar 26 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Better than ~isnan(A) might be isfinite(A), depending on how the other special values ought to be handled.
  
  – Ben Voigt
  Mar 26 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Great suggestions. Will update answer to include your notes w/ attribution.
  
  – SecretAgentMan
  Mar 26 at 13:05
  

  
  
  
  

add a comment | 

Option 1: toolbox free solution using sum and isnan from base MATLAB.

A = [1 2 3 4 5 6 7 8 9 NaN];
sum(A(~isnan(A))) % No toolbox required

Option 2: nansum (see this answer from OP)
Note: nansum requires the Statistics toolbox.

nansum(A) % Requires Statistics toolbox

Code tested using MATLAB R2018b.

---

Update from comments

Great suggestion from @Cris Luengo for those with more recent versions. Requires no toolbox.

sum(A,'omitnan') % No toolbox required

Another suggestion from @Ben Voigt for some applications. Also requires no toolbox.

sum(A(isfinite(A))) % No toolbox required

share|improve this answer

edited Mar 26 at 16:13

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

A = [1 2 3 4 5 6 7 8 9 NaN];
sum(A(~isnan(A))) % No toolbox required

nansum(A) % Requires Statistics toolbox

sum(A,'omitnan') % No toolbox required

sum(A(isfinite(A))) % No toolbox required

share|improve this answer

edited Mar 26 at 16:13

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

answered Mar 26 at 3:01

SecretAgentManSecretAgentMan

1,45099 silver badges2424 bronze badges

1

You can use the omitnan argument

A = [1 2 3 4 5 6 7 8 9 NaN];
s = sum( A, 'omitnan' )

Note, this is literally the same code as used by the nansum function from the Statistics toolbox, which was introduced before R2006a, so I would think compatibility is pretty good.

share|improve this answer

answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  When was it introduced? I don’t remember for sure, but I think it was after R2015a? (OP specifies this version)
  
  – Cris Luengo
  Mar 26 at 12:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just updated my answer based on @CrisLuengo 's comment then saw your answer here. My mistake. Should I delete that portion of my answer or just leave it @Wolfie?
  
  – SecretAgentMan
  Mar 26 at 13:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Cris I don't have access to that documentation, might be able to dig through change notes, or use doc sum in your version of choice to see whether it's present. As I say, it's literally the same code as nansum (see edit nansum), but I don't know if it always was.
  
  – Wolfie
  Mar 26 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It makes sense that they would rewrite nansum to use the new feature. It likely speeds things up. Core functions are more likely to be implemented in a compiled language, and toolbox functions are more likely to be implemented in M-files (though not always the case).
  
  – Cris Luengo
  Mar 26 at 13:30
  

  
  
  
  

add a comment | 

1

You can use the omitnan argument

A = [1 2 3 4 5 6 7 8 9 NaN];
s = sum( A, 'omitnan' )

Note, this is literally the same code as used by the nansum function from the Statistics toolbox, which was introduced before R2006a, so I would think compatibility is pretty good.

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answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges

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  When was it introduced? I don’t remember for sure, but I think it was after R2015a? (OP specifies this version)
  
  – Cris Luengo
  Mar 26 at 12:58
  

  
  
  
  


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  Just updated my answer based on @CrisLuengo 's comment then saw your answer here. My mistake. Should I delete that portion of my answer or just leave it @Wolfie?
  
  – SecretAgentMan
  Mar 26 at 13:04
  

  
  
  
  


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  @Cris I don't have access to that documentation, might be able to dig through change notes, or use doc sum in your version of choice to see whether it's present. As I say, it's literally the same code as nansum (see edit nansum), but I don't know if it always was.
  
  – Wolfie
  Mar 26 at 13:26
  

  
  
  
  


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  It makes sense that they would rewrite nansum to use the new feature. It likely speeds things up. Core functions are more likely to be implemented in a compiled language, and toolbox functions are more likely to be implemented in M-files (though not always the case).
  
  – Cris Luengo
  Mar 26 at 13:30
  

  
  
  
  

add a comment | 

You can use the omitnan argument

A = [1 2 3 4 5 6 7 8 9 NaN];
s = sum( A, 'omitnan' )

Note, this is literally the same code as used by the nansum function from the Statistics toolbox, which was introduced before R2006a, so I would think compatibility is pretty good.

share|improve this answer

answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges

A = [1 2 3 4 5 6 7 8 9 NaN];
s = sum( A, 'omitnan' )

share|improve this answer

answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges

answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges

answered Mar 26 at 10:14

WolfieWolfie

17.4k66 gold badges1717 silver badges4646 bronze badges