Difference between num2cell(1:4)' and {'1','2','3','4'}' ?

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Laurent Davenne 2017-11-21

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评论： Guillaume 2017-11-22

采纳的回答： Guillaume

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Hi,

I am having an error on creating a table with 'RowNames' property.

truc=array2table([-0.013 -0.470 0.109
                  -0.013 -0.454 0.107
                  -0.012 -0.437 0.104
                   0.009 0.077 -0.088],'VariableNames',{'Scx','SczF','SczR'},'RowNames',num2cell(1:4)');

Main error is: The RowNames property must be a cell array, with each element containing one nonempty character vector.

>> num2cell([1 2 3 4])'

ans =

4×1 cell array
    [1]
    [2]
    [3]
    [4]

>> {'1','2','3','4'}'

deux =

4×1 cell array
    '1'
    '2'
    '3'
    '4'

In both case it is a cell array, but the content is different.

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Stephen23 2017-11-21

编辑：Stephen23 2017-11-21

"In both case it is a cell array, but the content is different."

Yes. You just answered your own question: the content is indeed different. You put scalar numeric into one, and chars into the other.

John D'Errico 2017-11-21

Is it true that the number 1 is equal to the character string that contains the character '1'?

Of course not. So why are you surprised they are not the same?

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Answer 1

Guillaume 2017-11-21

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The simplest way to convert numbers to strings or char arrays since R2016b is:

string(1:15) %that's it!

In earlier versions:

sprintfc('%d', 1:15)  %undocumented function!
arrayfun(@num2str, 1:15, 'UniformOutput', false)  %using only documented functions

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Laurent Davenne 2017-11-21

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Thanks Guillaume,

last one works.

>> (arrayfun(@num2str, 3:0.5:34, 'UniformOutput', false))'

ans =

63×1 cell array
    '3'
    '3.5'
    '4'
    '4.5'
    '5'
    '5.5'...

But what if I need it to be with 1 decimal everywhere?

Guillaume 2017-11-22

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They all work! The last two in any version.

You can pass a format string to num2str:

arrayfun(@(n) num2str(n, '%.1f'), 3:0.5:34, 'UniformOutput', false)

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Answer 2

dpb 2017-11-21

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编辑：dpb 2017-11-21

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As the error message says, the cell content must be character string and Adam notes the difference between the two forms you wrote; the first is numeric values placed into a cell array, the second is the string value but you wrote manually..

To build the equivalent cellstr array for the purpose use

> cellstr(num2str([1:3].','%d'))
ans = 
  '1'
  '2'
  '3'
>>

or similar.

However I'd note there's not a lot of point in creating row names that are just ordinal values in sequence as you can address rows numerical simply by using the row number in array expression syntax.

ERRATUM Corrected the misplaced transpose operator discussed in following comments -- dpb

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Guillaume 2017-11-21

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Comment by Laurent Davenne mistakenly posted as an answer:

Humm....

Format is messed up when numbers go over 10

>> cellstr(num2str(1:14,'%d')')

ans =

27×1 cell array
    '1'
    ''
    '2'
    ''
    '3'
    ''
    '4'
    ''
    '5'
    ''
    '6'
    ''
    '7'
    ''
    '8'
    ''
    '9'
    '1'
    '0'
    '1'
    '1'
    '1'
    '2'
    '1'
    '3'
    '1'
    '4'

in fact I even need it to work for decimals

cellstr(num2str(3:0.5:34,'%d')')

I am quite lost with format spec... :(

dpb 2017-11-21

编辑：dpb 2017-11-21

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Misplaced location of the .' transpose operator to build the column vector, my bad. At the trailing location it takes the horizontal row vector of characters

>> num2str(1:14,'%d')
ans =
1 2 3 4 5 6 7 8 91011121314
>>

and transposes it to a column vector; hence the blank lines. What is needed and was intended is to transpose the numeric vector to column form, then convert it--

>> num2str([1:14].','%2d')
ans =
1
2
3
4
5
6
7
8
9
10
11
12
13
14
>> cellstr(num2str([1:14].','%2d'))
ans = 
  ' 1'
  ' 2'
  ' 3'
  ' 4'
  ' 5'
  ' 6'
  ' 7'
  ' 8'
  ' 9'
  '10'
  '11'
  '12'
  '13'
  '14'
>>

There will be some purists who'll nag me about using the [] where only () are needed, and it is true in a very deeply nested loop you might see a little performance hit, but I simply like the visual representation sufficiently to accept the hit.

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