Create a character vector and determine which characters are letters.

chr = '123 Maple Street'

chr = 
'123 Maple Street'

TF = isstrprop(chr,'alpha')

TF = 1x16 logical array

   0   0   0   0   1   1   1   1   1   0   1   1   1   1   1   1

Find indices for the letters in chr using TF.

idx = find(TF)

idx = 1×11

     5     6     7     8     9    11    12    13    14    15    16

chr(idx)

ans = 
'MapleStreet'

Create string arrays. Then determine which characters belong to various categories using the isstrprop function.

Create a string scalar and determine which of its characters are numeric digits. Starting in R2017a, you can create strings using double quotes.

str = "123 Maple Street"

str = 
"123 Maple Street"

TF = isstrprop(str,'digit')

TF = 1x16 logical array

   1   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0

Create a nonscalar string array. Determine which characters in each string are whitespace characters. isstrprop returns a cell array in which each cell contains results for a string in str.

str = ["123 Maple St.";"456 Oak St."]

str = 2x1 string
    "123 Maple St."
    "456 Oak St."

TF = isstrprop(str,'wspace')

TF=2×1 cell array
    {1x13 logical}
    {1x11 logical}

To display the results for the second string, str(2), index into TF{2}.

TF{2}

ans = 1x11 logical array

   0   0   0   1   0   0   0   1   0   0   0

Create a cell array of character vectors. Determine which characters are whitespace characters.

C = {'123 Maple St.';'456 Oak St.'}

C = 2x1 cell
    {'123 Maple St.'}
    {'456 Oak St.'  }

TF = isstrprop(C,'wspace')

TF=2×1 cell array
    {1x13 logical}
    {1x11 logical}

Find the punctuation characters in a character vector. isstrprop returns a logical vector indicating which characters belong to that category. Force isstrprop to return the logical vector in a cell array.

chr = 'A horse! A horse! My kingdom for a horse!'

chr = 
'A horse! A horse! My kingdom for a horse!'

TF = isstrprop(chr,'punct','ForceCellOutput',true)

TF = 1x1 cell array
    {1x41 logical}

Find indices for the punctuation marks in chr using TF{1}.

find(TF{1})

ans = 1×3

     8    17    41

Create a numeric array. Determine which numbers correspond to character codes for letters.

X = [77 65 84 76 65 66]

X = 1×6

    77    65    84    76    65    66

TF = isstrprop(X,'alpha')

TF = 1x6 logical array

   1   1   1   1   1   1

isstrprop identifies all the numbers as character codes for letters. Convert the numbers to their corresponding characters with the char function.

c = char(X)

c = 
'MATLAB'