To get the row and column of element in a matrix.

I have 3D matrices.

t(:,:,1) =
     1    10    19    28    37    46    55    64    73    82
     2    11    20    29    38    47    56    65    74    83
     3    12    21    30    39    48    57    66    75    84
     4    13    22    31    40    49    58    67    76    85
     5    14    23    32    41    50    59    68    77    86
     6    15    24    33    42    51    60    69    78    87
     7    16    25    34    43     1    61    70    79    88
     8    17    26    35    44    53     1     1    80    89
     9    18    27    36    45    54    63    72     1     1
     
     t(:,:,2) =
    91   100   109   118   127   136   145   154   163   172
    92   101   110   119   128   137   146   155   164   173
    93   102   111   120   129   138   147   156   165   174
    94   103   112   121   130   139   148   157   166   175
    95   104   113   122   131   140   149   158   167   176
    96   105   114   123   132   141   150   159   168   177
    97   106   115   124   133   142   151   160   169   178
    98   107   116   125   134   143   152   161   170   179
    99   108   117   126   135   144   153   162   171   180

How to know the row and column of 43 in t(:,:,1)?

The result that will come out will like this

a= (7,5,1)

Answer 1

[Walter Roberson]

 idx = find(t == 43);
 [r, c, p] = ind2sub(size(t), idx);
 a = [r, c, p];

[fyza affandi]

Thank you Sir

Answer 2

[madhan ravi]

[idx,idy]=find(ismember(t(:,:,1),43));
result=[idx idy]

[Walter Roberson]

Note that this particular syntax only works properly for 2D arrays, so you would not be able to apply it directly to all of t (but yes you can use it for t(:,:,1) )

[madhan ravi]

but this also works sir Walter

[idx,idy]=find(ismember(t,43))
result=[idx idy ]

[fyza affandi]

Thank you Sir

[madhan ravi]

Anytime :)

[Walter Roberson]

If the 43 is not on the first pane of the 3D array, then the value you get for idy will be tricky to interpret.

When you have a constant value to compare against, the ismember() is doing extra work for no gain. You might as well do

[idx, idy] = find(t==43);
result = [idx idy];

Either way, with or without ismember(), getting out a 2D index for a 3D search has to be used with care.

[madhan ravi]

Thank you sir Walter , you are a role model!

[fyza affandi]

a= [43 50 67]

Why cant I use this method instead of put 43 in it?

[idx, idy] = find(t==a(1));
result = [idx idy];

[Walter Roberson]

t1 = [202 69 224 2 187 248 40 102 184 206;128 11 101 153 42 47 129 148 78 202;142 76 117 169 72 238 145 5 118 73;161 142 54 148 67 12 49 148 13 18;25 248 194 233 140 62 83 238 99 15;63 176 140 163 139 3 183 28 93 163;158 184 92 135 201 172 141 187 74 109;78 143 179 67 222 231 37 248 209 231;196 137 28 14 201 146 97 156 115 107];
t2 = [40 18 61 45 56 130 254 39 140 220;138 47 8 30 223 159 84 35 54 115;239 147 168 45 55 188 35 181 201 167;169 48 41 160 214 59 99 119 135 78;101 75 205 215 220 6 144 29 146 155;67 118 105 131 134 36 162 179 108 72;217 89 84 43 122 197 139 46 184 204;241 82 191 183 227 248 81 205 19 204;97 118 191 232 17 99 41 132 152 244];
t = cat(3, t1, t2);
a= [43 50 67];
[idx, idy] = find(t==a(1));
result = [idx idy];

>> result
result =
     7    14

Now, what does that mean? Row 7, column 14? But there are only 10 columns in t.

In this example, the 43 is located at Row 7, Column 4, Pane 2.

>>  idx = find(t == 43);
 [r, c, p] = ind2sub(size(t), idx);
 a = [r, c, p];
>> a
a =
     7     4     2