1

我有一个A640x1 的单元格，其中每个单元格的值A(i,1)因行而异，例如：

A(1,1)=[]
A(2,1)=[1]
A(3,1)=[1,2,3]

我想要做的是转换A成这样：

A(1,1)=[]
A(2,1)=1
A(3,1)=1; A(3,2)=2; A(3,3)=3

这样每个单元格只有一个值。那么，如果有人可以帮助我，我该怎么做呢？这是谁A看起来像的一部分

A=
[]
0
0
0
145
[144;192]
[145;197;307]
4

1 回答 1

3

例子：

% first lets create some random cellarray containing data of different length
A = cell(10,1);
for i=1:numel(A)
    A{i} = rand(1,randi([0 5]));
end

% determine the maximum number of columns
colnum = max(cellfun(@numel, A));

% create new cellarray, and fill each row
B = cell(numel(A),colnum);
for i=1:numel(A)
    if isempty(A{i}), continue; end
    B(i,1:numel(A{i})) = num2cell(A{i});
end

所以我有第一个单元格数组：

>> celldisp(A)
A{1} =
    0.2217    0.1174    0.2967    0.3188
A{2} =
    0.5079    0.0855
A{3} =
    0.8010
A{4} =
     []
A{5} =
    0.7303    0.4886    0.5785    0.2373    0.4588
A{6} =
    0.5468    0.5211    0.2316    0.4889    0.6241
A{7} =
    0.3955    0.3674    0.9880    0.0377
A{8} =
    0.9133    0.7962    0.0987    0.2619    0.3354
A{9} =
    0.1366    0.7212    0.1068    0.6538
A{10} =
    0.7791    0.7150

变成：

>> B
B = 
    [0.2217]    [0.1174]    [0.2967]    [0.3188]          []
    [0.5079]    [0.0855]          []          []          []
    [0.8010]          []          []          []          []
          []          []          []          []          []
    [0.7303]    [0.4886]    [0.5785]    [0.2373]    [0.4588]
    [0.5468]    [0.5211]    [0.2316]    [0.4889]    [0.6241]
    [0.3955]    [0.3674]    [0.9880]    [0.0377]          []
    [0.9133]    [0.7962]    [0.0987]    [0.2619]    [0.3354]
    [0.1366]    [0.7212]    [0.1068]    [0.6538]          []
    [0.7791]    [0.7150]          []          []          []

我们将元素作为 2D 元胞数组访问：

>> B{5,3}
ans =
    0.5785

编辑：

这是一个稍微不同的实现：

len = cellfun(@numel, A);
padding = arrayfun(@(n)cell(1,n), max(len)-len, 'UniformOutput',false);
B = cellfun(@num2cell, A, 'UniformOutput',false);
B = cellfun(@(b,pad) [b{:} pad], B, padding, 'UniformOutput',false);
B = cat(1, B{:});

编辑2：

请注意，上面将返回一个 2D元胞数组。这是因为常规数字矩阵不能是锯齿状的（所有行必须具有相同的长度）。

如果要获得 2D numeric matrix，则必须插入NaN值作为填充符：

C = B;
C(cellfun(@isempty,C))= {NaN};
C = cell2mat(C)

输出：

>> C
C =
    0.2217    0.1174    0.2967    0.3188       NaN
    0.5079    0.0855       NaN       NaN       NaN
    0.8010       NaN       NaN       NaN       NaN
       NaN       NaN       NaN       NaN       NaN
    0.7303    0.4886    0.5785    0.2373    0.4588
    0.5468    0.5211    0.2316    0.4889    0.6241
    0.3955    0.3674    0.9880    0.0377       NaN
    0.9133    0.7962    0.0987    0.2619    0.3354
    0.1366    0.7212    0.1068    0.6538       NaN
    0.7791    0.7150       NaN       NaN       NaN
于 2013-08-26T05:21:54.800 回答