Question

我有一系列具有正值然后是NaN的列，例如：

12    16    10
11    13    9
8     10    7 
7     6     5
4     1     4
2     NaN   2
NaN   NaN   1
NaN   NaN   NaN

如果此矩阵为A，则可以使用以下内容找到每列中第一个NaN的（行）位置：sum(~isnan(A))

我现在想用零替换每个位置的NaN，例如：

12    16    10
11    13    9
8     10    7 
7     6     5
4     1     4
2     0     2
0     NaN   1
NaN   NaN   0

到目前为止，我尝试使用逻辑，索引和循环方法的组合失败了，创建了一个新的矩阵，其中所有NaN都为零，每个找到的行中的所有值都为零，或者仅在第一列或最后一列中的NaN为零。

实现这一目标的最佳方法是什么？感谢。

Answer 1

方法＃1

[m,n] = size(A) %// get size of input data
[v,ind] = max(isnan(A))  %// positions of first nans in each column
ind2 = bsxfun(@plus,ind,[0:n-1]*m)  %// linear indices of those positions
A(ind2(v))=0 %// set values of all those positions to zero

代码在示例输入上运行（稍微不同于有问题的示例） -

A (Input) =
    12    16    10
    11    13     9
     8    10     7
     7     6     5
     4     1     4
     2   NaN     2
     4   NaN     1
     1   NaN   NaN


A (Output) =
    12    16    10
    11    13     9
     8    10     7
     7     6     5
     4     1     4
     2     0     2
     4   NaN     1
     1   NaN     0

方法＃2

如果您想使用sum(~isnan(A)代码，这可能会形成另一种方法，但请注意，这假设一旦NaN元素开始出现在列中，就没有数值，因此＃1是更安全的。这是方法＃2的代码 -

[m,n] = size(A); %// get size of input data
ind = sum(~isnan(A))+1; %// positions of first nans in each column
v = ind<=m; %// position of valid ind values
ind2 = bsxfun(@plus,ind,[0:n-1]*m);  %// linear indices of those positions
A(ind2(v))=0 %// set values of all those positions to zero

Answer 2

基于accumarray的方法：

[ii jj] = find(isnan(A)); %// rows and columns of NaNs
ind = accumarray(jj,ii, [], @min); %// minimum row for each column with NaNs
ind = ind(ind~=0); %// a 0 indicates no NaNs in that column, so it should be removed
A(size(A,1)*(unique(jj)-1)+ind) = 0; %// set those entries to 0 using linear indexing

例如，

A = [ 12    16    10
      11    13     9
       8    10     7
       7     6     5
       4     1     4
       2   NaN     2
       4   NaN   NaN
       1     5   NaN ]

转化为

A = 
    12    16    10
    11    13     9
     8    10     7
     7     6     5
     4     1     4
     2     0     2
     4   NaN     0
     1     5   NaN

Answer 3

基于accumarray的方法：

[ii jj] = find(isnan(A)); %// rows and columns of NaNs
ind = accumarray(jj,ii, [], @min); %// minimum row for each column with NaNs
ind = ind(ind~=0); %// a 0 indicates no NaNs in that column, so it should be removed
A(size(A,1)*(unique(jj)-1)+ind) = 0; %// set those entries to 0 using linear indexing

例如，

A = [ 12    16    10
      11    13     9
       8    10     7
       7     6     5
       4     1     4
       2   NaN     2
       4   NaN   NaN
       1     5   NaN ]

转化为

A = 
    12    16    10
    11    13     9
     8    10     7
     7     6     5
     4     1     4
     2     0     2
     4   NaN     0
     1     5   NaN

MATLAB - 如何用零替换每列中的第一个NaN？

3 个答案: