根据序列前后的值替换NaN序列 [英] Replace NaN sequence according to values before and after the sequence

问题描述

如果有人可以帮助我解决这个问题，我将不胜感激...

I would appreciate if someone can help me with this problem...

我有一个向量

A = [NaN 1 1 1 1 NaN NaN NaN NaN NaN 2 2 2 NaN NaN NaN 2 NaN NaN 3 NaN NaN];

我想根据此逻辑填充NaN值.

I would like to fill the NaN values according to this logic.

1)如果NaN序列之前的值与序列后面的值不同=>将一半的NaN分配给第一个值，将一半的NaN分配给第二个值

1) if the value that precedes the sequence of NaN is different from the one that follows the sequence => assign half of the NaNs to the first value and half to the second value

2)如果NaN序列在2个相等的值之间=>用该值填充NaN.

2) if the NaN seqence is between 2 equal values => fill the NaN with that value.

A应为:

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

我把1设成小数点是因为我将该值分配给了前一半... NaN的序列是奇数.

I have put one 1 within brakets because I assigned that value to the first half...the sequence of NaNs is odd.

推荐答案

我在没有MATLAB的情况下在手机中键入了此信息-因此可能会出现一些问题.但这应该很接近:

I am typing this in my phone, without MATLAB - so there can be some issues. But this should be close:

t = 1:numel(A);
Anew = interp1(t(~isnan(A)),A(~isnan(A)),t,'nearest','extrap');

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