是否有可能找到一个向量的非纳米值,但也允许n个nans？例如,如果我有以下数据:
X = [18 3 nan nan 8 10 11 nan 9 14 6 1 4 23 24]; %// input array
thres = 1; % this is the number of nans to allow
我想只保留最长的非nans值序列,但允许在数据中保留'n'个nans.所以,说我愿意保持1纳米,我会有一个输出
X_out = [8 10 11 nan 9 14 6 1 4 23 24]; %// output array
那就是,开头的两个nans已被删除,因为它们超过了'thres'中的值,但是第三个nan本身就可以保存在数据中.我想开发一种方法,可以将thres定义为任何值.
我可以找到非纳米值
Y = ~isnan(X); %// convert to zeros and ones
有任何想法吗？
为了找到含有最多最长的序列threshold时间NaN,我们必须找到一开始说序列(S)的结束.
要生成所有可能的起点,我们可以使用hankel:
H = hankel(X)
H =
18 3 NaN NaN 8 10 11 NaN 9 14 6 1 4 23 24
3 NaN NaN 8 10 11 NaN 9 14 6 1 4 23 24 0
NaN NaN 8 10 11 NaN 9 14 6 1 4 23 24 0 0
NaN 8 10 11 NaN 9 14 6 1 4 23 24 0 0 0
8 10 11 NaN 9 14 6 1 4 23 24 0 0 0 0
10 11 NaN 9 14 6 1 4 23 24 0 0 0 0 0
11 NaN 9 14 6 1 4 23 24 0 0 0 0 0 0
NaN 9 14 6 1 4 23 24 0 0 0 0 0 0 0
9 14 6 1 4 23 24 0 0 0 0 0 0 0 0
14 6 1 4 23 24 0 0 0 0 0 0 0 0 0
6 1 4 23 24 0 0 0 0 0 0 0 0 0 0
1 4 23 24 0 0 0 0 0 0 0 0 0 0 0
4 23 24 0 0 0 0 0 0 0 0 0 0 0 0
23 24 0 0 0 0 0 0 0 0 0 0 0 0 0
24 0 0 0 0 0 0 0 0 0 0 0 0 0 0
现在我们需要找到每一行中的最后一个有效元素.为此,我们可以使用cumsum:
C = cumsum(isnan(H),2)
C =
0 0 1 2 2 2 2 3 3 3 3 3 3 3 3
0 1 2 2 2 2 3 3 3 3 3 3 3 3 3
1 2 2 2 2 3 3 3 3 3 3 3 3 3 3
1 1 1 1 2 2 2 2 2 2 2 2 2 2 2
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
每行的结束点是一个,其中相应的元素C最多threshold:
threshold = 1;
T = C<=threshold
T =
1 1 1 0 0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
最后一个有效元素使用:
[~,idx]=sort(T,2);
lastone=idx(:,end)
lastone =
3 2 1 4 15 15 15 15 15 15 15 15 15 15 15
我们必须确保每行的实际长度得到遵守:
lengths = length(X):-1:1;
real_length = min(lastone,lengths);
[max_length,max_idx] = max(real_length)
max_length =
11
max_idx =
5
如果有更多相等的最大长度的序列,我们只需取第一个并显示它:
selected_max_idx = max_idx(1);
H(selected_max_idx, 1:max_length)
ans =
8 10 11 NaN 9 14 6 1 4 23 24
完整的脚本
X = [18 3 nan nan 8 10 11 nan 9 14 6 1 4 23 24];
H = hankel(X);
C = cumsum(isnan(H),2);
threshold = 1;
T = C<=threshold;
[~,idx]=sort(T,2);
lastone=idx(:,end)';
lengths = length(X):-1:1;
real_length = min(lastone,lengths);
[max_length,max_idx] = max(real_length);
selected_max_idx = max_idx(1);
H(selected_max_idx, 1:max_length)
一种可能的方法是Y = double(~isnan(X));用一个窗口进行卷积n,其中n减少直到找到可接受的子序列."可接受"意味着子序列至少包含n-thres一个,即卷积至少给出n-thres.
Y = double(~isnan(X));
for n = numel(Y):-1:1 %// try all possible sequence lengths
w = find(conv(Y,ones(1,n),'valid')>=n-thres); %// is there any acceptable subsequence?
if ~isempty(w)
break
end
end
result = X(w:w+n-1);
卷积Y用的窗口n的人(如在方法1)等同于计算的累加值Y,然后采取与差异n间距.这在操作次数方面更有效.
Y = double(~isnan(X));
Z = cumsum(Y);
for n = numel(Y):-1:1
w = find([Z(n) Z(n+1:end)-Z(1:end-n)]>=n-thres);
if ~isempty(w)
break
end
end
result = X(w:w+n-1);
这基本上一次计算方法1中循环的所有迭代.
Y = double(~isnan(X));
z = conv2(Y, tril(ones(numel(Y))));
[nn, ww] = find(bsxfun(@ge, z, (1:numel(Y)).'-thres)); %'
[n, ind] = max(nn);
w = ww(ind)-n+1;
result = X(w:w+n-1);