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Matlab 循环不规则增量极限_Matlab_For Loop_Data Structures - Fatal编程技术网

Matlab 循环不规则增量极限

Matlab 循环不规则增量极限,matlab,for-loop,data-structures,Matlab,For Loop,Data Structures,我有一个向量‘a’,基于这个向量‘for loop’和‘filter’将看起来像是常数增量 A= [1:1:500]'; % random for i= min(A):10:max(A); filter = A>=i & A<=i+10 ; % upper limit here is according to constant increment end 现在“过滤器”应该是1到11,11到21,…,100到120,120到140,…,440到

我有一个向量‘a’,基于这个向量‘for loop’和‘filter’将看起来像是常数增量

 A= [1:1:500]'; % random         
for i= min(A):10:max(A);
    filter = A>=i  & A<=i+10 ; % upper limit here is according to constant increment
end

现在“过滤器”应该是1到11,11到21,…,100到120,120到140,…,440到480。

你可以这样做。但我相信有更好的办法

DefautIter = 10;
OldI = -9;

for i= [(min(A):10:100),(100:20:200),(200:40:max(A))];
    iteration  = i - OldI;
    OldI = i;
    filter = A>=i  & A<=i+iteration  ; 
end
DefautIter=10;
OldI=-9;
对于i=[(最小(A):10:100),(100:20:200),(200:40:max(A))];
迭代=i-OldI;
OldI=i;

filter=A>=i&A你可以这样做。但我相信有更好的办法

DefautIter = 10;
OldI = -9;

for i= [(min(A):10:100),(100:20:200),(200:40:max(A))];
    iteration  = i - OldI;
    OldI = i;
    filter = A>=i  & A<=i+iteration  ; 
end
DefautIter=10;
OldI=-9;
对于i=[(最小(A):10:100),(100:20:200),(200:40:max(A))];
迭代=i-OldI;
OldI=i;

filter=A>=i&A使用一个简单的循环索引,然后简单地索引到filter-limits数组的元素,怎么样

例如,像这样:

helper = [(min(A):10:100),(100:20:200),(200:40:max(A))]
for i=1:numel(helper)-1;
    filter = A>=helper(i) & A<=helper(i+1)  ; % i do not know how to set upper limit here according to increment
end
helper=[(最小(A):10:100)、(100:20:200)、(200:40:最大(A))]
对于i=1:numel(辅助对象)-1;

filter=A>=helper(i)&A使用一个简单的循环索引,然后简单地索引到filter limits数组的元素,怎么样

例如,像这样:

helper = [(min(A):10:100),(100:20:200),(200:40:max(A))]
for i=1:numel(helper)-1;
    filter = A>=helper(i) & A<=helper(i+1)  ; % i do not know how to set upper limit here according to increment
end
helper=[(最小(A):10:100)、(100:20:200)、(200:40:最大(A))]
对于i=1:numel(辅助对象)-1;

filter=A>=helper(i)&AIt在我看来,您应该使用while循环:
whileii
,并在每次迭代中重新计算新的
ii
。只是一个猜测,因为我并不真正理解您想要获得什么。在我看来,您应该使用while循环:
while ii
,并在每次迭代中重新计算新的
ii
。只是一个猜测,因为我真的不明白你想要得到什么。@bizou;谢谢,但我认为这是行不通的,因为我进一步扩展了上面的问题,以便更好地理解。@bizou;谢谢,但我认为这不会起作用,因为我进一步扩展了上面的问题,以便更好地理解。
helper = [(min(A):10:100),(100:20:200),(200:40:max(A))]
for i=1:numel(helper)-1;
    filter = A>=helper(i) & A<=helper(i+1)  ; % i do not know how to set upper limit here according to increment
end