我正在尝试生成一个输出,该输出将获取数组的最后一个值,然后使用该数组中找到的下一个最低值启动下一个数组。如果没有下一个最低值,我希望它只是结束循环参见我想要得到的答案示例。
9.0000 11.0000 5.0000 7.0000 3.0000 7.0100
7.0000 3.0000 7.0100 9.0000 11.0000 5.0000
3.0000 7.0100 9.0000 11.0000 5.0000 7.0000
我在下面使用的代码只能得到前两行的正确性,并在最后做任何想法如何解决这个问题。
代码:
clc
a=[9,11,5,7,3,7.01];
[a_sorted, a_idx] = sort(a, 2); %sorts array along with getting index values of numbers
a_sorted=a_sorted'; % sort into col
a_idx=a_idx'; % sort into col
a_val_idx=[a_sorted a_idx]; % combine array
loop_amount=length(find(a<a(end))) %how many values are less than the last value, loop this many times
for yy=1:loop_amount
a_val=find(a_val_idx(:,1)<a(end)); %find idx of next lowest value from end
nxt_low_idx_val=a_val_idx(a_val(end),2) %get idx of the next lowest value from end
b=circshift(a,[0 (length(a)-nxt_low_idx_val+1)])
a=b;
end
我得到的结果是
loop_amount = 3
a =
9.0000 11.0000 5.0000 7.0000 3.0000 7.0100
nxt_low_idx_val = 4
a =
7.0000 3.0000 7.0100 9.0000 11.0000 5.0000
nxt_low_idx_val = 5
a =
11.0000 5.0000 7.0000 3.0000 7.0100 9.0000
nxt_low_idx_val = 6
正如您所见,最后一行应为
nxt_low_idx_val = 2
3.0000 7.0100 9.0000 11.0000 5.0000 7.0000
任何想法如何解决这个问题?
由于
答案 0 :(得分:1)
太懒了,看看你的代码。怎么样?
a = [9,11,5,7,3,7.01];
disp(' ')
disp(a) % display original value
len = length(a);
loop_count = sum(a<a(end)); % as per your code
for count = 1:loop_count
b = a(1:end-1); % copy of a, will be overwritten
b(b>a(end)) = NaN; % these values do not count
if(all(isnan(b)))
break % exit if there are no lower values
end
[aux ind] = max(b); % max of the remaing values
perm = mod(ind+(0:len-1),len); % cyclic shift
perm(perm==0) = len; % correct zero to len
a = a(perm); % do the shift
disp(a) % display new value
end
答案 1 :(得分:0)
我只需要在for循环下移动一些东西
clc
a=[9,11,5,7,3,7.01];
loop_amount=length(find(a<a(end))) %how many values are less than the last value, loop this many times
for yy=1:loop_amount
[a_sorted, a_idx] = sort(a, 2); %sorts array along with getting index values of numbers
a_sorted=a_sorted'; % sort into col
a_idx=a_idx'; % sort into col
a_val_idx=[a_sorted a_idx]; % combine array
a_val=find(a_val_idx(:,1)<a(end)); %find idx of next lowest value from end
nxt_low_idx_val=a_val_idx(a_val(end),2) %get idx of the next lowest value from end
b=circshift(a,[0 (length(a)-nxt_low_idx_val+1)])
a=b;
end