如何根据条件替换某些列值?
我有一个矩阵 A
,
A=
4.0000 120.0000 92.0000 0 0 37.6000 0.1910 30.0000
10.0000 168.0000 74.0000 0 0 38.0000 0.5370 34.0000
10.0000 139.0000 80.0000 0 0 27.1000 1.4410 57.0000
1.0000 139.0000 60.0000 23.0000 846.0000 30.1000 0.3980 59.0000
5.0000 136.0000 72.0000 19.0000 175.0000 25.8000 0.5870 51.0000
7.0000 121.0000 0 0 0 30.0000 0.4840 32.0000
我想做两件事:
- 将第一列中大于 5 的值替换为 0。
- 在第二列中,如果值在 121- 范围内130,将其替换为 0。如果它们在 131-140 范围内,则将其替换为 1,将 141-150 替换为 2,将 151-160 替换为 3,依此类推。
因此,所需的结果矩阵将是:
A=
4.0000 0.0000 92.0000 0 0 37.6000 0.1910 30.0000
0.0000 4.0000 74.0000 0 0 38.0000 0.5370 34.0000
0.0000 1.0000 80.0000 0 0 27.1000 1.4410 57.0000
1.0000 1.0000 60.0000 23.0000 846.0000 30.1000 0.3980 59.0000
5.0000 1.0000 72.0000 19.0000 175.0000 25.8000 0.5870 51.0000
0.0000 0.0000 0 0 0 30.0000 0.4840 32.0000
我怎样才能完成此操作?
我正在尝试这样的事情:
counter=1;
for i = 1: rows
if A(i,1) > 5
A(i ,1) = 0;
end
if A(i,2) > 120 && A(i,2) < 130
A(i ,2) = 0;
end
counter = counter+1;
end
使用案例可以解决问题吗?
I have a matrix A
such that
A=
4.0000 120.0000 92.0000 0 0 37.6000 0.1910 30.0000
10.0000 168.0000 74.0000 0 0 38.0000 0.5370 34.0000
10.0000 139.0000 80.0000 0 0 27.1000 1.4410 57.0000
1.0000 139.0000 60.0000 23.0000 846.0000 30.1000 0.3980 59.0000
5.0000 136.0000 72.0000 19.0000 175.0000 25.8000 0.5870 51.0000
7.0000 121.0000 0 0 0 30.0000 0.4840 32.0000
I want to do two things:
- Replace the values of the first column that are greater than 5 by 0.
- In the second column, if the values are within the range 121-130, replace them by 0. If they are within the range 131-140, replace by 1, 141-150 by 2, 151-160 by 3, etc.
So the desired result matrix would be:
A=
4.0000 0.0000 92.0000 0 0 37.6000 0.1910 30.0000
0.0000 4.0000 74.0000 0 0 38.0000 0.5370 34.0000
0.0000 1.0000 80.0000 0 0 27.1000 1.4410 57.0000
1.0000 1.0000 60.0000 23.0000 846.0000 30.1000 0.3980 59.0000
5.0000 1.0000 72.0000 19.0000 175.0000 25.8000 0.5870 51.0000
0.0000 0.0000 0 0 0 30.0000 0.4840 32.0000
How can I accomplish this?
I was trying something like this:
counter=1;
for i = 1: rows
if A(i,1) > 5
A(i ,1) = 0;
end
if A(i,2) > 120 && A(i,2) < 130
A(i ,2) = 0;
end
counter = counter+1;
end
Would using a case do the trick?
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您可以像这样修改
A
的前 2 列:上面使用 矩阵索引和向量化运算避免 for 循环或 case 语句。
You can modify the first 2 columns of
A
like so:The above uses matrix indexing and vectorized operations to avoid for loops or case statements.