如果用做循环数组问题数组、问题

2023-09-11 02:47:54 作者：故事只说给懂的人听

这是我的code：

data INDAT8; set INDAT6;
Array myarray{24,27};
goodgroups=0;

do i=2 to 24 by 2;
   do j=2 to 27;

   if myarray[i,j] gt 1 then myarray[i+1,j] = 'bad';
       else if myarray[i,j] eq 1 and myarray[i+1,j] = 1 then myarray[i+1,j]= 'good';
   end;
end;
run;

PROC打印数据= INDAT8; 运行;

proc print data=INDAT8; run;

问题：

我有数据在这个格式化，这只是一个例子：n = 2的 X Y信息

I have the data in this format- it is just an example: n=2 X Y info

2      1       good
2      4       bad

3      2      good

4     1       bad
4      4      good

6       2     good
6       3     good

现在，上面的数据是按排序方式（共7行）。我需要的一组2，3或4行分别与生成的曲线图。另外，在上述的数据，我提出的一组2行。第三行被单独留下，因为这里在第三行没有其它列，以形成一个组。可仅在相同的行形成的基团。不与其它的行。

Now, the above data is in sorted manner (total 7 rows). I need to make a group of 2 , 3 or 4 rows separately and generate a graph. In the above data, I made a group of 2 rows. The third row is left alone as there is no other column in 3rd row to form a group. A group can be formed only within the same row. NOT with other rows.

现在，我将检查这两个行具有好在信息栏与否。如果这两个行具有好 - 构成的组还不错，否则不好。在上面的例子中，第三/最后一组是好的基团。其余的都是不好的组。有一次，我正在与所有的行做的，我会计算总没有。良好的团体形成/总数。的基团。

Now, I will check if both the rows have "good" in the info column or not. If both rows have "good" – the group formed is also good , otherwise bad. In the above example, 3rd /last group is "good" group. Rest are all bad group. Once I’m done with all the rows, I will calculate the total no. of Good groups formed/Total no. of groups.

在上面的例子中，输出将是：总无。优秀组/总数。基的=> 1/3

In the above example, the output will be: Total no. of good groups/Total no. of groups => 1/3.

这是n的情况下= 2（组大小）

This is the case of n=2(size of group)

现在，对于n = 3，我们使组3行对于n = 4，我们使一组4行，找到好的/坏的基团以类似的方式。如果在一个组中的所有行具有良好块的结果是好块，否则不好。

Now, for n=3, we make group of 3 rows and for n=4, we make a group of 4 rows and find the good /bad groups in a similar way. If all the rows in a group has "good" block—the result is good block, otherwise bad.

Example: n= 3

2      1       good
2      4       bad
2     6        good

3      2      good

4     1       good
4      4      good
4    6        good

6       2     good
6       3     good

在上述情况下，我离开了第4行和最后2行，因为我不能让3行与他们组。第一组的结果是坏和最后一组的结果是良好。输出：1 /

In the above case, I left the 4th row and last 2 rows as I can’t make group of 3 rows with them. The first group result is "bad" and last group result is "good". Output: 1/ 2

For n= 4:

2      1       good
2      4       good
2     6        good
2      7       good

3      2      good

4     1       good
4      4      good
4    6        good

6       2     good
6       3     good
6       4     good
6       5     good

在这种情况下，我让一组4，并发现的结果。第5，第6，第7，第8行被留下或忽略。我做2组，每组4行，无一不是好的块。输出：2/2

In this case, I make a group of 4 and finds the result. The 5th,6th,7th,8th row are left behind or ignored. I made 2 groups of 4 rows and both are "good" blocks. Output: 2/2

所以，从N = 2，N-3使用3的输出值，和n =后4我将绘制这些值的曲线图。

So, After getting 3 output values from n=2 , n-3, and n=4 I will plot a graph of these values.

如果您可以帮助任何使用数组，如果做循环的任何语言。这将是巨大的。

If you can help in any any language using array, if and do loop. it would be great.

我可以相应地改变我的code。

I can change my code accordingly.

更新

有此答案不必在SAS。因为它更算法相关的比什么都重要，我会接受建议的任何语言，只要他们展示如何做到这一点使用数组和不要。

The answer for this doesn't have to be in sas. Since it is more algorithm-related than anything, I will accept suggestions in any language as long as they show how to accomplish this using arrays and do.

如果用做循环数组问题数组、问题

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