C#打印出每个级别的卖家数量 [英] C# print out number of sellers for each level

问题描述

嗨！

我刚接触C#，并遇到了一些循环问题.

目前，我有一组卖家，需要根据所售商品的数量来计算.已通过以下条件将它们分配到各自的级别:

 如果(卖家[i] .articles <   50 )
    Seller [i] .level =  1 ;

如果(卖家[i] .articles >  =  50 &&卖方[i].文章<   100 )
    Seller [i] .level =  2 ;

如果(卖家[i] .articles >  =  100 &&卖方[i].文章<   200 )
    Seller [i] .level =  3 ;

如果(卖家[i] .articles >  =  200 )
    Seller [i] .level =  4 ;

但是，当我要计算多少个实际处于各自级别的卖家时，会出现一个小问题.
这可能会提示我要实现的目标:

  int  count =  0 ;
             for ( int  i =  0 ; i <   3 ; i ++)
            {
                如果(卖家[i] .level == i)
                    数++;
Console.WriteLine(" ，卖家[i ] .level，count);

            } 

此致
L

解决方案

我几乎很想做这样的事情..

// store the count of sellers at each level
// see the discussion about the number of elements in this array
int[] levels = new int[5] {0, 0, 0, 0, 0};

然后当你做

if (seller[i].articles < 50)
        seller[i].level = 1;

变成:-

if (seller[i].articles < 50) 
(
    seller[i].level = 1;
    levels[seller[i].level] += 1;
    // or levels[1] += 1;
)

...如此反复进行所有计算，与您建议的不同

我什至用我自己的方法遇到的唯一问题是，数组被索引为"0"，因此上面浪费了level [0](但是与您的1-4级对应)

我必须考虑，如果

int[] levels = new int[4] {0, 0, 0, 0};

并使用

// note the -1 below to map levels 1-4 to array 0-3
levels[seller[i].level -1] += 1;

更好.

我不知道是否有其他人同意，但是我讨厌多次遍历数据集，而一次通过却可以做到...

由于您仅指示4个级别，因此我认为数组是合理的-如果您具有任意数量的级别，则可能会使用哈希映射或诸如此类，其中Seller [i] .level -1是哈希中的索引-映射和计数是值

我仍然很担心(可能太多了，我在这里看不到什么-如果这是代码中唯一可以修改level []的地方，那么还可以，否则我会考虑检查是否是Seller [i] .level或Seller [i] .level -1在数组范围内，还取决于是否将您的1-4级映射到数组下标0-3

那就是我2美分的价值

Hi!

I''m very new to C# and is having a bit of a loop problem.

At the moment I have a group of sellers that needs to be counted by the amount of articles sold. They are already assigned to their respective level by the following conditions:

if (seller[i].articles < 50)
    seller[i].level = 1;

if (seller[i].articles >= 50 && seller[i].articles < 100)
    seller[i].level = 2;

if (seller[i].articles >= 100 && seller[i].articles < 200)
    seller[i].level = 3;

if (seller[i].articles >= 200)
    seller[i].level = 4;

However, a slight problem arises when I want to count how many of the sellers that actually are in respective level.
This might lend a clue to figure out what I''m trying to achieve:

            int count = 0;
            for (int i = 0; i < 3; i++)
            {
                if (seller[i].level == i) 
                    count++;
Console.WriteLine("Numbers of sellers in level {0}: {1}", seller[i].level, count);

            }

Regards,
L

解决方案

I''d almost be tempted to do something like this ..

// store the count of sellers at each level
// see the discussion about the number of elements in this array
int[] levels = new int[5] {0, 0, 0, 0, 0};

and then when you do

if (seller[i].articles < 50)
        seller[i].level = 1;

that becomes something like :-

if (seller[i].articles < 50) 
(
    seller[i].level = 1;
    levels[seller[i].level] += 1;
    // or levels[1] += 1;
)

...and so forth so all the calcs are done in one pass, unlike what you suggest

the only issue I have even with my own approach, is that arrays are indexed by ''0'', therefore levels[0] is wasted in the above (but there''s a correspondance to your levels 1-4)

I''d have to think if

int[] levels = new int[4] {0, 0, 0, 0};

and using

// note the -1 below to map levels 1-4 to array 0-3
levels[seller[i].level -1] += 1;

is better.

I don''t know if anyone else agrees or not, but I hate traversing data sets multiple times where a single pass will do ...

Since you only indicate 4 levels, I think an array is justified - if you had an arbitrary number of levels I''d likely use a hash-map or such where seller[i].level -1 is the index in the hash-map and count is the value

I still worry (probably too much about what I dont see here - if ths is the only place in the code that modifies levels[], then ok, else I''d be thinking about checking that either seller[i].level or seller[i].level -1 was in the range of the array bounds, also depending on the decision or not to map your levels 1-4 to the array subscripts 0-3

thats my 2 cents worth

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