-3

我有一个矩阵如下

`> y
       1         2         3
1  0.8802216 1.2277843 0.6875047
2  0.9381081 1.3189847 0.2046542
3  1.3245534 0.8221709 0.4630722
4  1.2006974 0.8890464 0.6710844
5  1.2344071 0.8354292 0.7259998
6  1.1670665 0.9214787 0.6826173
7  0.9670581 1.1070461 0.7742342
8  0.8867365 1.2160533 0.7024281
9  0.8235792 1.4424190 0.2030302
10 0.8821301 1.0541099 1.2279813
11 1.1958634 0.9708839 0.4297043
12 1.3542734 0.7747481 0.5119648
13 0.4385487 0.3588158 4.9167998
14 0.8530141 1.3578511 0.3698620
15 0.9651803 0.8426226 1.6132899
16 0.8854192 1.2272616 0.6715839
17 0.7779642 0.8132233 2.3386331
18 0.9936722 1.1629110 0.5083558
19 1.1235897 1.0018480 0.5764672
20 0.7887222 1.3101684 0.7373181
21 2.2276176 0.0000000 0.0000000`

我找到了一条线索，但它可以为整个矩阵提供位置，`

n<-read.table(file.choose(),header=T)

y<-n[,c("1","2","3")]

我的号码=1.12270420185886。

z<-abs(y-my.number)==min(abs(y-my.number)) which(z) [1] 19 `

我也想找到至少 5 个最接近的带有字母 & 列的值，换句话说，我想要矩阵中最接近的 5 个单个值及其位置。

4

1 回答 1

2

我不知道它是什么语言；是R吗？

在程序语言中，我会将所有值保存到映射 (val, (pos)) = (val (row, col); example (0.880..-> (1, 1))，然后按值排序。

然后迭代i<-pos（1到map.size-5），得到diff（pos（i），pos（i+5）），寻找最小值（diff），然后得到值和它们的位置.

这是Scala中的一个解决方案：

val matrix = """1  0.8802216 1.2277843 0.6875047
2  0.9381081 1.3189847 0.2046542
3  1.3245534 0.8221709 0.4630722
4  1.2006974 0.8890464 0.6710844
5  1.2344071 0.8354292 0.7259998
6  1.1670665 0.9214787 0.6826173
7  0.9670581 1.1070461 0.7742342
8  0.8867365 1.2160533 0.7024281
9  0.8235792 1.4424190 0.2030302
10 0.8821301 1.0541099 1.2279813
11 1.1958634 0.9708839 0.4297043
12 1.3542734 0.7747481 0.5119648
13 0.4385487 0.3588158 4.9167998
14 0.8530141 1.3578511 0.3698620
15 0.9651803 0.8426226 1.6132899
16 0.8854192 1.2272616 0.6715839
17 0.7779642 0.8132233 2.3386331
18 0.9936722 1.1629110 0.5083558
19 1.1235897 1.0018480 0.5764672
20 0.7887222 1.3101684 0.7373181
21 2.2276176 0.0000000 0.0000000"""

// split block of text into lines
val lines=matrix.split ("\n")
// split lines into words
val rows = lines.map (l => l.split (" \\+"))
// remove the index from the beginning (1, 2, ... 21) and 
// transform values from Strings to double numbers
// triples is: Array(Array(0.8802216, 1.2277843, 0.6875047), Array(0.9381081, 1.3189847, 0.2046542),
val triples = rows.map (_.tail).map(triple=> triple.map (_.toDouble))
// generate an own index for the rows and columns 
// elems is: elems: Array[Array[(Double, (Int, Int))]] = Array(Array((0.8802216,(0,0)), (1.2277843,(0,1)), (0.6875047,(0,2))), Array((0.9381081,(1,0)), ...
val elems = triples.zipWithIndex.map {t=> t._1.zipWithIndex.map (vc=> (vc._1 -> (t._2, vc._2)))}
// sorted = Array((0.0,(20,1)), (0.0,(20,2)), (0.2030302,(8,2)), (0.2046542,(1,2)), 
val sorted = elems.sortBy (e => e._1)
// delta5 = List(0.3588158, 0.369862, 0.2266741, 0.2338945, 0.10425639999999997, 0.1384938,
val delta5 = sorted.sliding (5, 1).map (q => q(4)._1-q(0)._1).toList
val minindex = delta5.indexOf (delta5.min) // minindex: Int = 29, delta5.min = 0.008824799999999966

// we found the smallest intervall of 5 values beginning at 29: 
(29 to 29 +5).map (sorted (_)) 
res568: scala.collection.immutable.IndexedSeq[(Double, (Int, Int))] = 
    Vector( (0.8802216,(0,0)), 
        (0.8821301,(9,0)), 
        (0.8854192,(15,0)), 
        (0.8867365,(7,0)), 
        (0.8890464,(3,1)), 
        (0.9214787,(5,1)))

由于 Scala 从 0 到 20 和从 0 到 2 计数，您的索引分别从 1 到 3 和 1 到 21 运行，因此您必须将 (1,1) 添加到每个位置=> (1,1), (10 ,1) 等等。

于 2012-04-26T14:04:39.103 回答