R实现的最小二乘lsfit函数学习

1.源码 

function (x, y, wt = NULL, intercept = TRUE, tolerance = 1e-07, 
          yname = NULL) 
{
  x <- as.matrix(x)
  y <- as.matrix(y)
  xnames <- colnames(x)#x的列名
  if (is.null(xnames)) {
    if (ncol(x) == 1L) #赋予列名
      xnames <- "X"
    else xnames <- paste0("X", 1L:ncol(x))
  }
  if (intercept) {#如果有截距的话,那么就多赋值一个列。
    x <- cbind(1, x)#按列合并,那么行数一样,即每行为一个样本
    xnames <- c("Intercept", xnames)
  }
  if (is.null(yname) && ncol(y) > 1) 
    yname <- paste0("Y", 1L:ncol(y))
  good <- complete.cases(x, y, wt)#去除NA值。这里暗示了x y行数是一样的?
  dimy <- dim(as.matrix(y))
  if (any(!good)) {#如果至少有一个是false
    warning(sprintf(ngettext(sum(!good), "%d missing value deleted", 
                             "%d missing values deleted"), sum(!good)), domain = NA)
    x <- as.matrix(x)[good, , drop = FALSE]
    y <- as.matrix(y)[good, , drop = FALSE]
    wt <- wt[good]
  }
  nrx <- NROW(x)
  ncx <- NCOL(x)
  nry <- NROW(y)
  ncy <- NCOL(y)
  nwts <- length(wt)
  if (nry != nrx) #如果x的样本数与y的样本数不相等
    stop(sprintf(paste0(ngettext(nrx, "'X' matrix has %d case (row)", 
                                 "'X' matrix has %d cases (rows)"), ", ", ngettext(nry, 
                                                                                   "'Y' has %d case (row)", "'Y' has %d cases (rows)")), 
                 nrx, nry), domain = NA)
  if (nry < ncx) 
    stop(sprintf(paste0(ngettext(nry, "only %d case", "only %d cases"), 
                        ", ", ngettext(ncx, "but %d variable", "but %d variables")), 
                 nry, ncx), domain = NA)
  if (!is.null(wt)) {#用于加权最小二乘
    if (any(wt < 0)) 
      stop("negative weights not allowed")
    if (nwts != nry) 
      stop(gettextf("number of weights = %d should equal %d (number of responses)", 
                    nwts, nry), domain = NA)
    wtmult <- sqrt(wt)
    if (any(wt == 0)) {
      xzero <- as.matrix(x)[wt == 0, ]
      yzero <- as.matrix(y)[wt == 0, ]
    }
    x <- x * wtmult
    y <- y * wtmult
    invmult <- 1/ifelse(wt == 0, 1, wtmult)
  }
  z <- .Call(C_Cdqrls, x, y, tolerance, FALSE)#调用C中的函数
  resids <- array(NA, dim = dimy)
  dim(z$residuals) <- c(nry, ncy)
  if (!is.null(wt)) {
    if (any(wt == 0)) {
      if (ncx == 1L) 
        fitted.zeros <- xzero * z$coefficients
      else fitted.zeros <- xzero %*% z$coefficients
      z$residuals[wt == 0, ] <- yzero - fitted.zeros
    }
    z$residuals <- z$residuals * invmult
  }
  resids[good, ] <- z$residuals#所有不含NA的行的残差被赋值为计算所得的残差
  if (dimy[2L] == 1 && is.null(yname)) {#如果y只有一列
    resids <- drop(resids)#此时会变成一个向量
    names(z$coefficients) <- xnames#系数被赋值为x的列名,
    #因为一列正好对应的是一个变量,行对应的是样本数
  }
  else {
    colnames(resids) <- yname
    colnames(z$effects) <- yname
    dim(z$coefficients) <- c(ncx, ncy)#是5行,100列,正好就是系数矩阵H
    dimnames(z$coefficients) <- list(xnames, yname)#确实是对应,
    #行是细胞类型的名字,列是基因名字。
  }
  z$qr <- as.matrix(z$qr)
  colnames(z$qr) <- xnames
  output <- list(coefficients = z$coefficients, residuals = resids)
  #输出中有系数和残差
  if (z$rank != ncx) {
    xnames <- xnames[z$pivot]
    dimnames(z$qr) <- list(NULL, xnames)
    warning("'X' matrix was collinear")
  }
  if (!is.null(wt)) {
    weights <- rep.int(NA, dimy[1L])
    weights[good] <- wt
    output <- c(output, list(wt = weights))
  }
  rqr <- list(qt = drop(z$effects), qr = z$qr, qraux = z$qraux, 
              rank = z$rank, pivot = z$pivot, tol = z$tol)
  oldClass(rqr) <- "qr"
  output <- c(output, list(intercept = intercept, qr = rqr))
  #这里intercept只是一个布尔值???
  return(output)
}

//感觉很坑啊,计算的部分是使用了

 z <- .Call(C_Cdqrls, x, y, tolerance, FALSE)#调用C中的函数

> stats:::C_Cdqrls
$`name`
[1] "Cdqrls"

$address
<pointer: 0x00000000025f96b0>
attr(,"class")
[1] "RegisteredNativeSymbol"

$dll
DLL name: stats
Filename: D:/RRsetup/R-3.5.1/library/stats/libs/x64/stats.dll
Dynamic lookup: FALSE

$numParameters
[1] 4

attr(,"class")
[1] "CallRoutine"      "NativeSymbolInfo"

这个就进行了最小二乘,具体的过程还是看不到....

//那似乎这个lsfit函数除了调用C函数计算之外,似乎也没什么了,就是去掉NA值,并且对结果进行整理,赋值列名行名之类的。

2.其中一些函数学习

2.1 paste0函数

> xy<-paste0("X",1:5)
> xy
[1] "X1" "X2" "X3" "X4" "X5"

 就是对其进行粘贴,形成一个字符串向量。

2.2 complete.cases函数——去除空值

转自:http://blog.sina.com.cn/s/blog_59990a450101qnvy.html

Value返回值为

A logical vector specifying which observations/rows have no missing values across the entire sequence.
一个逻辑向量,指明哪一行有缺失值NA

下面用实例来说明这两个函数的作用:

这是一个数据框final:
  gene hsap mmul mmus rnor cfam 
1 ENSG00000208234 0 NA NA NA NA 
2 ENSG00000199674 0 2 2 2 2 
3 ENSG00000221622 0 NA NA NA NA 
4 ENSG00000207604 0 NA NA 1 2 
5 ENSG00000207431 0 NA NA NA NA 
6 ENSG00000221312 0 1 2 3 2
如果要去除有NA的行,则可用:
final[complete.cases(final),] 
也可用 na.omit(final)

//对行进行检测,没有NA的行返回的是true,那么自然被保存。

上述运行结果如下:

  gene hsap mmul mmus rnor cfam 
2 ENSG00000199674 0 2 2 2 2 
6 ENSG00000221312 0 1 2 3 2

如果想过滤部分列:

final[complete.cases(final[,5:6]),]

即只对5、6列进行判断。

   gene hsap mmul mmus rnor cfam 
2 ENSG00000199674 0 2 2 2 2 
4 ENSG00000207604 0 NA NA 1 2 
6 ENSG00000221312 0 1 2 3 2

这样第四行含有空值,但是,我们的命令是只过滤掉第5列,第6列中含有NA的行。

2.3 any函数 

The value returned is TRUE if at least one of the values in x is TRUE, and FALSE if all of the values in x are FALSE 

//即如果有一个值为真,那么则返回真;如果全部为假,那么则返回假。

> range(x <- sort(round(stats::rnorm(10) - 1.2, 1)))
[1] -2.6  0.5
> if(any(x < 0)) cat("x contains negative values
")
x contains negative values
#

2.4 range函数

取向量中最大值与最小值。

> (r.x <- range(stats::rnorm(100)))
[1] -1.805257  3.410545

#产生100个服从正态分布的数,使用range函数取最值。

2.5 rnorm函数

rnorm(n, mean = 0, sd = 1)

产生n个符合生态分布的数,缺省的均值是0,标准差是1

2.6 ngettext函数

> for(n in 0:3)
+     print(sprintf(ngettext(n, "%d variable has missing values",
+                            "%d variables have missing values"),
+                   n))
[1] "0 variables have missing values"
[1] "1 variable has missing values"
[1] "2 variables have missing values"
[1] "3 variables have missing values"

//我认为是和C语言中的printf是比较像的,就是将其中的数进行匹配即可。

2.7 stop函数

> iter <- 12
> try(if(iter > 10) stop("too many iterations"))
Error in try(if (iter > 10) stop("too many iterations")) : 
  too many iterations

#能够终止函数的执行

2.8 drop函数

对于多维数据,去掉长度为1的维度。

> dim(drop(array(1:12, dim = c(1,3,1,1,2,1,2))))
[1] 3 2 2

上例中,即去掉第一维、第三四维、第六维这四个维度,那么还剩下的是这些。

> x<-drop(array(1:12, dim = c(1,3,1,1,2,1,2)))
> x
, , 1

     [,1] [,2]
[1,]    1    4
[2,]    2    5
[3,]    3    6

, , 2

     [,1] [,2]
[1,]    7   10
[2,]    8   11
[3,]    9   12

dim是给数组赋予维数:

> z<-c(1:6)
> dim(z)<-c(2,3)
> z
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6
> z<-c(1:6)
> dim(z)<-c(1,6)
> z
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    2    3    4    5    6
> drop(z)
[1] 1 2 3 4 5 6
> z<-c(1:6)
> dim(z)<-c(6,1)
> z
     [,1]
[1,]    1
[2,]    2
[3,]    3
[4,]    4
[5,]    5
[6,]    6
> drop(z)
[1] 1 2 3 4 5 6

#试验如果维度有一个为1的话,那么会是什么情况

那么似乎是去掉那个为1的维度,然后其他正常,此时

> z<-drop(z)
> dim(z)
NULL
原文地址:https://www.cnblogs.com/BlueBlueSea/p/10243059.html