【3.8】subset/which/[--筛选数据

[]、which与subset是常用的从数据框中提取数据的命令,本博文讨论一下其常规的一些用法

一、[ ]数据的提取

[ 用来提取对象相同的类型,可以包含不止一个元素

[[用来提取额 list或data frame里面的元素,一般只能提取单个的元素

$通过名字来提取 list或data frame的元素;

> x <- c("a", "b", "c", "c", "d", "a")
> x[1]
[1] "a"
> x[2]
[1] "b"
> x[1:4]
[1] "a" "b" "c" "c"
> x[x > "a"]
[1] "b" "c" "c" "d"
> u <- x > "a"
> u
[1] FALSE TRUE TRUE TRUE TRUE FALSE
> x[u]
[1] "b" "c" "c" "d"

矩阵通过 (i ; j)来提取

x <- matrix(1:6, 2, 3)
> x[1, 2]
[1] 3
> x[2, 1]
[1] 2
Indices can also be missing.
> x[1, ]
[1] 1 3 5
> x[, 2]
[1] 3 4
通过<span style="color: #ff0000;">drop = FALSE</span>,让提取出来的元素也是一个矩阵的形式。
> x <- matrix(1:6, 2, 3)
> x[1, 2]
[1] 3
> x[1, 2, drop = FALSE]
[,1]
[1,] 3
> x <- matrix(1:6, 2, 3)
> x[1, ]
[1] 1 3 5
> x[1, , drop = FALSE]
[,1] [,2] [,3]
[1,] 1 3 5

list的提取

> x <- list(foo = 1:4, bar = 0.6)
> x[1]
$foo
[1] 1 2 3 4
> x[[1]]
[1] 1 2 3 4
> x$bar
[1] 0.6
> x[["bar"]]
[1] 0.6
> x["bar"]
$bar
[1] 0.6

通过[]我们得到的是一个叫foo的向量,而通过[[]]

我们得到的仅仅是一个序列,不用告诉我们具体的位置

如果提取好几个对象,则可以这样,

> x <- list(foo = 1:4, bar = 0.6, baz = "hello")
> x[c(1, 3)]
$foo
[1] 1 2 3 4
$baz
[1] "hello"

$仅仅适用于原始的名字,如果改名字了,还是得用[[]]
> x <- list(foo = 1:4, bar = 0.6, baz = "hello")
> name <- "foo"
> x[[name]] ## computed index for `foo'
[1] 1 2 3 4
> x$name ## element `name' doesn't exist!
NULL
> x$foo
[1] 1 2 3 4 ## element `foo' does exist 
The [[ can take an integer sequence.
> x <- list(a = list(10, 12, 14), b = c(3.14, 2.81))
> x[[c(1, 3)]]
[1] 14
> x[[1]][[3]]
[1] 14
> x[[c(2, 1)]]
[1] 3.14


[[和$可以匹配部分名字.
> x <- list(aardvark = 1:5)
> x$a
[1] 1 2 3 4 5
> x[["a"]]
NULL
> x[["a", exact = FALSE]]
[1] 1 2 3 4 5

删掉丢失的数值(NAs).

> x <- c(1, 2, NA, 4, NA, 5)
> bad <- is.na(x)
> x[!bad]
[1] 1 2 4 5

许多不同类型的数据来去掉NA

> x <- c(1, 2, NA, 4, NA, 5)
> y <- c("a", "b", NA, "d", NA, "f")
> good <- complete.cases(x, y)
> good
[1] TRUE TRUE FALSE TRUE FALSE TRUE
> x[good]
[1] 1 2 4 5
> y[good]
[1] "a" "b" "d" "f"

> airquality[1:6, ]
Ozone Solar.R Wind Temp Month Day
1 41 190 7.4 67 5 1
2 36 118 8.0 72 5 2
3 12 149 12.6 74 5 3
4 18 313 11.5 62 5 4
5 NA NA 14.3 56 5 5
6 28 NA 14.9 66 5 6
> good <- complete.cases(airquality)
> airquality[good, ][1:6, ] ##非常强大的complete.cases
Ozone Solar.R Wind Temp Month Day
1 41 190 7.4 67 5 1
2 36 118 8.0 72 5 2
3 12 149 12.6 74 5 3
4 18 313 11.5 62 5 4
7 23 299 8.6 65 5 7

二、which的用法

> zz<-c(5,2,-3,8,12,1)
 > which(zz*zz>8)
 [1] 1 3 4 5
 which反馈回来的满足要求的索引号

 Ozone Solar.R Wind Temp Month Day
 1 41 190 7.4 67 5 1
 2 36 118 8.0 72 5 2
 3 12 149 12.6 74 5 3
 4 18 313 11.5 62 5 4
 5 NA NA 14.3 56 5 5
 6 28 NA 14.9 66 5 6
 Extract the subset of rows of the data frame where Ozone values are above 31 and Temp values are above 90
 > d<- a[which(a$Ozone>31 & a$Temp>90),]
 提取当Month等于6的列表,
 f<- a[which(a$Month==6),]

取只要有一个丰度值大于1%的otu

方法一:

arc_b_1<-arc_b[which(arc_b[,2]>0.01 |arc_b[,3]>0.01 |arc_b[,4]>0.01 |arc_b[,5]>0.01 |arc_b[,6]>0.01 |arc_b[,7]>0.01 |arc_b[,8]>0.01 |arc_b[,9]>0.01 |arc_b[,10]>0.01 |arc_b[,11]>0.01 |arc_b[,12]>0.01 |arc_b[,13]>0.01 |arc_b[,14]>0.01 |arc_b[,15]>0.01 |arc_b[,16]>0.01 |arc_b[,17]>0.01 |arc_b[,18]>0.01 |arc_b[,19]>0.01 |arc_b[,20]>0.01 |arc_b[,21]>0.01 |arc_b[,22]>0.01 |arc_b[,23]>0.01 |arc_b[,24]>0.01),]

方法二:

myfun<-function(x){sum(x>0.01)};
t80<-all_80_percent[,2:(nn+1)];
all_80_percent_1<-all_80_percent[as.logical(apply(t80,1,myfun)),]

大于0.01为存在,0.7为普遍存在
myfun<-function(x){sum(x>=0.01)};
m<-(147-99+1)*0.7;
cun_5<-bac_b[(apply(total,1,myfun)>=m),];

Selecting (Keeping) Variables
 # select variables v1, v2, v3
 myvars <- c("v1", "v2", "v3")
 newdata <- mydata[myvars]
 注意这里用的是[ ]

 # another method
 myvars <- <span style="color: #ff0000;">paste</span>("v", 1:3, sep="")
 newdata <- mydata[myvars]

 # select 1st and 5th thru 10th variables
 newdata <- mydata[c(1,5:10)]

 Excluding (DROPPING) Variables
 # exclude variables v1, v2, v3
 myvars <- names(mydata) %in% c("v1", "v2", "v3")
 newdata <- mydata[!myvars]
 # exclude 3rd and 5th variable
 newdata <- mydata[c(-3,-5)]
 注:符号和!表示“非”的意思
 # delete variables v3 and v5

 mydata$v3 <- mydata$v5 <- NULL
 Selecting Observations
 # first 5 observerations
 newdata <- mydata[1:5,]
 # based on variable values
 newdata <- mydata[ which(mydata$gender=='F'
 & mydata$age > 65), ]
 # or
 attach(newdata)
 newdata <- mydata[ which(gender=='F' & age > 65),]
 detach(newdata)
 Selection using the Subset Function

  重新提取数据的奇数行
 nii<-ni[seq(1,nrow(ni),2),]

判断某个数是否存在

aa$a = c(6,1,4,5,5,1)
if (length(which(aa$a ==100))) print("zzz")
if (length(which(aa$a ==6))) print("zzz")
if (!length(which(aa$a ==6))) print("zzz")

三 、Subset

#using subset function we select all rows that have a value of age greater than or equal to 20 or age less then 10. We keep the ID and Weight columns.

newdata <- subset(mydata, age >= 20 | age < 10,select=c(ID, Weight))

In the next example, we select all men over the age of 25 and we keep variables weight through income (weight, income and all columns between them).

using subset function (part 2)

newdata <- subset(mydata, sex=="m" & age > 25,select=weight:income)
 注:<span style="color: #ff0000;">&表示“和”,“|”表示或</span>

Random Samples Use the sample( ) function to take a random sample of size n from a dataset. #take a random sample of size 50 from a dataset mydata #sample without replacement

mysample <- mydata[sample(1:nrow(mydata), 50,replace=FALSE),]

library(gcookbook) # For the data set
 cdat <- subset(countries, Year==2009 &
 Name %in% c("Canada", "Ireland", "United Kingdom", "United States"))

library(gcookbook) # For the data set
c2009 <- subset(countries, Year==2009,
 select=c(Name, GDP, laborrate, healthexp, infmortality))

pairs(c2009[,2:5])

四、subset筛选与索引筛选的区别

索引筛选没法区分NA,不知道该不该删去

> x<-c(6,1:3,NA,12)
> x
[1] 6 1 2 3 NA 12
> x[x>5]
[1] 6 NA 12
> subset(x,x>5)
[1] 6 12

五、案例分析

问题:我们获得很多个样品,每个样品有很多个otu,在进行数据展示的时候,我没法将所有样品的所有Otu都层现出来,那么问题来了,我如何去层现?

1.测序公司一般给的思路就是针对一个Otu,算出他在所有样品的丰度之和,然后按照丰度大小,选取前100个Otu来层现(上海某家测序公司)

2.选缺只要在一个样品中丰度值大于0.01的这样的otu,这是我现在更倾向的问题,至于这个阈值是多少,可以人为的设定,那好,针对第二种策略用R如何实现呢??

策略1:

xx<-rnorm(10) > xx
 [1]  0.20129745 0.56048658 -0.03100652
 [4] -2.14841137 -0.33119550-0.98065025
 [7] -0.97097716 -0.53269168-0.16772913
[10]  0.53356301

> x1<-matrix(xx,5) > x1
            [,1]       [,2]
[1,]  0.20129745 -0.9806503
[2,]  0.56048658 -0.9709772
[3,] -0.03100652 -0.5326917
[4,] -2.14841137 -0.1677291
[5,] -0.33119550  0.5335630

> x2<-matrix(0.01,5,2) > x2
     [,1] [,2]
[1,] 0.01 0.01
[2,] 0.01 0.01
[3,] 0.01 0.01
[4,] 0.01 0.01
[5,] 0.01 0.01

> yyx1
      [,1]  [,2]
[1,]  TRUE FALSE
[2,]  TRUE FALSE
[3,] FALSE FALSE
[4,] FALSE FALSE
[5,] FALSE  TRUE

> colSums(yy)
[1] 2 1

> rowSums(yy)
[1] 1 1 0 0 1

> x3<-x1[as.logical(rowSums(yy)),] > x3
           [,1]       [,2]
[1,]  0.2012974 -0.9806503
[2,]  0.5604866 -0.9709772
[3,] -0.3311955  0.5335630

策略2:

myfun<-function(x){sum(x>0.01)}
> apply(x1,2,myfun)
[1] 2 1

> apply(x1,1,myfun)
[1] 1 1 0 0 1
x4<-x1[as.logical(apply(x1,1,myfun)),] > x4
           [,1]       [,2]
[1,]  0.2012974 -0.9806503
[2,]  0.5604866 -0.9709772
[3,] -0.3311955  0.5335630

策略3:

otus1<-otus[,2:24]; > nrow(otus1)
[1] 9143
> nrow(otus1)
[1] 9143

for (i in 1:23) {
for (m in 1:9143){
if (otus1[m,i]>0) otus1[m,i]

很明显策略2的方法更好一些,那么问题来了,我要知道在80%的样品中都存在的OTU,那如何弄呢??

nn=mnn),]

> x4
            [,1]       [,2]
[1,]  0.20129745 -0.9806503
[2,]  0.56048658 -0.9709772
[3,] -0.03100652 -0.5326917
[4,] -2.14841137 -0.1677291
[5,] -0.33119550  0.5335630

参考资料:

http://www.statmethods.net/management/subset.html

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