[]、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;
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> 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" |
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<strong>矩阵通过 (i ; j)来提取</strong> 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<strong><span style="color: #ff0000;">drop = FALSE</span>，让提取出来的元素也是一个矩阵的形式</strong>。 > 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 |
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<strong>list的提取</strong> > 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 $baz [1] "hello" |
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<strong>$仅仅适用于原始的名字，如果改名字了，还是得用[[]]</strong> > 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 |
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<strong>[[和$可以匹配部分名字.</strong> > x <- list(aardvark = 1:5) > x$a [1] 1 2 3 4 5 > x[["a"]] NULL > x[["a", exact = FALSE]] [1] 1 2 3 4 5 |
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<strong>删掉丢失的数值(NAs).</strong> > x <- c(1, 2, NA, 4, NA, 5) > bad <-<strong> is.na(x)</strong> > <strong>x[!bad]</strong> [1] 1 2 4 5 |
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<strong>许多不同类型的数据来去掉NA</strong> > 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" |
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><strong> airquality[1:6, ]</strong> 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 > <strong>good <- complete.cases(airquality)</strong> > <strong>airquality[good, ][1:6, ]</strong> ##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
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> 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 > <strong>d<- a[which(a$Ozone>31 & a$Temp>90),]</strong> 提取当Month等于6的列表， f<- <strong>a[which(a$Month==6),]</strong> <strong>取只要有一个丰度值大于1%的otu</strong> 方法一： arc_b_1<-arc_b[which(arc_b[,2]>0.01 |arc_b[,3]>0.01 |arc_b[,4]>0.01 |arc_b[,5]>0.01 |arc_b[,6]>0.01 |arc_b[,7]>0.01 |arc_b[,8]>0.01 |arc_b[,9]>0.01 |arc_b[,10]>0.01 |arc_b[,11]>0.01 |arc_b[,12]>0.01 |arc_b[,13]>0.01 |arc_b[,14]>0.01 |arc_b[,15]>0.01 |arc_b[,16]>0.01 |arc_b[,17]>0.01 |arc_b[,18]>0.01 |arc_b[,19]>0.01 |arc_b[,20]>0.01 |arc_b[,21]>0.01 |arc_b[,22]>0.01 |arc_b[,23]>0.01 |arc_b[,24]>0.01),] 方法二： myfun<-function(x){sum(x>0.01)}; t80<-all_80_percent[,2:(nn+1)]; all_80_percent_1<-all_80_percent[as.logical(apply(t80,1,myfun)),] 大于0.01为存在，0.7为普遍存在 myfun<-function(x){sum(x>=0.01)}; m<-(147-99+1)*0.7; cun_5<-bac_b[(apply(total,1,myfun)>=m),]; Selecting (Keeping) Variables # select variables v1, v2, v3 <strong>myvars <- c("v1", "v2", "v3")</strong> <strong> newdata <- mydata[myvars]</strong> 注意这里用的是[ ] # another method myvars <- <span style="color: #ff0000;">paste</span>("v", 1:3, sep="") newdata <- mydata[myvars] # select 1st and 5th thru 10th variables newdata <- mydata[c(1,5:10)] Excluding (DROPPING) Variables # exclude variables v1, v2, v3 myvars <- names(mydata) %in% c("v1", "v2", "v3") newdata <- mydata[!myvars] # exclude 3rd and 5th variable newdata <- mydata[c(-3,-5)] 注：符号和！表示“非”的意思 # delete variables v3 and v5 mydata$v3 <- mydata$v5 <- NULL Selecting Observations # first 5 observerations newdata <- mydata[1:5,] # based on variable values newdata <- mydata[ which(mydata$gender=='F' & mydata$age > 65), ] # or attach(newdata) newdata <- mydata[ which(gender=='F' & age > 65),] detach(newdata) Selection using the Subset Function <strong>重新提取数据的奇数行</strong> nii<-ni[seq(1,nrow(ni),2),] |
判断某个数是否存在
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aa$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.
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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)
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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
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mysample <- mydata[sample(1:nrow(mydata), 50,replace=FALSE),] |
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library(gcookbook) # For the data set cdat <- subset(countries, Year==2009 & Name %in% c("Canada", "Ireland", "United Kingdom", "United States")) |
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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：
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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：
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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：
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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？？
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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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