DAY7 生信技能树-数据挖掘第三期学习笔记

Ruizheng 的学习笔记

感谢 生信技能树 小洁老师

stringr

rm(list = ls())
if(!require(stringr))install.packages('stringr')
## Loading required package: stringr
library(stringr)

x <- "The birch canoe slid on the smooth planks."

1.检测字符串长度

length(x)
## [1] 1
str_length(x)
## [1] 42

2.字符串拆分与组合

str_split(x," ")
## [[1]]
## [1] "The"     "birch"   "canoe"   "slid"    "on"      "the"     "smooth" 
## [8] "planks."
x2 = str_split(x," ")[[1]]
str_c(x2,collapse = " ")
## [1] "The birch canoe slid on the smooth planks."
str_c(x2,1234,sep = "+")
## [1] "The+1234"     "birch+1234"   "canoe+1234"   "slid+1234"    "on+1234"     
## [6] "the+1234"     "smooth+1234"  "planks.+1234"

3.提取字符串的一部分

str_sub(x,5,9)
## [1] "birch"

4.大小写转换

str_to_upper(x2)
## [1] "THE"     "BIRCH"   "CANOE"   "SLID"    "ON"      "THE"     "SMOOTH" 
## [8] "PLANKS."
str_to_lower(x2)
## [1] "the"     "birch"   "canoe"   "slid"    "on"      "the"     "smooth" 
## [8] "planks."
str_to_title(x2)
## [1] "The"     "Birch"   "Canoe"   "Slid"    "On"      "The"     "Smooth" 
## [8] "Planks."

5.字符串排序

str_sort(x2)
## [1] "birch"   "canoe"   "on"      "planks." "slid"    "smooth"  "the"    
## [8] "The"

6.字符检测

str_detect(x2,"h")
## [1]  TRUE  TRUE FALSE FALSE FALSE  TRUE  TRUE FALSE
str_starts(x2,"T")
## [1]  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
str_ends(x2,"e")
## [1]  TRUE FALSE  TRUE FALSE FALSE  TRUE FALSE FALSE

与sum和mean连用,可以统计匹配的个数和比例

sum(str_detect(x2,"h"))
## [1] 4
mean(str_detect(x2,"h"))
## [1] 0.5

7.提取匹配到的字符串

str_subset(x2,"h")
## [1] "The"    "birch"  "the"    "smooth"

8.字符计数

str_count(x," ")
## [1] 7
str_count(x2,"o")
## [1] 0 0 1 0 1 0 2 0

9.字符串替换

str_replace(x2,"o","A")
## [1] "The"     "birch"   "canAe"   "slid"    "An"      "the"     "smAoth" 
## [8] "planks."
str_replace_all(x2,"o","A")
## [1] "The"     "birch"   "canAe"   "slid"    "An"      "the"     "smAAth" 
## [8] "planks."

结合正则表达式更加强大

练习6-2
#Bioinformatics is a new subject of genetic data collection,analysis and dissemination to the research community.
#1.将上面这句话作为一个长字符串,赋值给tmp
tmp <- "Bioinformatics is a new subject of genetic data collection,analysis and dissemination to the research community."

#2.拆分为一个由单词组成的向量,赋值给tmp2(注意标点符号)
tmp2 = tmp %>% 
  str_replace(","," ") %>%
  str_remove("[.]") %>% 
  str_split(" ")
tmp2 = tmp2[[1]]

#3.用函数返回这句话中有多少个单词。
length(tmp2)
## [1] 16
#4.用函数返回这句话中每个单词由多少个字母组成。
a <- str_count(tmp2);a
##  [1] 14  2  1  3  7  2  7  4 10  8  3 13  2  3  8  9
#5.统计tmp2有多少个单词中含有字母"e"
sum(str_detect(tmp2,"e"))
## [1] 7

str_detect(x, “h”) - 返回等长的逻辑值向量

条件语句

if条件语句:如果。。。就。。。,否则。。。

if(一个逻辑值){ 一段代码 } else { 一段代码 }

(1)只有if没有else,那么条件是FALSE时就什么都不做

i = -1
if (i<0) print('up')
## [1] "up"
if (i>0) print('up')

if(!require(tidyr)) install.packages("tidyr")
## Loading required package: tidyr

(2)有else

i =1
if (i>0){
  cat('+')   # cat 之间打出来里面的内容
} else {
  print("-") # print 加点东西
}
## +

ifelse 很重要

x=rnorm(10)
y=ifelse(x>0,"+","-")
y
##  [1] "+" "-" "-" "-" "-" "-" "-" "+" "+" "-"

(3)多个条件

i = 0
if (i>0){
  print('+')
} else if (i==0) {
  print('0')
} else if (i< 0){
  print('-')
}
## [1] "0"
ifelse(i>0,"+",ifelse((i<0),"-","0"))
## [1] "0"

2.switch()

cd = 3
foo <- switch(EXPR = cd, 
              #EXPR = "aa", 
              aa=c(3.4,1),
              bb=matrix(1:4,2,2),
              cc=matrix(c(T,T,F,T,F,F),3,2),
              dd="string here",
              ee=matrix(c("red","green","blue","yellow")))
foo
##       [,1]  [,2]
## [1,]  TRUE  TRUE
## [2,]  TRUE FALSE
## [3,] FALSE FALSE
dplyr::case_when() 解决无限套娃问题

1. For循环

循环中 中括号建议写两个
顺便看一下next和break
x <- c(5,6,0,3)
s=0
for (i in x){
  s=s+i
  #if(i == 0) next   跳到下一个循环
  #if (i == 0) break 跳出整个循环
  print(c(which(x==i),i,1/i,s))
}
## [1] 1.0 5.0 0.2 5.0
## [1]  2.0000000  6.0000000  0.1666667 11.0000000
## [1]   3   0 Inf  11
## [1]  4.0000000  3.0000000  0.3333333 14.0000000
x <- c(5,6,0,3)
s = 0
for (i in 1:length(x)){
  s=s+x[[i]]   ##  循环中 中括号建议写两个
  #if(i == 3) next
  #if (i == 3) break
  print(c(i,x[[i]],1/x[[i]],s))
}
## [1] 1.0 5.0 0.2 5.0
## [1]  2.0000000  6.0000000  0.1666667 11.0000000
## [1]   3   0 Inf  11
## [1]  4.0000000  3.0000000  0.3333333 14.0000000

如何将结果存下来?

s = 0
result = list()
for(i in 1:length(x)){
  s=s+x[[i]]
  result[[i]] = c(i,x[[i]],1/i,s)
}
do.call(cbind,result)
##      [,1] [,2]       [,3]  [,4]
## [1,]    1  2.0  3.0000000  4.00
## [2,]    5  6.0  0.0000000  3.00
## [3,]    1  0.5  0.3333333  0.25
## [4,]    5 11.0 11.0000000 14.00
练习6-3

注意最后aes()传参的问题

为什么for循环和aes八字不合

get()函数可以让字符变成变量名
#1.使用循环,查看"a",TRUE和3的数据类型
m <- list("a", TRUE, 3)
for (i in 1:3) {
  class(m[[i]])
}
#2.生成10个随机数,根据这10个随机数生成一个新向量,>中位数的值对应"A",<中位数的值对应"B"。
m <- rnorm(10)
m1 <- ifelse(m>median(m), "A", "B");m1
##  [1] "A" "A" "B" "A" "B" "A" "B" "B" "B" "A"
#3.根据上一练习题中的tmp2生成一个新向量,含有e的值对应"A",不含有e的值对应"B"
tmp = "Bioinformatics is a new subject of genetic data collection,analysis and dissemination to the research community."
library(stringr)
tmp2 = tmp %>% 
  str_replace(","," ") %>%
  str_remove("[.]") %>% 
  str_split(" ")
tmp2 = tmp2[[1]]
ifelse(str_detect(tmp2, "e"), "A", "B")
##  [1] "B" "B" "B" "A" "A" "B" "A" "B" "A" "B" "B" "A" "B" "A" "A" "B"
#2.生成一个随机数(rnorm)组成的10行6列的矩阵,列名为sample1,sample2….sample6,
# 行名为gene1,gene2…gene10,
# 分组为sample1、2、3属于A组,sample4、5、6属于B组。
# 用循环对每个基因画ggplot2箱线图。
set.seed(2020)
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
library(tidyr)
library(ggplot2)
library(cowplot)
## 
## ********************************************************
## Note: As of version 1.0.0, cowplot does not change the
##   default ggplot2 theme anymore. To recover the previous
##   behavior, execute:
##   theme_set(theme_cowplot())
## ********************************************************
library(patchwork)
## 
## Attaching package: 'patchwork'
## The following object is masked from 'package:cowplot':
## 
##     align_plots
exp = matrix(rnorm(60),nrow = 10)
colnames(exp) <- paste0("sample",1:6)
rownames(exp) <- paste0("gene",1:10)
exp[1:4,1:4]
##          sample1    sample2     sample3    sample4
## gene1  0.3769721 -0.8531228  2.17436525 -0.8125047
## gene2  0.3015484  0.9092592  1.09818265 -0.7437022
## gene3 -1.0980232  1.1963730  0.31822032  1.0953451
## gene4 -1.1304059 -0.3715839 -0.07314756  2.4353737
dat = data.frame(t(exp))
dat = mutate(dat,group = rep(c("A","B"),each = 3))%>% mutate(pair = rep(c("AA","BB","cc"),each = 2))

dat2 = gather(dat,key = "gene",value = "expression",-group,-pair)
ggplot(data = dat2)+
  geom_boxplot(aes(x = group,y = expression,color = group))+
  theme_bw()+
  facet_wrap(~gene,nrow = 2)
image.png
p <- list()
for (i in colnames(dat)[1:10]) {
  p[[which(colnames(dat)==i)]] = ggplot(data = dat, 
                   aes(x = group,y = !!dat[,i], color = group))+
    geom_boxplot()+
    ylab(paste0("Expression of ", i))
}

wrap_plots(p,nrow=2,guides = 'collect')
image.png

2.while 循环

i = 0

while (i < 5){
  print(c(i,i^2))
  i = i+1
}
## [1] 0 0
## [1] 1 1
## [1] 2 4
## [1] 3 9
## [1]  4 16

apply()族函数

1.apply 处理矩阵或数据框

apply(X, MARGIN, FUN, …)

其中X是数据框/矩阵名;

MARGIN为1表示取行,为2表示取列,FUN是函数

test<- iris[,1:4]

apply(test, 2, mean)
## Sepal.Length  Sepal.Width Petal.Length  Petal.Width 
##     5.843333     3.057333     3.758000     1.199333
a <- apply(test, 1, sum);a[1:3]
## [1] 10.2  9.5  9.4
res <- c()

for(i in 1:nrow(test)){
  res[[i]] <- sum(test[i,])
}
res[1:3]
## [[1]]
## [1] 10.2
## 
## [[2]]
## [1] 9.5
## 
## [[3]]
## [1] 9.4
练习:
# 1.加载test2.Rdata,求每一行的方差
# load(file = "test2.Rdata")
apply(test, 1, sd)[1:3]
## [1] 2.179449 2.036950 1.997498
# 2.加载class.Rdata,尝试将前6列转为数值型,得到一个新矩阵
# load(file = "class.Rdata")
# apply(y[,1:6], 2, as.numeric)

# 3.解析代码:
names(tail(sort(apply(test,1,sd)),1000))[1:10]
## NULL
# 求test每行的方差,从小到大排序,取后1000的名字

重点函数

  • sort
  • match
  • names
  • ifelse 和 str_detect
  • identical
  • arrange
  • merge 和 inner_join
  • unique 和 duplicated

重点知识点

  • 向量数据框、列表取子集
  • 数据框新增列
  • 文件读取
  • Rdata的加载与保存
  • 作图保存
  • R包安装和加载
  • 形式参数、实际参数、默认参数

R语言遍历、创建、删除文件夹

  • dir()
  • file.create()
  • file.exists(…)
  • file.remove()
  • file.rename(from, to)
  • file.append(file1, file2)
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