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R: 将向量划分为区间,并测试哪个整数落入哪个区间

cut intervals range r

Esben Eickhardt · 技术社区 · 9 年前

我有23条染色体及其长度

chromosome    length
1             249250621
2             243199373
3             198022430
4             191154276
5             180915260
6             171115067 
..            .........
Y             59373566

对于每条染色体,我想创建5000个大小相等的箱/间隔。

Chr1:
bin_number    start        end
1             1            49850
2             49851        99700
....          .....        .....
5000          249200771    249250621

为此,我试过同时使用“cut”和“cut2”。“cut2”无法处理染色体的长度并崩溃,而cut为每个位置提供了间隔(249250621间隔!)。

cut2(1:249250621, g=5000, onlycuts = TRUE)

cut(1:249250621, breaks=5000)

当我有区间时,我想指定每个区间的50000个变量。

我的数据(染色体1):

variant            chromosome    position
1:20000_G/A        1             20000
1:30000_C/CCCCT    1             30000
1:60000_G/T        1             60000
..............     ..            .......

我想要什么:

variant            chromosome    position    bin_number
1:20000_G/A        1             20000       1
1:30000_C/CCCCT    1             30000       1
1:60000_G/T        1             60000       2
..............     ..            .......     ...

我非常感谢任何有关将我的染色体分裂成间隔的方法的建议。当我有区间时,我需要能够快速测试变量属于哪个区间的方法。

3 回复 | 直到 9 年前

Ibon Tamayo 6 年前

如果我很好地理解你的算法,你会把每条染色体分成10000个箱。因此,每个系列都有不同的尺寸。我曾经应用此算法创建独立于染色体的相同大小范围。

chrSizes <- data.frame(chromosome = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "X", "Y"), 
                       length = c("249250621","243199373", "198022430", "191154276", "180915260", "171115067", "159138663", "146364022", "141213431", "135534747", "135006516", "133851895", "115169878", "107349540", "102531392", "90354753", "81195210", "78077248", "59128983", "63025520", "48129895", "51304566", "155270560", "59373566"), 
                       stringsAsFactors = FALSE)

sizerange <- 5000000
lastranges <- NA
h <- 0

for (i in 1:24) 
{
  thelast <- 1
  bynum <- format(sizerange, scientific = FALSE)
  chrlist <- c(paste0(chrSizes$chromosome[i],":1-",bynum))
  biggest <- chrSizes$length[i]
  while(thelast < biggest)
  {
    bynum1 <- format(as.numeric(bynum)+1, scientific = FALSE)
    bynum2 <- format(as.numeric(bynum1)+sizerange-1, scientific = FALSE)
    berria <- paste0(paste0(chrSizes$chromosome[i],":",bynum1,"-",as.character(bynum2)))
    chrlist <- c(chrlist,berria)
    thelast <- as.numeric(bynum2)+sizerange
    bynum <- format(as.numeric(bynum)+sizerange, scientific = FALSE)
  }
  azkenreg <- paste0(paste0(chrSizes$chromosome[i],":",bynum,"-",as.character(biggest)))
  chrlist <- c(chrlist,azkenreg)
  lastranges <- c(lastranges,chrlist)
}

lastranges <- lastranges[-1]

df <- data.frame(lastranges)
write.table(df,file = "fastacontigs_splited_bysize2.txt",quote = FALSE, row.names = FALSE, col.names = FALSE)

在这种情况下,结果是:

1:1-5000000
1:5000001-10000000
1:10000001-15000000
1:15000000-249250621
2:1-5000000
2:5000001-10000000
2:10000001-15000000
2:15000000-243199373
3:1-5000000
3:5000001-10000000
3:10000001-15000000
3:15000000-198022430
4:1-5000000
4:5000001-10000000
4:10000001-15000000
4:15000000-191154276
5:1-5000000
5:5000001-10000000
5:10000001-15000000
5:15000000-180915260
6:1-5000000
6:5000001-10000000
6:10000001-15000000
6:15000000-171115067
7:1-5000000
7:5000001-10000000
7:10000001-15000000
7:15000000-159138663
8:1-5000000
8:5000001-10000000
8:10000001-15000000
8:15000000-146364022
9:1-5000000
9:5000001-10000000
9:10000001-15000000
9:15000000-141213431
10:1-5000000
10:5000001-10000000
10:10000001-15000000
10:15000000-135534747
11:1-5000000
11:5000001-10000000
11:10000001-15000000
11:15000000-135006516
12:1-5000000
12:5000001-10000000
12:10000001-15000000
12:15000000-133851895
13:1-5000000
13:5000001-10000000
13:10000001-15000000
13:15000000-115169878
14:1-5000000
14:5000001-10000000
14:10000001-15000000
14:15000000-107349540
15:1-5000000
15:5000001-10000000
15:10000001-15000000
15:15000000-102531392
16:1-5000000
16:5000001-10000000
16:10000001-15000000
16:15000001-20000000
16:20000001-25000000
16:25000001-30000000
16:30000001-35000000
16:35000001-40000000
16:40000001-45000000
16:45000001-50000000
16:50000001-55000000
16:55000001-60000000
16:60000001-65000000
16:65000001-70000000
16:70000001-75000000
16:75000001-80000000
16:80000001-85000000
16:85000000-90354753
17:1-5000000
17:5000001-10000000
17:10000001-15000000
17:15000001-20000000
17:20000001-25000000
17:25000001-30000000
17:30000001-35000000
17:35000001-40000000
17:40000001-45000000
17:45000001-50000000
17:50000001-55000000
17:55000001-60000000
17:60000001-65000000
17:65000001-70000000
17:70000001-75000000
17:75000000-81195210
18:1-5000000
18:5000001-10000000
18:10000001-15000000
18:15000001-20000000
18:20000001-25000000
18:25000001-30000000
18:30000001-35000000
18:35000001-40000000
18:40000001-45000000
18:45000001-50000000
18:50000001-55000000
18:55000001-60000000
18:60000001-65000000
18:65000000-78077248
19:1-5000000
19:5000001-10000000
19:10000001-15000000
19:15000001-20000000
19:20000001-25000000
19:25000001-30000000
19:30000001-35000000
19:35000001-40000000
19:40000001-45000000
19:45000000-59128983
20:1-5000000
20:5000001-10000000
20:10000001-15000000
20:15000001-20000000
20:20000001-25000000
20:25000001-30000000
20:30000001-35000000
20:35000001-40000000
20:40000001-45000000
20:45000001-50000000
20:50000001-55000000
20:55000000-63025520
21:1-5000000
21:5000001-10000000
21:10000001-15000000
21:15000001-20000000
21:20000001-25000000
21:25000001-30000000
21:30000001-35000000
21:35000000-48129895
22:1-5000000
22:5000001-10000000
22:10000001-15000000
22:15000001-20000000
22:20000001-25000000
22:25000001-30000000
22:30000001-35000000
22:35000001-40000000
22:40000001-45000000
22:45000000-51304566
X:1-5000000
X:5000001-10000000
X:10000001-15000000
X:15000000-155270560
Y:1-5000000
Y:5000001-10000000
Y:10000001-15000000
Y:15000001-20000000
Y:20000001-25000000
Y:25000001-30000000
Y:30000001-35000000
Y:35000001-40000000
Y:40000001-45000000
Y:45000000-59373566

Kota Mori 9 年前

如果bin范围是恒定的,则此操作有效:

mydata <- data.frame(position = c(20000, 30000, 60000, 
                              49850, 49851, 99700, 99701))
mydata$bin <- ceiling(mydata$position / 49850)

更一般地,如果bin范围不是恒定的,但您已经定义了切割点,则可以使用 cut 将其作为 breaks .

cutpoints <- c(0, 49850, 99700, 149550)
mydata$bin2 <- cut(mydata$position, breaks = cutpoints)

您可以稍加调整来编辑标签。

mydata$bin3 <- cut(mydata$position, breaks = cutpoints,
               labels = seq(length(cutpoints)-1))

Esben Eickhardt 9 年前

感谢您的输入。我选择使用一个简单的循环来创建间隔,以确保间隔具有所需的大小。

我创建了一个数据。染色体大小的框架

chrSizes <- data.frame(chromosome = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "X", "Y"), length = c("249250621","243199373", "198022430", "191154276", "180915260", "171115067", "159138663", "146364022", "141213431", "135534747", "135006516", "133851895", "115169878", "107349540", "102531392", "90354753", "81195210", "78077248", "59128983", "63025520", "48129895", "51304566", "155270560", "59373566"), stringsAsFactors = FALSE)

然后,我通过找到精确的块大小,然后向下取整,在每个染色体上循环创建间隔。然后,余数被用于将一个添加到最初的多个间隔中。

numOfBins <- 10000
chrBinList <- list()
for (i in 1:24) {
  chrBins <- c()
  chrLength <- as.numeric(chrSizes[i,2])
  chunkSize <- floor(chrLength/numOfBins)
  remainder <- chrLength %% chunkSize
  counter <- 1

  # Adding remainder to the first intervals
  for (j in 1:(remainder-1)) {
    chrBins <- c(chrBins, counter)
    counter <- counter + chunkSize + 1
    chrBins <- c(chrBins, counter)
  }

  # Adding normal sized chunks to remaining intervals
  for (k in remainder:numOfBins) {
    chrBins <- c(chrBins, counter)
    counter <- counter + chunkSize
    chrBins <- c(chrBins, counter)
  }

  # Creating a data.frame with intervals
  interval.df <- as.data.frame(matrix(chrBins,ncol = 2, byrow = TRUE))
  colnames(interval.df) <- c("start", "end")

  # Adding to list
  chrBinList[[chrSizes[i,1]]] <- interval.df
}

至于测试值是否落入不同的容器中,我使用apply提出了一个缓慢的解决方案。然而,我目前正在研究并行应用函数。