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，我想创建相同大小的5000个区间/间隔。
Chr1:
bin_number start end
1 1 49850
2 49851 99700
.... ..... .....
5000 249200771 249250621
为此，我尝试过使用“cut”和“cut2”。 “cut2”无法处理染色体和崩溃的长度，而cut为每个地方提供一个间隔（249250621个间隔！）。
cut2(1:249250621, g=5000, onlycuts = TRUE)
cut(1:249250621, breaks=5000)
当我有间隔时，我想分配哪个bin / interval 50.000变量都在其中。
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
.............. .. .......
我想要的是什么：
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
.............. .. ....... ...
对于将染色体分成间隔相关的方法，我将不胜感激。当我有间隔时，我需要能够快速测试变体属于哪个区间的方法。
答案 0 :(得分：1)
如果我对您的算法很了解，那么您会将每个染色体分成10000个bin。因此，您将获得每个范围的不同大小。我曾经应用此算法来创建独立于染色体的相同大小的范围。
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，结果为：
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
答案 1 :(得分：0)
如果bin范围是常量，则可以：
mydata <- data.frame(position = c(20000, 30000, 60000,
49850, 49851, 99700, 99701))
mydata$bin <- ceiling(mydata$position / 49850)
更一般地说，如果bin范围不是常数但您已经定义了切割点，则可以使用cut作为breaks。
cutpoints <- c(0, 49850, 99700, 149550)
mydata$bin2 <- cut(mydata$position, breaks = cutpoints)
您可以通过一些调整来编辑标签。
mydata$bin3 <- cut(mydata$position, breaks = cutpoints,
labels = seq(length(cutpoints)-1))
答案 2 :(得分：0)
感谢您的投入。我选择使用简单的循环创建间隔，以确保间隔具有所需的大小。
我创建了一个染色体大小的数据框
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)
然后我通过每个染色体循环创建间隔，找到精确的块大小然后向下舍入。然后剩余部分用于在第一个间隔中添加一个。
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
}
至于测试值是否属于不同的箱子，我使用apply提出了一个缓慢的解决方案。但是，我现在正在研究并行应用函数。