R 两个逗号分隔的字符串之间的匹配
R match between two comma-separated strings
我正在尝试寻找一种优雅的方法来查找数据框中以下两个字符列之间的匹配项。复杂的部分是任一字符串都可以包含逗号分隔的列表，如果一个列表的成员与另一个列表的任何成员匹配，则整个条目将被视为匹配。我不确定我对此解释得如何，所以这里是示例数据和输出：
Alt1:
- 在
- 一个
- G
- CGTCC,AT
- CGC
备选方案 2：
- AA
- 一个
- GG
- AT,GGT
- CG
每行预期匹配：
- 第 1 行 = none
- 第 2 行 = A
- 第 3 行 = none
- 第 4 行 = AT
- 第 5 行 = none
无效解决方案：
第一次尝试：按所需列合并整个数据框，然后匹配上面显示的替代列：
match1 = data.frame(merge(vcf.df, ref.df, by=c("chr", "start", "end", "ref")))
matches = unique(match1[unlist(sapply(match1$Alt1 grep, match1$Alt2, fixed=TRUE)),])
第二种方法，使用 VariantAnnoatation/Granges 中的 findoverlaps 功能：
findoverlaps(ranges(vcf1), ranges(vcf2))
如有任何建议，我们将不胜感激！谢谢！
解决方案
感谢@Marat Talipov 在下面的回答，以下解决方案可以比较两个逗号分隔的字符串：
> ##read in edited kaviar vcf and human ref
> ref <- readVcfAsVRanges("ref.vcf.gz", humie_ref)
Warning message:
In .vcf_usertag(map, tag, ...) :
ScanVcfParam ‘geno’ fields not present: ‘AD’
> ##rename chromosomes to match with vcf files
> ref <- renameSeqlevels(ref, c("1"="chr1"))
> ##################################
> ## Gather VCF files to process ##
> ##################################
> ##data frame *.vcf.gz files in directory path
> vcf_path <- data.frame(path=list.files(vcf_dir, pattern="*.vcf.gz$", full=TRUE))
> ##read in everything but sample data for speediness
> vcf_param = ScanVcfParam(samples=NA)
> vcf <- readVcfAsVRanges("test.vcf.gz", humie_ref, param=vcf_param)
> #################
> ## Match SNP's ##
> #################
> ##create data frames of info to match on
> vcf.df = data.frame(chr =as.character(seqnames(vcf)), start = start(vcf), end = end(vcf), ref = as.character(ref(vcf)),
+ alt=alt(vcf), stringsAsFactors=FALSE)
> ref.df = data.frame(chr =as.character(seqnames(ref)), start = start(ref), end = end(ref),
+ ref = as.character(ref(ref)), alt=alt(ref), stringsAsFactors=FALSE)
>
> ##merge based on all positional fields except vcf
> col_match = data.frame(merge(vcf.df, ref.df, by=c("chr", "start", "end", "ref")))
> library(stringi)
> ##split each alt column by comma and bind together
> M1 <- stri_list2matrix(sapply(col_match$alt.x,strsplit,','))
> M2 <- stri_list2matrix(sapply(col_match$alt.y,strsplit,','))
> M <- rbind(M1,M2)
> ##compare results
> result <- apply(M,2,function(z) unique(na.omit(z[duplicated(z)])))
> ##add results column to col_match df for checking/subsetting
> col_match$match = result
> head(col_match)
chr start end ref alt.x alt.y match
1 chr1 39998059 39998059 A G G G
2 chr1 39998059 39998059 A G G G
3 chr1 39998084 39998084 C A A A
4 chr1 39998084 39998084 C A A A
5 chr1 39998085 39998085 G A A A
6 chr1 39998085 39998085 G A A A
如果输入列表的长度相等并且您想以成对的方式比较列表元素，您可以使用此解决方案：
library(stringi)
M1 <- stri_list2matrix(sapply(Alt1,strsplit,','))
M2 <- stri_list2matrix(sapply(Alt2,strsplit,','))
M <- rbind(M1,M2)
result <- apply(M,2,function(z) unique(na.omit(z[duplicated(z)])))
示例输入：
Alt1 <- list('AT','A','G','CGTCC,AT','CGC','GG,CC')
Alt2 <- list('AA','A','GG','AT,GGT','CG','GG,CC')
输出：
# [[1]]
# character(0)
#
# [[2]]
# [1] "A"
#
# [[3]]
# character(0)
#
# [[4]]
# [1] "AT"
#
# [[5]]
# character(0)
#
# [[6]]
# [1] "GG" "CC"
你可以这样做：
Alt1 <- list('AT','A','G',c('CGTCC','AT'),'CGC')
Alt2 <- list('AA','A','GG',c('AT','GGT'),'CG')
# make sure you change the lists within in the lists into vectors
matchlist <- list()
for (i in 1:length(Alt1)){
matchlist[[i]] <- ifelse(Alt1[[i]] %in% Alt2[[i]],
paste("Row",i,"=",c(Alt1[[i]],Alt2[[i]])[duplicated(c(Alt1[[i]],Alt2[[i]]))],sep=" "),
paste("Row",i,"= none",sep=" "))
}
print(matchlist)
坚持使用 stringi 包，您可以使用 Marat 的回答中的 Alt1 和 Alt2 数据来做这样的事情。
library(stringi)
f <- function(x, y) {
ssf <- stri_split_fixed(c(x, y), ",", simplify = TRUE)
if(any(sd <- stri_duplicated(ssf))) ssf[sd] else NA_character_
}
Map(f, Alt1, Alt2)
# [[1]]
# [1] NA
#
# [[2]]
# [1] "A"
#
# [[3]]
# [1] NA
#
# [[4]]
# [1] "AT"
#
# [[5]]
# [1] NA
#
# [[6]]
# [1] "GG" "CC"
或者在 base R 中，我们可以使用 scan() 用逗号分隔字符串。
g <- function(x, y, sep = ",") {
s <- scan(text = c(x, y), what = "", sep = sep, quiet = TRUE)
s[duplicated(s)]
}
Map(g, Alt1, Alt2)
我正在尝试寻找一种优雅的方法来查找数据框中以下两个字符列之间的匹配项。复杂的部分是任一字符串都可以包含逗号分隔的列表，如果一个列表的成员与另一个列表的任何成员匹配，则整个条目将被视为匹配。我不确定我对此解释得如何，所以这里是示例数据和输出：
Alt1:
- 在
- 一个
- G
- CGTCC,AT
- CGC
备选方案 2：
- AA
- 一个
- GG
- AT,GGT
- CG
每行预期匹配：
- 第 1 行 = none
- 第 2 行 = A
- 第 3 行 = none
- 第 4 行 = AT
- 第 5 行 = none
无效解决方案：
第一次尝试：按所需列合并整个数据框，然后匹配上面显示的替代列：
match1 = data.frame(merge(vcf.df, ref.df, by=c("chr", "start", "end", "ref")))
matches = unique(match1[unlist(sapply(match1$Alt1 grep, match1$Alt2, fixed=TRUE)),])
第二种方法，使用 VariantAnnoatation/Granges 中的 findoverlaps 功能：
findoverlaps(ranges(vcf1), ranges(vcf2))
如有任何建议，我们将不胜感激！谢谢！
解决方案 感谢@Marat Talipov 在下面的回答，以下解决方案可以比较两个逗号分隔的字符串：
> ##read in edited kaviar vcf and human ref
> ref <- readVcfAsVRanges("ref.vcf.gz", humie_ref)
Warning message:
In .vcf_usertag(map, tag, ...) :
ScanVcfParam ‘geno’ fields not present: ‘AD’
> ##rename chromosomes to match with vcf files
> ref <- renameSeqlevels(ref, c("1"="chr1"))
> ##################################
> ## Gather VCF files to process ##
> ##################################
> ##data frame *.vcf.gz files in directory path
> vcf_path <- data.frame(path=list.files(vcf_dir, pattern="*.vcf.gz$", full=TRUE))
> ##read in everything but sample data for speediness
> vcf_param = ScanVcfParam(samples=NA)
> vcf <- readVcfAsVRanges("test.vcf.gz", humie_ref, param=vcf_param)
> #################
> ## Match SNP's ##
> #################
> ##create data frames of info to match on
> vcf.df = data.frame(chr =as.character(seqnames(vcf)), start = start(vcf), end = end(vcf), ref = as.character(ref(vcf)),
+ alt=alt(vcf), stringsAsFactors=FALSE)
> ref.df = data.frame(chr =as.character(seqnames(ref)), start = start(ref), end = end(ref),
+ ref = as.character(ref(ref)), alt=alt(ref), stringsAsFactors=FALSE)
>
> ##merge based on all positional fields except vcf
> col_match = data.frame(merge(vcf.df, ref.df, by=c("chr", "start", "end", "ref")))
> library(stringi)
> ##split each alt column by comma and bind together
> M1 <- stri_list2matrix(sapply(col_match$alt.x,strsplit,','))
> M2 <- stri_list2matrix(sapply(col_match$alt.y,strsplit,','))
> M <- rbind(M1,M2)
> ##compare results
> result <- apply(M,2,function(z) unique(na.omit(z[duplicated(z)])))
> ##add results column to col_match df for checking/subsetting
> col_match$match = result
> head(col_match)
chr start end ref alt.x alt.y match
1 chr1 39998059 39998059 A G G G
2 chr1 39998059 39998059 A G G G
3 chr1 39998084 39998084 C A A A
4 chr1 39998084 39998084 C A A A
5 chr1 39998085 39998085 G A A A
6 chr1 39998085 39998085 G A A A
如果输入列表的长度相等并且您想以成对的方式比较列表元素，您可以使用此解决方案：
library(stringi)
M1 <- stri_list2matrix(sapply(Alt1,strsplit,','))
M2 <- stri_list2matrix(sapply(Alt2,strsplit,','))
M <- rbind(M1,M2)
result <- apply(M,2,function(z) unique(na.omit(z[duplicated(z)])))
示例输入：
Alt1 <- list('AT','A','G','CGTCC,AT','CGC','GG,CC')
Alt2 <- list('AA','A','GG','AT,GGT','CG','GG,CC')
输出：
# [[1]]
# character(0)
#
# [[2]]
# [1] "A"
#
# [[3]]
# character(0)
#
# [[4]]
# [1] "AT"
#
# [[5]]
# character(0)
#
# [[6]]
# [1] "GG" "CC"
你可以这样做：
Alt1 <- list('AT','A','G',c('CGTCC','AT'),'CGC')
Alt2 <- list('AA','A','GG',c('AT','GGT'),'CG')
# make sure you change the lists within in the lists into vectors
matchlist <- list()
for (i in 1:length(Alt1)){
matchlist[[i]] <- ifelse(Alt1[[i]] %in% Alt2[[i]],
paste("Row",i,"=",c(Alt1[[i]],Alt2[[i]])[duplicated(c(Alt1[[i]],Alt2[[i]]))],sep=" "),
paste("Row",i,"= none",sep=" "))
}
print(matchlist)
坚持使用 stringi 包，您可以使用 Marat 的回答中的 Alt1 和 Alt2 数据来做这样的事情。
library(stringi)
f <- function(x, y) {
ssf <- stri_split_fixed(c(x, y), ",", simplify = TRUE)
if(any(sd <- stri_duplicated(ssf))) ssf[sd] else NA_character_
}
Map(f, Alt1, Alt2)
# [[1]]
# [1] NA
#
# [[2]]
# [1] "A"
#
# [[3]]
# [1] NA
#
# [[4]]
# [1] "AT"
#
# [[5]]
# [1] NA
#
# [[6]]
# [1] "GG" "CC"
或者在 base R 中，我们可以使用 scan() 用逗号分隔字符串。
g <- function(x, y, sep = ",") {
s <- scan(text = c(x, y), what = "", sep = sep, quiet = TRUE)
s[duplicated(s)]
}
Map(g, Alt1, Alt2)