Question

我有这个矩阵（它的大小很大）＆＃34; mymat＆＃34;。我需要复制具有＆＃34; /＆＃34;的列。在他们的列名称匹配＆＃34; /＆＃34;并制作一个＆＃34; resmatrix＆＃34;。我怎样才能在R中完成这项工作？

mymat

 a   b   IID:WE:G12D/V    GH:SQ:p.R172W/G   c
 1   3               4                  2   4
22   4               2                  2   4
 2   3               2                  2   4

resmatrix

 a   b   IID:WE:G12D   IID:WE:G12V    GH:SQ:p.R172W    GH:SQ:p.R172G   c
 1   3             4             4                2                2   4
22   4             2             2                2                2   4
 2   3             2             2                2                2   4

Answer 1

找出哪些列具有“/”并复制它们，然后重命名。要计算新名称，只需在/上拆分并替换第二个名称的最后一个字母。

# which columns have '/' in them?
which.slash <- grep('/', names(mymat), value=T)
new.names <- unlist(lapply(strsplit(which.slash, '/'),
       function (bits) {
         # bits[1] is e.g. IID:WE:G12D and bits[2] is the V
         # take bits[1] and replace the last letter for the second colname
         c(bits[1], sub('.$', bits[2], bits[1]))
       }))

# make resmat by copying the appropriate columns
resmat <- cbind(mymat, mymat[, which.slash])
# order the columns to make sure the names replace properly
resmat <- resmat[, order(names(resmat))]
# put the new names in
names(resmat)[grep('/', names(resmat))] <- sort(new.names)

resmat看起来像这样

#    a b c GH:SQ:p.R172G GH:SQ:p.R172W IID:WE:G12D IID:WE:G12V
# 1  1 3 4             2             2           4           4
# 2 22 4 4             2             2           2           2
# 3  2 3 4             2             2           2           2

Answer 2

您可以使用grep获取包含/（'nm1'）的列名称索引，使用sub/scan复制'nm1'中的列名称以创建'nm2' 。然后，cbind不是'nm1'的列，使用复制的列（'nm1'），将列名更改为'nm2'，如果需要order，则更改列。

 #get the column index with grep
 nm1 <- grepl('/', names(df1))
 #used regex to rearrange the substrings in the nm1 column names
 #removed the `/` and use `scan` to split at the space delimiter
 nm2 <- scan(text=gsub('([^/]+)(.)/(.*)', '\\1\\2 \\1\\3', 
           names(df1)[nm1]), what='', quiet=TRUE)
 #cbind the columns that are not in nm1, with the replicate nm1 columns 
 df2 <- cbind(df1[!nm1], setNames(df1[rep(which(nm1), each= 2)], nm2))
 #create another index to find the starting position of nm1 columns
 nm3 <- names(df1)[1:(which(nm1)[1L]-1)] 
 #we concatenate the nm3, nm2, and the rest of the columns to match 
 #the expected output order
 df2N <- df2[c(nm3, nm2, setdiff(names(df1)[!nm1], nm3))]
 df2N 
 #   a b IID:WE:G12D IID:WE:G12V GH:SQ:p.R172W GH:SQ:p.R172G c
 #1  1 3           4           4             2             2 4
 #2 22 4           2           2             2             2 4
 #3  2 3           2           2             2             2 4

数据

df1 <-  structure(list(a = c(1L, 22L, 2L), b = c(3L, 4L, 3L),
`IID:WE:G12D/V` = c(4L, 
2L, 2L), `GH:SQ:p.R172W/G` = c(2L, 2L, 2L), c = c(4L, 4L, 4L)),
.Names = c("a", "b", "IID:WE:G12D/V", "GH:SQ:p.R172W/G", "c"),
class = "data.frame", row.names = c(NA, -3L))

如何复制列名，将它们分隔符＆＃39; /＆＃39;分成R中的多个列名？

2 个答案:

数据