R 从列名中删除字符串的一部分
这是一个数据:R 从列名中删除字符串的一部分,r,regex,R,Regex,这是一个数据: structure(list(Fasta.headers = c("Person01050.1", "Person01080.1", "Person01090.1", "Person01100.4", "Person01140.1", "Person01220.1"), ToRemove.Gr_1 = c(0, 1107200, 17096000, 0, 0, 0), ToRemo
structure(list(Fasta.headers = c("Person01050.1", "Person01080.1",
"Person01090.1", "Person01100.4", "Person01140.1", "Person01220.1"),
ToRemove.Gr_1 = c(0, 1107200, 17096000, 0, 0, 0), ToRemove.Gr_10 = c(0,
37259000, 1104800000, 783870, 0, 1308600), ToRemove.Gr_11 = c(1835800,
53909000, 623960000, 0, 0, 0), ToRemove.Gr_12 = c(0, 19117000,
808600000, 0, 0, 719400), ToRemove.Gr_13 = c(2544200, 2461400,
418770000, 0, 0, 0), ToRemove.Gr_14 = c(5120400, 1373700,
117330000, 0, 0, 0), ToRemove.Gr_15 = c(6623500, 0, 73336000,
0, 0, 0), ToRemove.Gr_16 = c(0, 0, 31761000, 0, 0, 0), ToRemove.Gr_17 = c(13475000,
0, 29387000, 0, 0, 0), ToRemove.Gr_18 = c(7883300, 0, 27476000,
0, 0, 0), ToRemove.Gr_19 = c(82339000, 3254700, 50825000,
0, 0, 0), ToRemove.Gr_2 = c(1584100, 84847000, 5219500000,
6860700, 0, 8337700), ToRemove.Gr_20 = c(205860000, 0, 67685000,
0, 0, 0), ToRemove.Gr_21 = c(867120000, 1984400, 2.26e+08,
0, 0, 10502000)), .Names = c("Fasta.headers", "ToRemove.Gr_1",
"ToRemove.Gr_10", "ToRemove.Gr_11", "ToRemove.Gr_12", "ToRemove.Gr_13",
"ToRemove.Gr_14", "ToRemove.Gr_15", "ToRemove.Gr_16", "ToRemove.Gr_17",
"ToRemove.Gr_18", "ToRemove.Gr_19", "ToRemove.Gr_2", "ToRemove.Gr_20",
"ToRemove.Gr_21"), row.names = c(NA, 6L), class = "data.frame")
正如列名已经建议从名称中删除部分“ToRemove”,并且只有Gr_*应该保留
对于这个问题,我希望有两种解决办法。首先,基于指定的字符串,它应该删除列名的一部分或基于特定字符,例如
。它应该在点之前或之后删除整个部分。我们可以使用sub
names(df1)[-1] <- sub(".*\\.", "", names(df1)[-1])
为了精确匹配模式,我们还可以从字符串的开头(
^
)匹配零个或多个非do t([^.]*
)的字符,后跟一个点(\.
-将点转义为表示任何字符的元字符),并将其替换为空白(“”
)
正如上面已经提到的“ToRemove”
sub("ToRemove.", "", names(df1)[-1], fixed = TRUE)
另外,如果我们需要删除所有字符,包括
names(df1)[-1] <- sub(".*\\.", ".", names(df1)[-1])
sub("\\..*", "", names(df1)[-1])
相关想法:
sub("\\..*", "", names(df1)[-1])