我已经创建了一个数据帧的子集,用于计算。我现在留下了许多子集,我想将它们合并到一列中。子集如下所示:
> E
$`1`
[1] "AAAaaa" "TTTaaa" "CCCaaa" "GGGaaa" "AAAttt" "TTTttt" "CCCttt" "GGGttt"
[9] "AAAccc" "TTTccc" "CCCccc" "GGGccc" "AAAggg" "TTTggg" "CCCggg" "GGGggg"
$`2`
[1] "ATAata" "TATata" "CGCata" "GCGata" "BBBata" "ATAtat" "TATtat" "CGCtat"
[9] "GCGtat" "BBBtat" "ATAcgc" "TATcgc" "CGCcgc" "GCGcgc" "BBBcgc" "ATAgcg"
[17] "TATgcg" "CGCgcg" "GCGgcg" "BBBgcg" "ATAbbb" "TATbbb" "CGCbbb" "GCGbbb"
[25] "BBBbbb"
我试过:
A=vector()
cbind(A,ExonJunction,deparse.level = 1)
A
但这给我留下了
E
1 Character,16
2 Character,25
我希望字符列表在一列中。我该怎么做?
也可以在c
函数中尝试recursive
参数,例如
c(E, recursive = TRUE, use.names = FALSE)
# [1] "AAAaaa" "TTTaaa" "CCCaaa" "GGGaaa" "AAAttt" "TTTttt" "CCCttt" "GGGttt" "AAAccc" "TTTccc" "CCCccc" "GGGccc" "AAAggg" "TTTggg" "CCCggg" "GGGggg" "ATAata"
# [18] "TATata" "CGCata" "GCGata" "BBBata" "ATAtat" "TATtat" "CGCtat" "GCGtat" "BBBtat" "ATAcgc" "TATcgc" "CGCcgc" "GCGcgc" "BBBcgc" "ATAgcg" "TATgcg" "CGCgcg"
# [35] "GCGgcg" "BBBgcg" "ATAbbb" "TATbbb" "CGCbbb" "GCGbbb" "BBBbbb"
或者,如果您希望将其作为数据框中的列,可以尝试
df <- data.frame(Res = c(E, recursive = TRUE))
您可以unlist
list
并使用data.frame
创建单列数据框
dat <- data.frame(Col1=unlist(E, use.names=FALSE), stringsAsFactors=FALSE)
数据
E <- structure(list(`1` = c("AAAaaa", "TTTaaa", "CCCaaa", "GGGaaa",
"AAAttt", "TTTttt", "CCCttt", "GGGttt", "AAAccc", "TTTccc", "CCCccc",
"GGGccc", "AAAggg", "TTTggg", "CCCggg", "GGGggg"), `2` = c("ATAata",
"TATata", "CGCata", "GCGata", "BBBata", "ATAtat", "TATtat", "CGCtat",
"GCGtat", "BBBtat", "ATAcgc", "TATcgc", "CGCcgc", "GCGcgc", "BBBcgc",
"ATAgcg", "TATgcg", "CGCgcg", "GCGgcg", "BBBgcg", "ATAbbb", "TATbbb",
"CGCbbb", "GCGbbb", "BBBbbb")), .Names = c("1", "2"))
你也可以
使用 stack
,像这样(前提是你正在处理一个命名列表,就像你一样):
stack(E)
一个不错的功能是名称成为"ind"列,因此该过程很容易逆转。
head(stack(E))
# values ind
# 1 AAAaaa 1
# 2 TTTaaa 1
# 3 CCCaaa 1
# 4 GGGaaa 1
# 5 AAAttt 1
# 6 TTTttt 1
tail(stack(E))
# values ind
# 36 BBBgcg 2
# 37 ATAbbb 2
# 38 TATbbb 2
# 39 CGCbbb 2
# 40 GCGbbb 2
# 41 BBBbbb 2