Question

我有一个df：

structure(list(chrom = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("2.11E+14", 
"2L", "2R", "3L", "3R", "mitochondrion_genome", "rDNA", "Unmapped_Scaffold_58_D1862", 
"X", "X3X4_mapped_Scaffold_14_D1732", "X3X4_mapped_Scaffold_6_D1712"
), class = "factor"), pos = c(32476L, 56682L, 95341L, 161983L, 
190298L, 214264L, 344985L, 424235L, 424262L, 424295L, 431370L, 
444713L, 449545L, 464397L, 464407L, 465414L, 465416L, 465424L, 
483832L, 523201L), nFreq = c(48, 61, 43, 53, 38, 52, 44, 55, 
55, 56, 64, 42, 31, 31, 30, 44, 47, 45, 53, 40), n_dist = c(2, 
11, 7, 3, 12, 2, 6, 5, 5, 6, 14, 8, 19, 19, 20, 6, 3, 5, 3, 10
), tFreq = c(90, 100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100, 86, 
87, 0, 0, 0, 100, 100), t_dist = c(40, 50, 50, 50, 50, 50, 50, 
50, 50, 50, 50, 50, 50, 36, 37, 50, 50, 50, 50, 50)), .Names = c("chrom", 
"pos", "nFreq", "n_dist", "tFreq", "t_dist"), row.names = c(NA, 
20L), class = "data.frame")

看起来像这样：

chrom    pos nFreq n_dist tFreq t_dist
2L  32476    48      2    90     40
2L  56682    61     11   100     50
2L  95341    43      7     0     50
2L 161983    53      3     0     50
2L 190298    38     12     0     50
2L 214264    52      2     0     50

我正在尝试使用gather将其转换为如下所示的长数据框：

chrom   pos freq    dist    condition
2L  32476   48  2   norm
2L  32476   90  40  tum
2L  56682   61  11  norm
2L  56682   100 50  tum
2L  95341   43  7   norm
2L  95341   0   50  tum

我不知道如何添加两个以上的将来的列标题，这是我正在尝试的方法：

library(tidyr)
long_df <- df %>% 
    tidyr::gather(freq, dist, nFreq:t_dist, factor_key=TRUE)

哪个会产生：

  chrom    pos aFreq dist
1    2L  32476 nFreq   48
2    2L  56682 nFreq   61
3    2L  95341 nFreq   43
4    2L 161983 nFreq   53
5    2L 190298 nFreq   38
6    2L 214264 nFreq   52

如何调整此值以实现所需的输出？

Answer 1

这里是一种解决方案，先收集所有列，然后将其分为两部分（范数和tum），然后进行扩展和合并。

library(tidyr)
library(dplyr)

#make data into long format
new<-gather(df, "key", "value", -c(1:2))

#split into 2 parts into wide format 
norm<-spread(new[new$key=="nFreq"|new$key=="n_dist",], key, value) %>% mutate(condition="norm")
tum<-spread(new[new$key=="tFreq"|new$key=="t_dist",], key, value) %>% mutate(condition="tum")

#Ensure columns names are the same 
names(norm)<-c("chrom", "pos", "freq", "dist", "condition")
names(tum)<-c("chrom", "pos", "freq", "dist", "condition")

#combine and reorder
answer<-rbind(norm, tum)
answer<-answer[order(answer$pos),]

这可行，但我相信可以简化。

将广泛的数据收集到多列中

1 个答案: