我在下面有此数据。我想将一列中的所有No_of.reads列融化，并将另一列中的所有_contamination_列融化。因此，最终数据帧将具有diluted_sample，No_of_reads和_contamination_列。我尝试分两步执行此操作，但这会使我反复观察。正确的方法是什么？
代码：
test.dput.melted <- melt(test.dput, id = 1:3, measure = 4:7)
test.dput.melted <- melt(test.dput.melted, id = c(1,4,5), measure = 2:3)
数据：
test.dput<- structure(list(diluted_sample = c("100%", "95%", "90%", "85%",
"80%", "75%"), No_of_reads_from_NA12878 = c("15,000,000", "14,250,000",
"13,500,000", "12,750,000", "12,000,000", "11,250,000"), No_of_reads_from_NA12877 = c("0",
"750,000", "1,500,000", "2,250,000", "3,000,000", "3,750,000"
), tEst_contamination_of_NA12878 = c("99.60%", "99.10%", "96.80%",
"92.60%", "88%", "82.60%"), pair_contamination_of_NA12878 = c("100.00%",
"94.15%", "88.72%", "83.36%", "78.20%", "73.08%"), tEst_contamination_of_NA12877 = c("0.10%",
"7%", "13.60%", "20.10%", "26.20%", "32.10%"), pair_contamination_of_NA12877 = c("0.10%",
"5.21%", "10.50%", "15.85%", "20.92%", "26.04%")), .Names = c("diluted_sample",
"No_of_reads_from_NA12878", "No_of_reads_from_NA12877", "tEst_contamination_of_NA12878",
"pair_contamination_of_NA12878", "tEst_contamination_of_NA12877",
"pair_contamination_of_NA12877"), row.names = c(NA, 6L), class = "data.frame")
答案 0 :(得分：2)
自从您标记了data.table和melt
library(magrittr)
library(data.table)
setDT(test.dput)
n.reads <-
test.dput[, grep('diluted|reads', names(test.dput)), with = F] %>%
melt(1, variable.name = 'Which_No_of_reads',
value.name = 'No_of_reads') %>%
.[, Which_No_of_reads := gsub('No_of_reads_from_', '', Which_No_of_reads)]
contam <-
test.dput[, grep('diluted|contamination', names(test.dput)), with = F] %>%
melt(1, variable.name = 'Which_contamination',
value.name = '_contamination_') %>%
.[, Which_contamination := gsub('contamination_of_', '', Which_contamination)]
cbind(n.reads, contam) %>%
.[, unique(names(.)), with = F]
# diluted_sample Which_No_of_reads No_of_reads Which_contamination _contamination_
# 1: 100% NA12878 15,000,000 tEst_NA12878 99.60%
# 2: 95% NA12878 14,250,000 tEst_NA12878 99.10%
# 3: 90% NA12878 13,500,000 tEst_NA12878 96.80%
# 4: 85% NA12878 12,750,000 tEst_NA12878 92.60%
# 5: 80% NA12878 12,000,000 tEst_NA12878 88%
# 6: 75% NA12878 11,250,000 tEst_NA12878 82.60%
# 7: 100% NA12877 0 pair_NA12878 100.00%
# 8: 95% NA12877 750,000 pair_NA12878 94.15%
# 9: 90% NA12877 1,500,000 pair_NA12878 88.72%
# 10: 85% NA12877 2,250,000 pair_NA12878 83.36%
# 11: 80% NA12877 3,000,000 pair_NA12878 78.20%
# 12: 75% NA12877 3,750,000 pair_NA12878 73.08%
# 13: 100% NA12878 15,000,000 tEst_NA12877 0.10%
# 14: 95% NA12878 14,250,000 tEst_NA12877 7%
# 15: 90% NA12878 13,500,000 tEst_NA12877 13.60%
# 16: 85% NA12878 12,750,000 tEst_NA12877 20.10%
# 17: 80% NA12878 12,000,000 tEst_NA12877 26.20%
# 18: 75% NA12878 11,250,000 tEst_NA12877 32.10%
# 19: 100% NA12877 0 pair_NA12877 0.10%
# 20: 95% NA12877 750,000 pair_NA12877 5.21%
# 21: 90% NA12877 1,500,000 pair_NA12877 10.50%
# 22: 85% NA12877 2,250,000 pair_NA12877 15.85%
# 23: 80% NA12877 3,000,000 pair_NA12877 20.92%
# 24: 75% NA12877 3,750,000 pair_NA12877 26.04%
答案 1 :(得分：1)
使用tidyr::gather和dplyr：
test.melted<-gather(test.dput,key="reads_source",value="reads",starts_with("No_of_reads"))
test.melted.NA12878<-test.melted[test.melted$reads_source=="No_of_reads_from_NA12878",] %>%
gather(key="contamination_type",value="contamination",
contains("contamination_of_NA12878"))
test.melted.NA12877<-test.melted[test.melted$reads_source=="No_of_reads_from_NA12877",] %>%
gather(key="contamination_type",value="contamination",
contains("contamination_of_NA12877"))
test.melted.full<-rbind(test.melted.NA12877[,c(-2:-3)],test.melted.NA12878[,c(-2:-3)])
该解决方案显然仅适用于此特定数据集。如果您有更多阅读站点，也可以使用greping并可能使用for循环来进行管理。