以下是要重现的数据帧的最小示例。
df <- structure(list(Gene = structure(c(147L, 147L, 148L, 148L, 148L,
87L, 87L, 87L, 87L, 87L), .Label = c("genome", "k141_1189_101",
"k141_1189_104", "k141_1189_105", "k141_1189_116", "k141_1189_13",
"k141_1189_14", "k141_1189_146", "k141_1189_150", "k141_1189_18",
"k141_1189_190", "k141_1189_194", "k141_1189_215", "k141_1189_248",
"k141_1189_251", "k141_1189_252", "k141_1189_259", "k141_1189_274",
"k141_1189_283", "k141_1189_308", "k141_1189_314", "k141_1189_322",
"k141_1189_353", "k141_1189_356", "k141_1189_372", "k141_1189_373",
"k141_1189_43", "k141_1189_45", "k141_1189_72", "k141_1597_15",
"k141_1597_18", "k141_1597_23", "k141_1597_41", "k141_1597_55",
"k141_1597_66", "k141_1597_67", "k141_1597_68", "k141_1597_69",
"k141_2409_34", "k141_2409_8", "k141_3390_69", "k141_3390_83",
"k141_3390_84", "k141_3726_25", "k141_3726_31", "k141_3726_49",
"k141_3726_50", "k141_3726_62", "k141_3726_8", "k141_3726_80",
"k141_3790_1", "k141_3993_114", "k141_3993_122", "k141_3993_162",
"k141_3993_172", "k141_3993_183", "k141_3993_186", "k141_3993_188",
"k141_3993_24", "k141_3993_25", "k141_3993_28", "k141_3993_32",
"k141_3993_44", "k141_3993_47", "k141_3993_53", "k141_3993_57",
"k141_3993_68", "k141_4255_80", "k141_4255_81", "k141_4255_87",
"k141_5079_107", "k141_5079_110", "k141_5079_130", "k141_5079_14",
"k141_5079_141", "k141_5079_16", "k141_5079_184", "k141_5079_185",
"k141_5079_202", "k141_5079_24", "k141_5079_39", "k141_5079_63",
"k141_5079_65", "k141_5079_70", "k141_5079_77", "k141_5079_87",
"k141_5079_9", "k141_5313_16", "k141_5313_17", "k141_5313_20",
"k141_5313_23", "k141_5313_39", "k141_5313_5", "k141_5313_51",
"k141_5313_52", "k141_5313_78", "k141_5545_101", "k141_5545_103",
"k141_5545_104", "k141_5545_105", "k141_5545_106", "k141_5545_107",
"k141_5545_108", "k141_5545_109", "k141_5545_110", "k141_5545_111",
"k141_5545_112", "k141_5545_113", "k141_5545_114", "k141_5545_119",
"k141_5545_128", "k141_5545_130", "k141_5545_139", "k141_5545_141",
"k141_5545_145", "k141_5545_16", "k141_5545_169", "k141_5545_17",
"k141_5545_172", "k141_5545_6", "k141_5545_60", "k141_5545_62",
"k141_5545_63", "k141_5545_86", "k141_5545_87", "k141_5545_88",
"k141_5545_89", "k141_5545_91", "k141_5545_92", "k141_5545_93",
"k141_5545_94", "k141_5545_96", "k141_5545_97", "k141_5545_98",
"k141_5545_99", "k141_5734_13", "k141_5734_2", "k141_5734_4",
"k141_5734_5", "k141_5734_6", "k141_6014_124", "k141_6014_2",
"k141_6014_34", "k141_6014_75", "k141_6014_96", "k141_908_14",
"k141_908_2", "k141_908_5", "k141_957_126", "k141_957_135", "k141_957_136",
"k141_957_14", "k141_957_140", "k141_957_141", "k141_957_148",
"k141_957_179", "k141_957_191", "k141_957_35", "k141_957_47",
"k141_957_55", "k141_957_57", "k141_957_59", "k141_957_6", "k141_957_63",
"k141_957_65", "k141_957_68", "k141_957_77", "k141_957_95"), class = "factor"),
depth = c(9L, 10L, 9L, 10L, 11L, 14L, 15L, 16L, 17L, 18L),
bases_covered = c(6L, 3L, 4L, 7L, 4L, 59L, 54L, 70L, 34L,
17L), gene_length = c(1140L, 1140L, 591L, 591L, 591L, 690L,
690L, 690L, 690L, 690L), regioncoverage = c(54L, 30L, 36L,
70L, 44L, 826L, 810L, 1120L, 578L, 306L)), .Names = c("Gene",
"depth", "bases_covered", "gene_length", "regioncoverage"), row.names = c(1L,
2L, 33L, 34L, 35L, 78L, 79L, 80L, 81L, 82L), class = "data.frame")
数据框如下所示：
Gene depth bases_covered gene_length regioncoverage
1 k141_908_2 9 6 1140 54
2 k141_908_2 10 3 1140 30
33 k141_908_5 9 4 591 36
34 k141_908_5 10 7 591 70
35 k141_908_5 11 4 591 44
78 k141_5079_9 14 59 690 826
79 k141_5079_9 15 54 690 810
80 k141_5079_9 16 70 690 1120
81 k141_5079_9 17 34 690 578
82 k141_5079_9 18 17 690 306
我想要的是，对于每个基因（例如k141_908_2），我想要对区域覆盖进行求和并除以唯一（基因长度）。事实上，每个基因的基因长度总是相同的。
例如对于Gene K141_908_2，我会这样做：（54 + 30）/ 1140 = 0.07 例如对于Gene K141_908_5，我会这样做：（36 + 70 + 44）/ 591 = 0.25
最终的数据框应报告两列。
Gene Newcoverage
1 k141_908_2 0.07
2 k141_908_5 0.25
3 ......
等等。
感谢您的帮助
答案 0 :(得分：2)
使用VerticalAlignment="{TemplateBinding VerticalAlignment}"
HorizontalAlignment="{TemplateBinding HorizontalAlignment}"
：
dplyr答案 1 :(得分：1)
我需要将第一列设置为character而将其他列设置为numeric。但之后你可以通过基因分割df，然后进行必要的计算。
df[,2:5] = lapply(df[,2:5], as.numeric)
df$Gene = as.character(df$Gene)
sapply(split(df, df$Gene), function(x) sum(x[,5]/x[1,4]))
#k141_5079_9 k141_908_2 k141_908_5
# 5.27536232 0.07368421 0.25380711
答案 2 :(得分：1)
我们可以使用tidyverse
library(tidyverse)
df %>%
group_by(Gene) %>%
summarise(Newcoverage = sum(regioncoverage)/gene_length[1])
# A tibble: 3 × 2
# Gene Newcoverage
# <fctr> <dbl>
#1 k141_5079_9 5.27536232
#2 k141_908_2 0.07368421
#3 k141_908_5 0.25380711
或base R选项
by(df[4:5], list(as.character(df[,'Gene'])), FUN= function(x) sum(x[,2])/x[1,1])
答案 3 :(得分：0)
快速方法
require(data.table)
DT <- setDT(df)
#just to output unique rows
DT[, .(New_Coverage = unique(sum(regioncoverage)/gene_length)), by = .(Gene)]
输出
Gene New_Coverage
1: k141_908_2 0.07368421
2: k141_908_5 0.25380711
3: k141_5079_9 5.27536232
答案 4 :(得分：0)
我经常使用dplyr。所以这是一种方式：
library(dplyr)
df %>%
group_by(Gene) %>%
mutate(Newcoverage=sum(regioncoverage)/unique(gene_length))
如果您只想要每个基因的唯一值：
df %>%
group_by(Gene) %>%
transmute(Newcoverage=sum(regioncoverage)/unique(gene_length)) %>%
unique()