我想将一个数据框的每一列与另一数据框列进行比较,并将每个结果重叠部分打印到单独的文件中。
我从两个测试数据集开始:
df1 <- data.frame("x" = c("a_b", "c_d", "e_f/c_f", "g_h"),
"y" = c(9,2,1,4),
"z" = c(7,5,8,5))
df2 <- data.frame("m" = c("c_f", "x_y"),
"n" = c("a_b", "x_y"))
并使用for循环获取结果。
for (i in colnames(df2)){
ccc<-df1[grep(paste(df2[,i], collapse = "|"), df1$x), ]
write.csv(ccc, file = paste(i, ".csv", sep=""))
}
一切都很好。
现在,我正在尝试在完整的数据集中使用相同的循环(以下是对df1和df2的修改):
df1<- structure(list(BGC_Accession = structure(c(1L, 1L, 1L, 2L), .Label = c("BGC0000647",
"BGC0000984"), class = "factor"), Genbank_ID = structure(c(1L,
3L, 2L, 4L), .Label = c("GCA_000202835", "GCA_000219295", "GCA_000964345",
"GCA_003029685"), class = "factor"), BGC_Class = structure(c(2L,
2L, 2L, 1L), .Label = c("NRP/Polyketide", "Terpene"), class = "factor"),
BGC_Start = c(2093957L, 1L, 1L, 2656134L), BGC_End = c(2115021L,
4440L, 4186L, 2721658L), Product = structure(c(1L, 1L, 1L,
2L), .Label = c("Carotenoid", "Delftibactin"), class = "factor"),
Similarity = structure(c(1L, 1L, 1L, 1L), .Label = "100%", class = "factor"),
Species_name = structure(c(1L, 4L, 2L, 3L), .Label = c("Acidiphilium_multivorum",
"Acidiphilium_sp_PM", "Acidovorax_avenae/Acidovorax_avene",
"Acinetobacter_baumannii"), class = "factor"), Kingdom = structure(c(1L,
1L, 1L, 1L), .Label = "k__Bacteria", class = "factor"), Phylum = structure(c(1L,
1L, 1L, 1L), .Label = "p__Proteobacteria", class = "factor"),
Class = structure(c(1L, 1L, 1L, 2L), .Label = c("c__Alphaproteobacteria",
"c__Betaproteobacteria"), class = "factor"), Order = structure(c(2L,
2L, 2L, 1L), .Label = c("o__Burkholderiales", "o__Rhodospirillales"
), class = "factor"), Family = structure(c(1L, 1L, 1L, 2L
), .Label = c("f__Acetobacteraceae", "f__Comamonadaceae"), class = "factor"),
Genus = structure(c(1L, 1L, 1L, 2L), .Label = c("g__Acidiphilium",
"g__Acidovorax"), class = "factor"), Species = structure(c(1L,
1L, 2L, 3L), .Label = c("s__Acidiphilium_multivorum", "s__Acidiphilium_sp_PM",
"s__Acidovorax_avenae"), class = "factor")), class = "data.frame", row.names = c(NA,
-4L))
df2<- structure(list(Gut_SRS011111 = structure(c(2L, 1L, 1L), .Label = c("",
"Actinobaculum_unclassified"), class = "factor"), Gut_SRS011269 = structure(c(3L,
1L, 2L), .Label = c("Acidiphilium_multivorum", "Acinetobacter_baumannii",
"Clostridium_citroniae"), class = "factor"), Gut_SRS011355 = structure(c(2L,
3L, 1L), .Label = c("", "Acidovorax_avene", "Streptococcus_gordonii"
), class = "factor")), class = "data.frame", row.names = c(NA,
-3L))
使用上面的脚本:
for (i in colnames(df2)){
overlap_data<-df1[grep(paste(df2[,i], collapse = "|"), df1$Species_name), ]
write.csv(overlap_data, file = paste(i, ".csv", sep=""))
}
似乎三个重叠列中只有一个(在df2中)给出正确的结果。 例如,在df2的第一列中,与df1没有重叠,因此应提供空白结果文件。第二列输出文件看起来正常。在第三个文件中,我应该得到一个重叠,而不是输出文件中给出的四个重叠。
我在做什么错了?
感谢您的耐心配合!
答案 0 :(得分:1)
问题似乎是空的""
单元格,应该是NA
。
df2[df2 == ""] <- NA
现在,grep
应该可以工作了。我在这里使用lapply
而不是for
循环:
invisible(lapply(names(df2), function(x) {
rr <- df1[grep(paste0(df2[,x], collapse= "|"), df1$Species_name), ]
write.csv(rr, file = paste(x, ".csv", sep=""))
}))
(invisible
避免了不必要和无聊的输出到控制台,您也可以将其忽略。)