将一个数据框的每一列与另一个数据框的每一列进行比较,并将每个结果的重叠打印到单独的文件中。

问题描述 投票:0回答:1

我想将一个数据框的每一列与另一个数据框的每一列进行比较,并将每个结果的重叠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循环来得到结果。

for (i in colnames(df2)){ 
  ccc<-df1[grep(paste(df2[,i], collapse = "|"), df1$x), ]
  write.csv(ccc, file = paste(i, ".csv", sep=""))
}

一切看起来都很好。

现在我尝试在我的完整数据集中使用同样的循环(下面是修改后的df1和df2)。

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=""))
}

似乎只有三列重叠的一列(在df2中)给出了正确的结果。例如,在df2的第一列中,没有与df1重叠,它应该给出一个空白的结果文件。第二列的输出文件看起来不错。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。

我到底做错了什么?

谢谢您的耐心等待

r
1个回答
1
投票

问题似乎是空的 "" 细胞,应该是 NA.

df2[df2 == ""] <- NA

现在... grep 应该可以。我用的是 lapply 而不是 for 循环在这里。

invisible(lapply(names(df2), function(x) {
  rr <- df1[grep(paste0(df2[,x], collapse= "|"), df1$Species_name), ]
  write.csv(rr, file = paste(x, ".csv", sep=""))
}))

(The invisible 避免了不必要的、枯燥无味的输出到控制台,你也可以不使用它)。)

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