提高 R 中 3 for 循环的速度

Question

我正在使用 206 行 x 196 列的矩阵 set_onco，我有一个向量 genes_100（它是一个矩阵，但我只取第一个列），有 101 个名称。 这是他们的外观的sn-p

> set_onco[1:10,1:10]
                             V2       V3        V4        V5      V6     V7     V8      V9     V10      V11
GLI1_UP.V1_DN             COPZ1 C10orf46 C20orf118   TMEM181   CCNL2  YIPF1  GTDC1    OPN3   RSAD2  SLC22A1
GLI1_UP.V1_UP            IGFBP6 HLA-DQB1     CCND2     PTH1R TXNDC12   M6PR   PPT2   STAU1     IGJ    TMOD3
E2F1_UP.V1_DN           TGFB1I1    CXCL5    POU5F1    SAMD10    KLF2  STAT6 ENTPD6    VCAN  HMGCS1    ANXA8
E2F1_UP.V1_UP             RRP1B     HES1     ADCY6    CHAF1B  VPS37B  GRSF1   TLX2  SSX2IP    DNA2     CMA1
EGFR_UP.V1_DN             NPY1R    PDZK1     GFRA1     GREB1    MSMB   DLC1    MYB SLC6A14   IFI44   IFI44L
EGFR_UP.V1_UP               FGG     GBP1 TNFRSF11B       FGB    GJA1  DUSP6 S100A9     ADM   ITGB6    DUSP4
ERB2_UP.V1_DN             NPY1R    PDZK1     ANXA3     GREB1   HSPB8   DLC1  NRIP1    FHL2    EGR3    IFI44
FAM18B1                                                                                                    
ERB2_UP.V1_UP            CYP1A1  CEACAM5   FAM129A TNFRSF11B   DUSP4 CYP1B1   UPK2    DAB2 CEACAM6 KIAA1199
GCNP_SHH_UP_EARLY.V1_DN   SRRM2 KIAA1217     DEFA1      DLK1   PITX2   CCL2  UPK3B    SEZ6   TAF15     EMP1

genes_100[1:10,1]
 [1] AL591845.1   B3GALT6      RAP1GAP      HSPG2        BX293535.1   RP1-159A19.1 IFI6         FAM76A       FAM176B      CSF3R       
101 Levels: 5_8S_rRNA AC018470.1 AC091179.2 AC103702.3 AC138972.1 ACVR1B AL049829.5 AL137797.2 AL139260.2 AL450326.2 AL591845.1 AL607122.2 B3GALT6 BX293535.1 ... ZNF678

我要做的是解析矩阵并计算每行包含genes_100中的名称的频率

为此，我创建了 3 个 for 循环：第一个同时向下移动一行，第二个移入该行，第三个循环遍历列表 genes_100 检查匹配项。 最后，我将genes_100 与每行中的术语匹配的次数保存在矩阵中，同时保存矩阵中的行名（以便我知道哪个是哪个）

代码可以正常工作并给我正确的输出...但它真的很慢！！

输出的一个sn-p是：

head(result_matrix_100)

                    freq_100
[1,] "GLI1_UP.V1_DN" "0"     
[2,] "GLI1_UP.V1_UP" "0"     
[3,] "E2F1_UP.V1_DN" "0"     
[4,] "E2F1_UP.V1_UP" "0"     
[5,] "EGFR_UP.V1_DN" "0"     
[6,] "EGFR_UP.V1_UP" "0" 

我使用了system.time()，我得到了：

  user  system elapsed 
 525.38    0.06  530.34

这太慢了，因为我要解析更大的矩阵，在某些情况下我必须重复这 10k 次！！！

代码是：

result_matrix_100 <- matrix(nrow=0, ncol=2)

for (q in seq(1,nrow(set_onco),1)) {
  for (j in seq(1, length(set_onco[q,]),1)) {
    for (x in seq(1,101,1)) {
      if (as.character(genes_100[x,1]) == as.character(set_onco[q,j])) {
        freq_100 <- freq_100+1
      }
    }
  }
  result_matrix_100 <- rbind(result_matrix_100, cbind(row.names(set_onco)[q], freq_100))
}

你有什么建议？

提前感谢:)

Answer 1

@joran 可能会更快，尽管它可能不是“因素安全的”。您的set_onco 值可能被编码为因子变量（因为您的genes_100 对象显然是。）这会更安全：

set_onco[] <- lapply(set_onco, as.character)
# that converts a data.frame with factor columns to character valued
# at that point @joran's solution could be used safely
freq100 <- apply(set_onco, 1, function(x) sum(x %in% genes_100) )
# that does a row-by-row count of the number of matches to genes_100
freq100
          GLI1_UP.V1_DN           GLI1_UP.V1_UP           E2F1_UP.V1_DN 
                      0                       0                       0 
          E2F1_UP.V1_UP           EGFR_UP.V1_DN           EGFR_UP.V1_UP 
                      0                       0                       0 
          ERB2_UP.V1_DN           ERB2_UP.V1_UP GCNP_SHH_UP_EARLY.V1_DN 
                      0                       0                       0 

您的数据集的大小（206 行 x 196 列）非常小，因此几乎可以立即完成。这些 dput 语句和输出可用于构造我认为您的对象内部的样子：

dput(set_onco)
structure(list(V2 = structure(c(1L, 4L, 8L, 6L, 5L, 3L, 5L, 2L, 
7L), .Label = c("COPZ1", "CYP1A1", "FGG", "IGFBP6", "NPY1R", 
"RRP1B", "SRRM2", "TGFB1I1"), class = "factor"), V3 = structure(c(1L, 
6L, 3L, 5L, 8L, 4L, 8L, 2L, 7L), .Label = c("C10orf46", "CEACAM5", 
"CXCL5", "GBP1", "HES1", "HLA-DQB1", "KIAA1217", "PDZK1"), class = "factor"), 
    V4 = structure(c(3L, 4L, 8L, 1L, 7L, 9L, 2L, 6L, 5L), .Label = c("ADCY6", 
    "ANXA3", "C20orf118", "CCND2", "DEFA1", "FAM129A", "GFRA1", 
    "POU5F1", "TNFRSF11B"), class = "factor"), V5 = structure(c(7L, 
    5L, 6L, 1L, 4L, 3L, 4L, 8L, 2L), .Label = c("CHAF1B", "DLK1", 
    "FGB", "GREB1", "PTH1R", "SAMD10", "TMEM181", "TNFRSF11B"
    ), class = "factor"), V6 = structure(c(1L, 8L, 5L, 9L, 6L, 
    3L, 4L, 2L, 7L), .Label = c("CCNL2", "DUSP4", "GJA1", "HSPB8", 
    "KLF2", "MSMB", "PITX2", "TXNDC12", "VPS37B"), class = "factor"), 
    V7 = structure(c(8L, 6L, 7L, 5L, 3L, 4L, 3L, 2L, 1L), .Label = c("CCL2", 
    "CYP1B1", "DLC1", "DUSP6", "GRSF1", "M6PR", "STAT6", "YIPF1"
    ), class = "factor"), V8 = structure(c(2L, 5L, 1L, 7L, 3L, 
    6L, 4L, 8L, 9L), .Label = c("ENTPD6", "GTDC1", "MYB", "NRIP1", 
    "PPT2", "S100A9", "TLX2", "UPK2", "UPK3B"), class = "factor"), 
    V9 = structure(c(4L, 8L, 9L, 7L, 6L, 1L, 3L, 2L, 5L), .Label = c("ADM", 
    "DAB2", "FHL2", "OPN3", "SEZ6", "SLC6A14", "SSX2IP", "STAU1", 
    "VCAN"), class = "factor"), V10 = structure(c(8L, 6L, 4L, 
    2L, 5L, 7L, 3L, 1L, 9L), .Label = c("CEACAM6", "DNA2", "EGR3", 
    "HMGCS1", "IFI44", "IGJ", "ITGB6", "RSAD2", "TAF15"), class = "factor"), 
    V11 = structure(c(8L, 9L, 1L, 2L, 6L, 3L, 5L, 7L, 4L), .Label = c("ANXA8", 
    "CMA1", "DUSP4", "EMP1", "IFI44", "IFI44L", "KIAA1199", "SLC22A1", 
    "TMOD3"), class = "factor")), .Names = c("V2", "V3", "V4", 
"V5", "V6", "V7", "V8", "V9", "V10", "V11"), class = "data.frame", row.names = c("GLI1_UP.V1_DN", 
"GLI1_UP.V1_UP", "E2F1_UP.V1_DN", "E2F1_UP.V1_UP", "EGFR_UP.V1_DN", 
"EGFR_UP.V1_UP", "ERB2_UP.V1_DN", "ERB2_UP.V1_UP", "GCNP_SHH_UP_EARLY.V1_DN"
))

dput(factor(genes_100) )
structure(c(1L, 2L, 9L, 7L, 3L, 10L, 8L, 6L, 5L, 4L), .Label = c("AL591845.1", 
"B3GALT6", "BX293535.1", "CSF3R", "FAM176B", "FAM76A", "HSPG2", 
"IFI6", "RAP1GAP", "RP1-159A19.1"), class = "factor")

Answer 2

这样的事情可能会很快：

#Sample data
m <- matrix(sample(letters,206*196,replace = TRUE),206,196)
genes_100 <- letters[1:5]

m1 <- matrix(m %in% genes_100,206,196)
rowSums(m1)