r - 基于列或相邻列中的值的数据框子集

Question

我想找出哪一列包含最多的 1。数字 1 每行应该只出现一次。一旦找到具有最高编号 1 的列，脚本还应检查相邻的列 (+1+/-1)，如果有的话其中包含数字 1，也应将其选中。所有这些行都应保留在子集函数中。

我们放一部分原始数据:

structure(list(   `10` = c(0, 0, 0, 0),  `34` = c(0, 0, 0, 0),
                  `59` = c(0, 0, 0, 0),  `84` = c(0, 0, 0, 0),
                 `110` = c(0, 0, 0, 0), `134` = c(0, 0, 0, 0),
                 `165` = c(0, 0, 0, 0), `199` = c(0, 0, 0, 0),
                 `234` = c(0, 0, 0, 0),
                 `257` = c(0.0160178986200301, 0, 0.0409772658686249, 0.0289710439505515),
                 `362` = c(0.0679054515644214, 0.126933274414494, 0.0855598028367368, 0.0596214721268868),
                 `433` = c(0.490914059297718, 0.604765061128296, 0.813348757670254, 1),
                 `506` = c(1, 1, 1, 0.971410482822965),
                 `581` = c(0.198244295668807, 0.234158197083517, 0.269655970224324, 0.195318383259472),
                 `652` = c(0.271177756524115, 0.223018854028576, 0.301352982597324, 0.142584385725234),
                 `733` = c(0.212426561005602, 0.212778023272942, 0.228513228045468, 0),
                 `818` = c(0.213816778248395, 0.168570481661511, 0.264465345538678, 0),
                 `896` = c(0.137102063123377, 0, 0.320234382858867, 0),
                 `972` = c(0.108932231179123, 0, 0.179106729705261, 0),
                `1039` = c(0.101762535865555, 0, 0, 0),
                   EOD = c("Peter", "Peter", "Peter", "Peter"),
               Complex = c(""FT team", "FT team", "FT team", "FT team")),
          .Names = c("10", "34", "59", "84", "110", "134", "165", "199",
                     "234", "257", "362", "433", "506", "581", "652", "733",
                     "818", "896", "972", "1039", "EOD", "Complex"),
          row.names = c("Peter_1_Rep_1_E", "Peter_1_Rep_2_E",
                        "Peter_1_Rep_3_E", "Peter_1_Rep_4_E"),
          class = "data.frame")

正如您在原始数据中清楚看到的那样，506 列应该被选为包含最多 1 的列，并且数据应该基于它进行子集化。但是，输出将完全相同，因为在此数据中，相邻分数 (-1, 433) 也包含 1。这是一个简单的例子。

情况可能更复杂，比如在那种情况下:

structure(list(    `10` = c(0, 0, 0, 0, 0, 0, 0, 0),
                   `34` = c(0, 0, 0, 0, 0, 0, 0, 0),
                   `59` = c(0, 0, 0, 0, 0, 0, 0, 0),
                   `84` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `110` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `134` = c(0.168783347110543, 0, 0.382618775924215, 0, 0.530638724516877, 0, 0.169526042048202, 0),
                  `165` = c(1, 0.36380544964196, 1, 0.13979454361738, 1, 0.239652477288689, 1, 0.240341578327444),
                  `199` = c(0.355158938904336, 1, 0.646724265971128, 1, 0.582637073151552, 1, 0.20319390520841, 1),
                  `234` = c(0.0963628165627114, 0.575436312346942, 0.229853828180188, 0.433555069046817, 0.247567185011894, 0.508529485059242, 0.138356164383562, 0.389880251276011),
                  `257` = c(0, 0.17393595585728, 0, 0.127787133715056, 0, 0.117147323350173, 0, 0),
                  `362` = c(0, 0, 0, 0.0919333108790839, 0, 0, 0, 0),
                  `433` = c(0, 0, 0, 0.0745570899292691, 0, 0, 0, 0),
                  `506` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `581` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `652` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `733` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `818` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `896` = c(0, 0, 0, 0, 0, 0, 0, 0),
                  `972` = c(0, 0, 0, 0, 0, 0, 0, 0),
                 `1039` = c(0, 0, 0, 0, 0, 0, 0, 0),
                    EOD = c("Paul", "Paul", "Paul", "Paul", "Paul", "Paul", "Paul", "Paul"),
                Complex = c("GG Team", "GG Team", "GG Team", "GG Team", "GG Team", "GG Team", "GG Team", "GG Team")),
          .Names = c("10", "34", "59", "84", "110", "134", "165", "199", "234", "257", "362", "433", "506", "581", "652", "733", "818", "896", "972", "1039", "EOD", "Complex"),
          row.names = c("PaulG_1_Rep_1_E", "Paul_1_Rep_1_E", "PaulN_1_Rep_2_E", "PaulG_1_Rep_2_E", "Paul_1_Rep_3_E", "PaulC_1_Rep_3_E", "PaulC_1_Rep_4_E", "Paul_1_Rep_4_E"),
          class = "data.frame")

在这种情况下，有两列包含相同数量的 1。在这种情况下，应选择具有更大 colsum 的列。

Answer 1

让 df1 成为您的输入:

df_num <- df1[,sapply(df1,is.numeric)]            # keep only numeric columns to build filter
n1 <- colSums(df_num == 1)                        # number of 1s per column
i  <- which(n1 == max(n1))                        # index of cols with max 1s
if(length(i) > 1){
  max_cs <- which.max(colSums(df_num[,i]))        # index of col with max colsum among results
  i <- i[max_cs]                                  # our column index
}
filter <- rowSums(df_num[,seq(max(i-1,0),min(i+1,ncol(df_num)))]==1) >0    # filter is true if chosen column is 1 or if any neighbour is 1

df1[filter,] # your result

在你的两个例子中，所有的行都被保留