Question

我想找到哪个列包含1的最大数量。数字1每行最多只能出现一次。只要找到编号为1最高的列，脚本就会检查相邻列（+1+ / -1），如果其中任何一列包含数字1，则应该是也选择了。所有这些行都应保留在子集函数中。

让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可以清楚地看到的那样，应该选择列506作为包含最大数量1的列，并且应该根据它来对数据进行子目录。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，就像那种情况一样：

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1s。在这种情况下，应选择较大colsum的列。

Answer 1

让extendObservable成为您的输入：

df1

在两个示例中，所有行都保留

Answer 2

我使用tidyverse将其转换为长格式，然后拉入列总和以确定第一个（具有最大总和）的位置：

library(tidyverse)

# add rownames to the data frame
df2$id  <- rownames(df2)

# make a data frame of each column's sum
thecolsums  <- colSums(df2[,map_lgl(df2, is.numeric)]) %>% 
  enframe(name = "colname", value = "colsum")

# change the data frame to long format
dflong  <- df2 %>% 
  mutate(rowid = row_number()) %>% 
  gather(colname, val, -rowid)

# which column has the first 1 value
whichcol  <- dflong %>% 
  group_by(colname) %>% 
  filter(val ==1) %>% 
  summarize(
    firstone = min(rowid, na.rm = T)
  ) %>% 
  left_join(thecolsums, by = 'colname') %>% 
  filter(colsum == max(colsum)) %>% 
  pluck('colname')

# what's the numerical index of the column
whichcolindex  <- which(names(df2) == whichcol)

# get previous and next columns if they exist
prevcolindex  <- ifelse(whichcolindex < 1, F, whichcolindex -1)
nextcolindex  <- ifelse(whichcolindex == ncol(df2) , F, whichcolindex +1)

# do the previous and next columns have 1s in them?
prevcolhasone  <- any(df2[,prevcolindex] == 1)
nextcolhasone  <- any(df2[,nextcolindex] == 1)

# create a vector with 1, 2 or 3 column indexes
finalindex  <- c(
    prevcolindex[prevcolhasone]
  , whichcolindex
  , nextcolindex[nextcolhasone]
)

# subset the original data frame, only preserving the columns in question
results  <- df2[, finalindex]

数据框的子集基于列或相邻列中的值

2 个答案: