Question

我正在尝试更改列表的多个特定列。我想使某些行（特别是2,3,5,6）中的所有值都相同。即，列中的值对于每一行是相同的。我想将第4和第7列从NA＆＃39;更改为零（0）

我可以使用以下内容向列表中的每个数据框添加一个带零的列：

lapply(df1, function(x) cbind(x,replace(x$efficiency, is.na(x$efficiency), "0")))

我一直试图用这个：

lapply(df1, na.locf,df1$Receiver)

作为一种解决方法，但进展甚微

示例数据：

df1<-list(structure(list(dt15 = structure(c(1457929800, 1457930700, 
1457931600, 1457932500, 1457933400, 1457934300), class = c("POSIXct", 
"POSIXt"), tzone = ""), Receiver = c(480432, 480432, NA, NA, 
NA, NA), Code = c(62431, 62431, NA, NA, NA, NA), detections = c(6, 
1, NA, NA, NA, NA), distance = c(168.948559873358, 168.948559873358, 
NA, NA, NA, NA), Repeat_Rate = c(90L, 90L, NA, NA, NA, NA), efficiency =     c("60", 
"10", NA, NA, NA, NA)), .Names = c("dt15", "Receiver", "Code", 
"detections", "distance", "Repeat_Rate", "efficiency"), row.names = 635:640,     class = "data.frame"), 
structure(list(dt15 = structure(c(1457956800, 1457957700, 
1457958600, 1457959500, 1457960400, 1457961300, 1457962200, 
1457963100), class = c("POSIXct", "POSIXt"), tzone = ""), 
    Receiver = c(480422, 480422, NA, NA, NA, NA, 480422, 
    480422), Code = c(62427, 62427, NA, NA, NA, NA, 62427, 
    62427), detections = c(2, 2, NA, NA, NA, NA, 1, 3), distance =   c(301.8128773339, 
    301.8128773339, NA, NA, NA, NA, 301.8128773339, 301.8128773339
    ), Repeat_Rate = c(90L, 90L, NA, NA, NA, NA, 90L, 90L
    ), efficiency = c("20", "20", NA, NA, NA, NA, "10", "30"
    )), .Names = c("dt15", "Receiver", "Code", "detections", 
    "distance", "Repeat_Rate", "efficiency"), row.names = 665:672, class =    "data.frame"))

期望的输出：

list(structure(list(dt15 = structure(c(1457929800, 1457930700, 
1457931600, 1457932500, 1457933400, 1457934300), class = c("POSIXct", 
"POSIXt"), tzone = ""), Receiver = c(480432, 480432, 480432, 
480432, 480432, 480432), Code = c(62431, 62431, 62431, 62431, 
62431, 62431), detections = c(6, 1, 0, 0, 0, 0), distance =  c(168.948559873358, 
168.948559873358, 168.948559873358, 168.948559873358, 168.948559873358, 
168.948559873358), Repeat_Rate = c(90L, 90L, 90L, 90L, 90L, 90L
), efficiency = c("60", "10", "0", "0", "0", "0")), .Names = c("dt15", 
"Receiver", "Code", "detections", "distance", "Repeat_Rate", 
"efficiency"), row.names = 635:640, class = "data.frame"), structure(list(
dt15 = structure(c(1457956800, 1457957700, 1457958600, 1457959500, 
1457960400, 1457961300, 1457962200, 1457963100), class = c("POSIXct", 
"POSIXt"), tzone = ""), Receiver = c(480422, 480422, 480422, 
480422, 480422, 480422, 480422, 480422), Code = c(62427, 
62427, 62427, 62427, 62427, 62427, 62427, 62427), detections = c(2, 
2, 0, 0, 0, 0, 1, 3), distance = c(301.8128773339, 301.8128773339, 
301.8128773339, 301.8128773339, 301.8128773339, 301.8128773339, 
301.8128773339, 301.8128773339), Repeat_Rate = c(90L, 90L, 
90L, 90L, 90L, 90L, 90L, 90L), efficiency = c("20", "20", 
"0", "0", "0", "0", "10", "30")), .Names = c("dt15", "Receiver", 
"Code", "detections", "distance", "Repeat_Rate", "efficiency"
), row.names = 665:672, class = "data.frame"))

Answer 1

我不太确定你想做什么，但这应该有所帮助。创建一个修复单个数据框的函数

fix_data_frame = function(x) {
  x[is.na(x[,7]),7] = 0
  x[is.na(x[,4]),4] = 0
  # What ever else you want to do

  return(x)
}

测试：

fix_data_frame(df[[1]])
fix_data_frame(df[[2]])

一旦有效，只需将其贴在lapply

中即可

lapply(df1, fix_data_frame)

lapply特定列中的数据框列表

1 个答案: