我有一些数据的形式：
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11
1. 14342667 4336 1.5 2015-10-03 \\N 2015-10-03 804 2.0 2.0 0 3.0
2. 14342667 4336 0.8 2015-06-13 \\N 2015-06-11 2912 2.0 2.0 0 \\N
3. 14342667 4336 0.5 2016-01-02 \\N 2015-12-27 1618 0.0 0.0 1 \\N
4. 14342667 4336 0.7 2015-08-22 \\N 2015-08-22 1780 2.0 2.0 0 \\N
5. 14342667 4336 0.9 2015-02-21 1.2 2015-02-17 1548 0.0 0.0 1 \\N
6. 14342667 4336 1.0 2015-08-08 \\N 2015-08-06 1538 2.0 2.0 0 2.25
7. 14342667 4336 0.9 2015-03-28 \\N 2015-03-24 2129 7.0 7.0 0 \\N
8. 14342667 4336 0.8 2015-04-11 \\N 2015-04-11 2316 1.0 2.0 0 \\N
我想\\N使用R和data.table将NA的实例替换为NA。当我将数据收集到HIVE中时，我把这些留在了空白处。
我已经尝试过类似的事情，data1 <- data1[, lapply(.SD, recode, '"\\N"=NA')]并data1 <- data1[, lapply(.SD, recode, '"\N"=NA')]使用了CAR包，该包可以很好地用于其他替换，但在当前情况下失败并出现错误：
FUN（X [[1L]]，...）中的错误：在重新编码项中：“ \ N” = NA消息：错误：'\ N'是无法识别的以“” \ N“开头的字符串的转义
我也读过类似的解决方案，x[x=="\\N"] <- NA但是我不能成功地使用data.table来应用那些解决方案。我也研究了HIVE表端解决方案，但显然regexp_replace仅适用于一列，而不适用于所有列中的所有实例。
我们可以找出具有此模式的列，grep然后as.numeric在这些列上使用
library(data.table)
nm1 <- names(df1)[colSums(sapply(df1, grepl, pattern = "\\\\"))!=0]
setDT(df1)[,(nm1):= lapply(.SD, as.numeric) , .SDcols= nm1]
df1
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11
#1: 14342667 4336 1.5 2015-10-03 NA 2015-10-03 804 2 2 0 3.00
#2: 14342667 4336 0.8 2015-06-13 NA 2015-06-11 2912 2 2 0 NA
#3: 14342667 4336 0.5 2016-01-02 NA 2015-12-27 1618 0 0 1 NA
#4: 14342667 4336 0.7 2015-08-22 NA 2015-08-22 1780 2 2 0 NA
#5: 14342667 4336 0.9 2015-02-21 1.2 2015-02-17 1548 0 0 1 NA
#6: 14342667 4336 1.0 2015-08-08 NA 2015-08-06 1538 2 2 0 2.25
#7: 14342667 4336 0.9 2015-03-28 NA 2015-03-24 2129 7 7 0 NA
#8: 14342667 4336 0.8 2015-04-11 NA 2015-04-11 2316 1 2 0 NA
df1 <- structure(list(V1 = c(14342667L, 14342667L, 14342667L, 14342667L,
14342667L, 14342667L, 14342667L, 14342667L), V2 = c(4336L, 4336L,
4336L, 4336L, 4336L, 4336L, 4336L, 4336L), V3 = c(1.5, 0.8, 0.5,
0.7, 0.9, 1, 0.9, 0.8), V4 = c("2015-10-03", "2015-06-13", "2016-01-02",
"2015-08-22", "2015-02-21", "2015-08-08", "2015-03-28", "2015-04-11"
), V5 = c("\\\\N", "\\\\N", "\\\\N", "\\\\N", "1.2", "\\\\N",
"\\\\N", "\\\\N"), V6 = c("2015-10-03", "2015-06-11", "2015-12-27",
"2015-08-22", "2015-02-17", "2015-08-06", "2015-03-24", "2015-04-11"
), V7 = c(804L, 2912L, 1618L, 1780L, 1548L, 1538L, 2129L, 2316L
), V8 = c(2, 2, 0, 2, 0, 2, 7, 1), V9 = c(2, 2, 0, 2, 0, 2, 7,
2), V10 = c(0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L), V11 = c("3.0", "\\\\N",
"\\\\N", "\\\\N", "\\\\N", "2.25", "\\\\N", "\\\\N")), .Names = c("V1",
"V2", "V3", "V4", "V5", "V6", "V7", "V8", "V9", "V10", "V11"),
class = "data.frame", row.names = c("1.",
"2.", "3.", "4.", "5.", "6.", "7.", "8."))
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句