我正在尝试使用grep过滤我的数据,但还要在结果中包括NA,由于它们与grep表达式不匹配,因此目前已将其删除。
platform x86_64-w64-mingw32
version.string R version 3.5.3 (2019-03-11)
value expected_result actual_result
1 10001 Pass Pass
2 0 Pass Pass
3 6 Pass Pass
4 20004 Pass Pass
5 NA Pass Fail
6 4829 Fail Fail
7 521 Fail Fail
8 89 Fail Fail
9 40012 Fail Fail
10 47321 Fail Fail
df <- df[grep("(\\b\\d{1}\\b)|([0-9]{1}[0]{3}[0-9]{1})", df$value),]
1)该值将包含0到5个数字字符。
2)应该保留的三个值是:
a)一位数据。 (示例值2和3)
b)没有数据或不适用(示例值5)
c)数据的五位数,但中间的三位数必须全部为零。 (示例值1和4)
答案 0 :(得分:3)
要包括NA
行,请在is.na
上用|
创建第二个条件,并与grepl
(OR)联接起来
df[grepl("(\\b\\d{1}\\b)|([0-9]{1}[0]{3}[0-9]{1})", df$value)|is.na(df$value),]
# value expected_result actual_result
#1 10001 Pass Pass
#2 0 Pass Pass
#3 6 Pass Pass
#4 20004 Pass Pass
#5 NA Pass Fail
或者使其更紧凑
grepl("^\\d$|^([1-9]0{3}[1-9]$)", df$value)|is.na(df$value)
df <- structure(list(value = c(10001L, 0L, 6L, 20004L, NA, 4829L, 521L,
89L, 40012L, 47321L), expected_result = c("Pass", "Pass", "Pass",
"Pass", "Pass", "Fail", "Fail", "Fail", "Fail", "Fail"), actual_result = c("Pass",
"Pass", "Pass", "Pass", "Fail", "Fail", "Fail", "Fail", "Fail",
"Fail")), class = "data.frame", row.names = c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10"))