查找具有一系列连续列值的行

Question

假设我有一个类似于下面的数据框,我需要识别每个行,其中一个或多个缺失值(NA)后面跟着至少一个有效值(任何数字).你能帮助我吗？

a <- c(1, 'S06.4', 6.7, 7.0, 6.5, 7.0, 7.2, NA, NA, 6.6,6.7) 
b <- c(2 ,'S06.2' ,5.0, NA, 4.9, 7.8, 9.3, 8.0, 7.8, 8.0,NA)
c <- c(3, 'S06.5', 7.0, 5.5, NA, NA, 7.2, 8.0, 7.6, NA,6.7) 
d <- c(4, 'S06.5', 7.0, 7.0, 7.0, 6.9, 6.8, 9.0, 6.0, 6.6,6.7) 
e <- c(5, 'S06.1', 6.7, NA, NA, NA, NA, NA, NA, NA,NA) 

df <- data.frame(rbind(a,b,c,d,e))
colnames(df) <- c('id','dx','dia01','dia02','dia03','dia04','dia05','dia06','dia07','dia08','dia09')

Answer 1

附:

df[rowSums(is.na(df[,3:10]) * !is.na(df[,4:11])) > 0,]

你得到:

  id    dx dia01 dia02 dia03 dia04 dia05 dia06 dia07 dia08 dia09
a  1 S06.4   6.7     7   6.5     7   7.2  <NA>  <NA>   6.6   6.7
b  2 S06.2     5  <NA>   4.9   7.8   9.3     8   7.8     8  <NA>
c  3 S06.5     7   5.5  <NA>  <NA>   7.2     8   7.6  <NA>   6.7

这是做什么的:

• is.na(df[,3:10])检查dia01to dia08列中的哪个值NA并返回逻辑矩阵.
• !is.na(df[,4:11])对每行中的下一个值执行相同操作,df[,3:10]并返回逻辑矩阵
• 将这两个矩阵相乘得出所需条件的逻辑矩阵.
• 随着rowSums你检查的条件是否每一行中至少遇到一次.

在回复您的评论时:如果您想确保NA后面跟一个数值,您可以将以上解决方案更改为:

# first convert the 'dia*''-columns to numeric
df[-c(1,2)] <- lapply(df[-c(1,2)], function(x) as.numeric(as.character(x)))
# then do the same because values that can't converted to numeric will give NA
df[rowSums(is.na(df[,3:10]) * !is.na(df[,4:11])) > 0,]

或者首先不转换为数字:

df[rowSums(is.na(df[,3:10]) * !is.na(sapply(df[4:11], function(x) as.numeric(as.character(x))))) > 0,]

注意:

使用您用于构建示例数据的方法,您将得到所有因子列.其中我想你不想要那个.

可能格式正确的示例数据集将是:

df <- structure(list(id = c("1", "2", "3", "4", "5"), 
                     dx = c("S06.4", "S06.2", "S06.5", "S06.5", "S06.1"), 
                     dia01 = c(6.7, 5, 7, 7, 6.7),
                     dia02 = c(7, NA, 5.5, 7, NA), 
                     dia03 = c(6.5, 4.9, NA, 7, NA),
                     dia04 = c(7, 7.8, NA, 6.9, NA),
                     dia05 = c(7.2, 9.3, 7.2, 6.8, NA),
                     dia06 = c(NA, 8, 8, 9, NA),
                     dia07 = c(NA, 7.8, 7.6, 6, NA),
                     dia08 = c(6.6, 8, NA, 6.6, NA),
                     dia09 = c(6.7, NA, 6.7, 6.7, NA)),
                .Names = c("id", "dx", "dia01", "dia02", "dia03", "dia04", "dia05", "dia06", "dia07", "dia08", "dia09"),
                row.names = c("a", "b", "c", "d", "e"),
                class = "data.frame")

所提出的方法也适用于此.

正如@Frank在评论中所指出的,最好以长格式存储数据.附:

library(data.table)
setDT(df)[, 3:11 := lapply(.SD, function(x) as.numeric(as.character(x))), .SDcols = 3:11][]
melt(df, id = 1:2)[, if(any(is.na(value) & !is.na(shift(value, type = 'lead')))) .SD, by = .(id, dx)]

你得到:

    id    dx variable value
 1:  1 S06.4    dia01   6.7
 2:  1 S06.4    dia02   7.0
 3:  1 S06.4    dia03   6.5
 4:  1 S06.4    dia04   7.0
 5:  1 S06.4    dia05   7.2
 6:  1 S06.4    dia06    NA
 7:  1 S06.4    dia07    NA
 8:  1 S06.4    dia08   6.6
 9:  1 S06.4    dia09   6.7
10:  2 S06.2    dia01   5.0
11:  2 S06.2    dia02    NA
12:  2 S06.2    dia03   4.9
13:  2 S06.2    dia04   7.8
14:  2 S06.2    dia05   9.3
15:  2 S06.2    dia06   8.0
16:  2 S06.2    dia07   7.8
17:  2 S06.2    dia08   8.0
18:  2 S06.2    dia09    NA
19:  3 S06.5    dia01   7.0
20:  3 S06.5    dia02   5.5
21:  3 S06.5    dia03    NA
22:  3 S06.5    dia04    NA
23:  3 S06.5    dia05   7.2
24:  3 S06.5    dia06   8.0
25:  3 S06.5    dia07   7.6
26:  3 S06.5    dia08    NA
27:  3 S06.5    dia09   6.7

另一种选择是:

setDT(df)[, 3:11 := lapply(.SD, function(x) as.numeric(as.character(x))), .SDcols = 3:11][]
df[unique(melt(df, id = 1:2)[, .I[is.na(value) & !is.na(shift(value, type = 'lead'))], by = .(id, dx)], by = 'id')[,'id'], on = 'id']

然而,这种方法的结果仍然是如本答案第一部分所述的宽格式.