这是我的数据快照:
structure(list(CPUBID = c(1000001L, 1000002L, 1000003L, 10001L,
1000201L, 1000203L, 10003L, 1000801L, 1000802L, 1000803L, 1001L,
1001101L, 1001102L, 1001601L, 1002401L, 1002402L, 1002403L, 1002601L,
1002602L, 1002604L), MPUBID = c(10000L, 10000L, 10000L, 100L,
10002L, 10002L, 100L, 10008L, 10008L, 10008L, 10L, 10011L, 10011L,
10016L, 10024L, 10024L, 10024L, 10026L, 10026L, 10026L), CYRB = c(1982L,
1984L, 1988L, 1985L, 1986L, 1992L, 1993L, 1984L, 1986L, 1988L,
1983L, 1987L, 1992L, 1977L, 1981L, 1984L, 1998L, 1980L, 1981L,
1984L), twinfam = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), SAMESEX = c(1L, 1L, 1L,
1L, 0L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L,
0L), top25 = c(0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0,
0, 0, 0, 0), top5 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0), quantity = c(1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1)), .Names = c("CPUBID", "MPUBID",
"CYRB", "twinfam", "SAMESEX", "top25", "top5", "quantity"), row.names = c(NA,
20L), class = "data.frame")
我正在尝试使用twinfam(家庭中的双胞胎)和SAMESEX(前两个同性出生的 child )二进制变量来创建第四个变量，该变量采用4个可能的值:
玩了一段时间后，我尝试使用:
df <- df %>% mutate(both = for (i in 1:nrow(PIATmathreg6)) {
if(twinfam[i] == 0 & SAMESEX[i] == 0) both = 1
else if(twinfam[i] == 0 & SAMESEX[i] == 1) both = 2
else if(twinfam[i] == 1 & SAMESEX[i] == 0) both = 3
else both = 4})
但是我得到了错误:
Error: Unsupported type NILSXP for column "both"
并且似乎无法解决此错误。对于为什么会出现此错误以及如何解决该错误的任何建议，将不胜感激!
最佳答案
最好创建一个键/值数据集并执行left_join
library(dplyr)
df2 <- data.frame(SAMESEX = c(0, 1, 0, 1), twinfam = c(0, 0, 1, 1), both = 1:4)
left_join(df, df2, by = c("SAMESEX", "twinfam"))
# CPUBID MPUBID CYRB twinfam SAMESEX top25 top5 quantity both
#1 1000001 10000 1982 0 1 0 0 1 2
#2 1000002 10000 1984 0 1 0 0 1 2
#3 1000003 10000 1988 0 1 0 0 1 2
#4 10001 100 1985 0 1 0 0 1 2
#5 1000201 10002 1986 0 0 0 0 1 1
#6 1000203 10002 1992 0 0 1 0 1 1
#7 10003 100 1993 0 1 0 0 1 2
#8 1000801 10008 1984 0 0 0 0 1 1
#9 1000802 10008 1986 0 0 0 0 1 1
#10 1000803 10008 1988 0 0 0 0 1 1
#11 1001 10 1983 0 1 1 0 0 2
#12 1001101 10011 1987 0 0 0 0 0 1
#13 1001102 10011 1992 0 0 0 0 0 1
#14 1001601 10016 1977 0 1 0 0 1 2
#15 1002401 10024 1981 0 0 1 0 1 1
#16 1002402 10024 1984 0 0 0 0 1 1
#17 1002403 10024 1998 0 0 0 0 1 1
#18 1002601 10026 1980 0 0 0 0 1 1
#19 1002602 10026 1981 0 0 0 0 1 1
#20 1002604 10026 1984 0 0 0 0 1 1
关于R，mutate和 "Unsupported type NILSXP for column"，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/39183359/