Question

这是我的数据的快照：

  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")

我试图使用twinfam（家庭中的双胞胎）和SAMESEX（前两个孩子同性别）二元变量来创建第四个变量，它具有4个可能的值：

1如果SAMESEX == 0＆amp; twinfam == 0
2如果SAMESEX == 1＆amp; twinfam == 0
3如果SAMESEX == 0＆amp; twinfam == 1
4如果SAMESEX == 1＆amp; twinfam == 1

玩了一下之后我尝试使用：

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})

但是我收到了错误：

Error: Unsupported type NILSXP for column "both"

似乎无法解决此错误。关于我为什么会遇到这个错误以及如何解决这个错误的任何建议都将不胜感激！

Answer 1

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

Answer 2

如果您想使用dplyr，可以尝试以下方法。您的数据框在此处称为mydf。您可以使用case_when()并创建both。如果我没有弄错，你还不能使用mutate()中的函数。因此，您想要创建一个向量并最后使用cbind()。

both <- case_when(mydf$SAMESEX == 0 & mydf$twinfam == 0 ~ 1,
                  mydf$SAMESEX == 1 & mydf$twinfam == 0 ~ 2,
                  mydf$SAMESEX == 0 & mydf$twinfam == 1 ~ 3,
                  mydf$SAMESEX == 1 & mydf$twinfam == 1 ~ 4)

cbind(mydf, both)

正如akrun所评论的，现在您可以在case_when()中使用mutate()。

mydf %>%
mutate(both = case_when(.$SAMESEX == 0 & .$twinfam == 0 ~ 1,
                        .$SAMESEX == 1 & .$twinfam == 0 ~ 2,
                        .$SAMESEX == 0 & .$twinfam == 1 ~ 3,
                        .$SAMESEX == 1 & .$twinfam == 1 ~ 4))

#    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

R，mutate和＆＃34;不支持的类型NILSXP用于列＆＃34;

2 个答案: