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我想绑定行。但是，data.frames 的少数列具有不同的属性。喜欢df1$caseid和具有与和df1$v001 不同的属性。想知道我怎样才能在那里绑定data.frames。df2$caseiddf2$v001

library(tidyverse)
library(tidytable)
#> 
#> Attaching package: 'tidytable'
#> The following object is masked from 'package:stats':
#> 
#>     dt

df1 <- 
  structure(list(caseid = structure(c("   11 1  1 1  2", "   11 1  1 1  2", 
"   11 1  1 1  2", "   11 1  1 1  2", "   11 1  1 1  2", "   11 1  1 2  2"
), label = "case identification", class = c("labelled", "character"
), format = "%15s"), bidx = structure(c(1L, 2L, 3L, 4L, 5L, 1L
), label = "birth column number", class = c("labelled", "integer"
), format = "%8.0g"), v000 = structure(c("PK2", "PK2", "PK2", 
"PK2", "PK2", "PK2"), label = "country code and phase", class = c("labelled", 
"character"), format = "%3s"), v001 = structure(c(1101001L, 1101001L, 
1101001L, 1101001L, 1101001L, 1101001L), label = "cluster number", class = c("labelled", 
"integer"), format = "%12.0g"), v002 = structure(c(1L, 1L, 1L, 
1L, 1L, 2L), label = "household number", class = c("labelled", 
"integer"), format = "%8.0g")), row.names = c(NA, -6L), class = "data.frame")

df2 <- 
  structure(list(caseid = structure(c(1L, 1L, 1L, 1L, 1L, 2L), .Label = c("       1   1  2", 
"       1   4  1"), class = "factor"), bidx = structure(c(1L, 
2L, 3L, 4L, 5L, 1L), label = c(BIDX = "Birth column number"), class = c("labelled", 
"numeric")), v000 = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "PK7", class = "factor"), 
    v001 = structure(c(1L, 1L, 1L, 1L, 1L, 1L), label = c(V001 = "Cluster number"), class = c("labelled", 
    "numeric")), v002 = structure(c(1L, 1L, 1L, 1L, 1L, 4L), label = c(V002 = "Household number"), class = c("labelled", 
    "numeric"))), row.names = c(NA, -6L), class = "data.frame")

rbind(df1, df2)
#>             caseid bidx v000    v001 v002
#> 1     11 1  1 1  2    1  PK2 1101001    1
#> 2     11 1  1 1  2    2  PK2 1101001    1
#> 3     11 1  1 1  2    3  PK2 1101001    1
#> 4     11 1  1 1  2    4  PK2 1101001    1
#> 5     11 1  1 1  2    5  PK2 1101001    1
#> 6     11 1  1 2  2    1  PK2 1101001    2
#> 7         1   1  2    1  PK7       1    1
#> 8         1   1  2    2  PK7       1    1
#> 9         1   1  2    3  PK7       1    1
#> 10        1   1  2    4  PK7       1    1
#> 11        1   1  2    5  PK7       1    1
#> 12        1   4  1    1  PK7       1    4

bind_rows(df1, df2)
#> Error: Can't combine `..1$caseid` <labelled> and `..2$caseid` <factor<da793>>.

bind_rows.(df1, df2)
#> Error in rbindlist(dots, idcol = .id, use.names = .use_names, fill = .fill): Class attribute on column 2 of item 2 does not match with column 2 of item 1.
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1 回答 1

0

听起来您无论如何都需要修复列类以匹配。如果数字列始终是整数，则将类更改df2为integer。

i <- sapply(df2, is.numeric)
df2[i] <- lapply(df2[i], as.integer)

将因子转换为字符向量。如果两者都是因素，但水平不同，bind_rows仍然会失败。

i <- sapply(df2, is.factor)
df2[i] <- lapply(df2[i], as.character)

如果您需要将这些列作为因子，请在绑定行后重构它们。

于 2020-08-07T12:16:37.623 回答