有效绑定不同属性的行答案

【问题标题】：Efficiently Binding rows of different attributes有效绑定不同属性的行
【发布时间】：2020-08-07 11:18:17
【问题描述】：

我想绑定行。但是，data.frames 的少数列具有不同的属性。像df1$caseid 和df1$v001 具有与df2$caseid 和df2$v001 不同的属性。想知道如何才能在那里绑定 data.frames。

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.

【问题讨论】：

rbind 对你有用，对吧？问题是什么？
@RonakShah 是的 rbind 适用于这些小型 data.frames，但不适用于大型数据集。同样来自tidyverse 的bind_rows 和来自tidytable 的bind_rows. 即使对于这些小的data.frames 也不起作用。所以寻找一种有效的方法。
rbind 会发生什么？它会给出错误（它是什么？）还是很慢？当班级不同时，bind_rows 将不起作用。它总是会报错。
@RonakShah：使用我的实际数据集rbind 会抛出以下错误消息Error in rbindlist(l, use.names, fill, idcol) : Class attribute on column 2 of item 2 does not match with column 2 of item 1。
试试base::rbind

标签： r tidyverse tidytable

【解决方案1】：

听起来您无论如何都需要修复列类以匹配。如果数字列始终为整数，请将df2 中的类更改为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，但级别不同，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。

【讨论】：