I have 2 columns of data with the same type of data (Strings).
我有2列数据具有相同类型的数据（字符串）。
I want to join the levels of the columns. ie. we have:
我想加入列的级别。即。我们有：
col1 col2
Bob John
Tom Bob
Frank Jane
Jim Bob
Tom Bob
... ... (and so on)
now col1 has 4 levels (Bob, Tom Frank, Jim) and col2 has 3 levels (John, Jane, Bob)
现在col1有4级（Bob，Tom Frank，Jim），col2有3级（John，Jane，Bob）
But I want both columns to have all the factor levels (Bob, Tom, Frank, Jim, Jane, John), as to later replace each of the 'names' with a unique id, such that the final output would be:
但是我希望两个列都具有所有因子级别（Bob，Tom，Frank，Jim，Jane，John），以便稍后用唯一的id替换每个'names'，这样最终的输出将是：
col1 col2
1 5
2 1
3 6
4 1
2 1
that is Bob -> 1, Tom -> 2, etc. in both columns.
这两个列中的Bob - > 1，Tom - > 2等。
Any ideas :) ?
有任何想法吗 ：） ？
edit: Thanks all for the wonderful answers! You are all awesome as far as I know :)
编辑：谢谢大家的精彩答案！据我所知，你们都很棒:)
3 个解决方案
#1
6
You want the factors to include all the unique names from both columns.
您希望因子包含两列中的所有唯一名称。
col1 <- factor(c("Bob", "Tom", "Frank", "Jim", "Tom"))
col2 <- factor(c("John", "Bob", "Jane", "Bob", "Bob"))
mynames <- unique(c(levels(col1), levels(col2)))
fcol1 <- factor(col1, levels = mynames)
fcol2 <- factor(col2, levels = mynames)
EDIT: a little nicer if you replace the third line with this:
编辑：如果你用这个替换第三行更好一点：
mynames <- union(levels(col1), levels(col2))
#2
11
x <- structure(list(col1 = structure(c(1L, 4L, 2L, 3L, 4L), .Label = c("Bob", "Frank", "Jim", "Tom"), class = "factor"), col2 = structure(c(3L, 1L, 2L, 1L, 1L), .Label = c("Bob", "Jane", "John"), class = "factor")), .Names = c("col1", "col2"), class = "data.frame", row.names = c(NA, -5L))
Make a simple union of factor names:
建立因子名称的简单联合：
both <- union(levels(x$col1), levels(x$col2))
And relevel the two factors:
并重新考虑这两个因素：
x$col1 <- factor(x$col1, levels=both)
x$col2 <- factor(x$col2, levels=both)
After editing: added example to make numeric values from factors
编辑后：添加了从因子中生成数值的示例
You could simply transform the factor levels to numeric values, e.g.:
您可以简单地将因子级别转换为数值，例如：
as.numeric(x$col1)
Or a more simpler, nicer solution based on @Gavin Simpson's hint below in one step:
或者更简单，更好的解决方案基于@Gavin Simpson的一步提示：
data.matrix(x)
#3
2
Could have sworn this didn't work when I was writing the abomination below, but it does now:
当我在下面写下令人憎恶的事情时，我可以发誓这不起作用，但它现在做了：
## self contained example:
txt <- "col1 col2
Bob John
Tom Bob
Frank Jane
Jim Bob
Tom Bob"
dat <- read.table(textConnection(txt), header = TRUE)
Just compute unique set of levels and coerce each colX to a factor:
只计算一组唯一的级别并将每个colX强制转换为一个因子：
> dat3 <- dat
> lev <- as.character(unique(unlist(sapply(dat, levels))))
> dat3 <- within(dat3, col1 <- factor(col1, levels = lev))
> dat3 <- within(dat3, col2 <- factor(col2, levels = lev))
> str(dat3)
'data.frame': 5 obs. of 2 variables:
$ col1: Factor w/ 6 levels "Bob","Tom","Frank",..: 1 2 3 4 2
$ col2: Factor w/ 6 levels "Bob","Tom","Frank",..: 5 1 6 1 1
> data.matrix(dat3)
col1 col2
[1,] 1 5
[2,] 2 1
[3,] 3 6
[4,] 4 1
[5,] 2 1
[Original: to show how stupidly complex and obfuscated one can write R code it one tries really hard!] Not sure this is particularly elegant (and it isn't), but...
[原文：为了表明一个人可以编写R代码是如此愚蠢复杂和混淆，人们会非常努力！]不确定这是否特别优雅（并且它不是），但......
We first unlist the data:
我们首先取消列出数据：
tmp <- unlist(dat)
then compute the unique levels
然后计算独特的水平
lev <- as.character(unique(tmp))
and then restructure tmp (from above) back into the same dimensions as the original data, convert to data.frame (preserving the strings), lapply over this data frame, creating a factor with levels lev computed above, and finally coerce to a data frame.
然后将tmp（从上面）重新调整回与原始数据相同的维度，转换为data.frame（保留字符串），在此数据帧上进行填充，创建具有上面计算的级别lev的因子，最后强制转换为数据帧。
dat2 <- data.frame(lapply(data.frame(matrix(tmp, ncol = ncol(dat)),
stringsAsFactors = FALSE),
FUN = factor, levels = lev))
Which gives:
这使：
> dat2
X1 X2
1 Bob John
2 Tom Bob
3 Frank Jane
4 Jim Bob
5 Tom Bob
> sapply(dat2, levels)
X1 X2
[1,] "Bob" "Bob"
[2,] "Tom" "Tom"
[3,] "Frank" "Frank"
[4,] "Jim" "Jim"
[5,] "John" "John"
[6,] "Jane" "Jane"
> data.matrix(dat2)
X1 X2
[1,] 1 5
[2,] 2 1
[3,] 3 6
[4,] 4 1
[5,] 2 1
#1
6
You want the factors to include all the unique names from both columns.
您希望因子包含两列中的所有唯一名称。
col1 <- factor(c("Bob", "Tom", "Frank", "Jim", "Tom"))
col2 <- factor(c("John", "Bob", "Jane", "Bob", "Bob"))
mynames <- unique(c(levels(col1), levels(col2)))
fcol1 <- factor(col1, levels = mynames)
fcol2 <- factor(col2, levels = mynames)
EDIT: a little nicer if you replace the third line with this:
编辑：如果你用这个替换第三行更好一点：
mynames <- union(levels(col1), levels(col2))
#2
11
x <- structure(list(col1 = structure(c(1L, 4L, 2L, 3L, 4L), .Label = c("Bob", "Frank", "Jim", "Tom"), class = "factor"), col2 = structure(c(3L, 1L, 2L, 1L, 1L), .Label = c("Bob", "Jane", "John"), class = "factor")), .Names = c("col1", "col2"), class = "data.frame", row.names = c(NA, -5L))
Make a simple union of factor names:
建立因子名称的简单联合：
both <- union(levels(x$col1), levels(x$col2))
And relevel the two factors:
并重新考虑这两个因素：
x$col1 <- factor(x$col1, levels=both)
x$col2 <- factor(x$col2, levels=both)
After editing: added example to make numeric values from factors
编辑后：添加了从因子中生成数值的示例
You could simply transform the factor levels to numeric values, e.g.:
您可以简单地将因子级别转换为数值，例如：
as.numeric(x$col1)
Or a more simpler, nicer solution based on @Gavin Simpson's hint below in one step:
或者更简单，更好的解决方案基于@Gavin Simpson的一步提示：
data.matrix(x)
#3
2
Could have sworn this didn't work when I was writing the abomination below, but it does now:
当我在下面写下令人憎恶的事情时，我可以发誓这不起作用，但它现在做了：
## self contained example:
txt <- "col1 col2
Bob John
Tom Bob
Frank Jane
Jim Bob
Tom Bob"
dat <- read.table(textConnection(txt), header = TRUE)
Just compute unique set of levels and coerce each colX to a factor:
只计算一组唯一的级别并将每个colX强制转换为一个因子：
> dat3 <- dat
> lev <- as.character(unique(unlist(sapply(dat, levels))))
> dat3 <- within(dat3, col1 <- factor(col1, levels = lev))
> dat3 <- within(dat3, col2 <- factor(col2, levels = lev))
> str(dat3)
'data.frame': 5 obs. of 2 variables:
$ col1: Factor w/ 6 levels "Bob","Tom","Frank",..: 1 2 3 4 2
$ col2: Factor w/ 6 levels "Bob","Tom","Frank",..: 5 1 6 1 1
> data.matrix(dat3)
col1 col2
[1,] 1 5
[2,] 2 1
[3,] 3 6
[4,] 4 1
[5,] 2 1
[Original: to show how stupidly complex and obfuscated one can write R code it one tries really hard!] Not sure this is particularly elegant (and it isn't), but...
[原文：为了表明一个人可以编写R代码是如此愚蠢复杂和混淆，人们会非常努力！]不确定这是否特别优雅（并且它不是），但......
We first unlist the data:
我们首先取消列出数据：
tmp <- unlist(dat)
then compute the unique levels
然后计算独特的水平
lev <- as.character(unique(tmp))
and then restructure tmp (from above) back into the same dimensions as the original data, convert to data.frame (preserving the strings), lapply over this data frame, creating a factor with levels lev computed above, and finally coerce to a data frame.
然后将tmp（从上面）重新调整回与原始数据相同的维度，转换为data.frame（保留字符串），在此数据帧上进行填充，创建具有上面计算的级别lev的因子，最后强制转换为数据帧。
dat2 <- data.frame(lapply(data.frame(matrix(tmp, ncol = ncol(dat)),
stringsAsFactors = FALSE),
FUN = factor, levels = lev))
Which gives:
这使：
> dat2
X1 X2
1 Bob John
2 Tom Bob
3 Frank Jane
4 Jim Bob
5 Tom Bob
> sapply(dat2, levels)
X1 X2
[1,] "Bob" "Bob"
[2,] "Tom" "Tom"
[3,] "Frank" "Frank"
[4,] "Jim" "Jim"
[5,] "John" "John"
[6,] "Jane" "Jane"
> data.matrix(dat2)
X1 X2
[1,] 1 5
[2,] 2 1
[3,] 3 6
[4,] 4 1
[5,] 2 1