我们将'Class'创建为factor并使用xtabs

df1$Class <- factor(df1$Class, levels = 1:7)

 xtabs(Val ~ SEZ + Class, df1)

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• 输出

Class
SEZ     1 2 3 4 5 6 7
  1_1_1 2 0 0 0 2 0 0
  1_1_2 0 0 0 0 2 0 0
  1_1_3 1 0 0 0 2 0 0
  1_1_4 1 0 0 0 0 0 0
  1_1_5 0 1 0 0 0 0 0
  1_2_1 2 0 0 0 2 0 0

型
如果我们需要data.frame输出，

out <- as.data.frame.matrix( xtabs(Val ~ SEZ + Class, df1))
out$SEZ <- row.names(out)
row.names(out) <- NULL

型

数据

df1 <- structure(list(SEZ = c("1_1_1", "1_1_1", "1_1_2", "1_1_3", "1_1_3", 
"1_1_4", "1_1_5", "1_2_1", "1_2_1"), Class = c(1L, 5L, 5L, 1L, 
5L, 1L, 2L, 1L, 5L), Val = c(2L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 
2L)), row.names = c(NA, -9L), class = "data.frame")

型

另一个使用reshape + merge的基本R选项

reshape(
  merge(df,
    expand.grid(
      SEZ = unique(df$SEZ),
      Class = 1:7
    ),
    all = TRUE
  ),
  direction = "wide",
  idvar = "SEZ",
  timevar = "Class"
)

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给

SEZ Val.1 Val.2 Val.3 Val.4 Val.5 Val.6 Val.7
1  1_1_1     2    NA    NA    NA     2    NA    NA
8  1_1_2    NA    NA    NA    NA     2    NA    NA
15 1_1_3     1    NA    NA    NA     2    NA    NA
22 1_1_4     1    NA    NA    NA    NA    NA    NA
29 1_1_5    NA     1    NA    NA    NA    NA    NA
36 1_2_1     2    NA    NA    NA     2    NA    NA

型

当“Values”列包含整数时会发生什么？
我尝试将@akran解决方案（上面发布的）应用到这个表（碰巧在“Values”列中有字符串而不是整数）。

数据

"1_1_4", "1_1_5", "1_2_1", "1_2_1"), Class = c(1L, 5L, 5L, 1L, 
5L, 1L, 2L, 1L, 5L), Val = c(2L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 
2L)), row.names = c(NA, -9L), class = "data.frame")

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运行此命令将导致错误：

xtabs(Val ~ SEZ + Class, df1)

型
x1c 0d1x的数据
我们需要将输出列“瓦尔”转换为数值，然后将其返回为文本。此脚本完成了这项工作（R版本4.3.1（2023-06-16）&djr 1.1.2）：

# create list to translate
list_mapTclass <- seq(1:length(unique(df1$Val)))
print(length(list_mapTclass))
names(list_mapTclass) <-unique(df1$Val)

list_mapTclass[1:3]

df1$idxVal <- sapply(df1$Val, function(x) list_mapTclass[[x]] )

#essential! conver to factor
#df1$idxVal <- factor(df1$idxVal)

#df1

#xtabs(idxVal ~ SEZ + Class, df1)
wide_df_matrix <-xtabs(idxVal ~ SEZ + Class, df1)
 


#print(list_mapTclassInverted[1:3])
#--- add subid
wide_df0 <- as.data.frame.matrix( wide_df_matrix) %>% 
  setNames(paste0('col_', names(.))) #%>%

wide_df0
# reconvert values into the therapyclass content

# inverted list:

list_mapTclassInverted <- setNames(names(list_mapTclass), list_mapTclass)
# sometimes if you dont have 0s , you have to include it in the "list_mapTclassInverted" , as "Unknown"
#list_mapTclassInverted[['0']] <- 'Unknown'
print(list_mapTclassInverted)
wide_df0


# Next piece of code is to go through the full table and convert all the integers into their text
### VERY USEFUL! (Thanks to: @Martin Morgan - https://stackoverflow.com/questions/7547597/dictionary-style-replace-multiple-items )

wide_df00 <- wide_df0
wide_df00[] <- list_mapTclassInverted[unlist(wide_df00)]

wide_df00
## convert to a dataframe

wide_df0$SEZ <- row.names(wide_df0) 
row.names(wide_df0) <- NULL

# See the output
wide_df0

型
测试结果：

我将写另一篇文章（here）与此问题的情况下，这一个不完全匹配的问题.希望这有帮助！有时是耗时的事情在R基地，但正如问题中提到的，当你不得不使用没有更新的技术，你不是IT管理员，没有其他方法（特别是如果你必须在明天交付！）.