Question

我有以下数据框;

Fruit <- c("orange", "orange", "apple", "pineapple", "lemon", "apple", "orange") 

Name <- c("julius", "julius", "john", "mary", "kathy", "john", "julius")

df <- data.frame(Fruit, Name);df

我的目标是得到每个人吃掉的所有水果的总和，以便最后得到下表;

         orange apple    pineapple  lemon
julius    2      1      
john             2      
mary                        1   
kathy     1                           1

我正在尝试聚合功能，但只能设法输出每个人吃掉的水果总数，如下所示;

df2 <- aggregate(Fruit~Name,df,length); df2

输出是;

    Name Fruit
1   john     2
2 julius     3
3  kathy     1
4   mary     1

任何帮助将不胜感激。感谢

Answer 1

选项1

library(reshape2)
dcast(df, Name~Fruit)
    Name apple lemon orange pineapple
1   john     2     0      0         0
2 julius     0     0      3         0
3  kathy     0     1      0         0
4   mary     0     0      0         1

选项2

table(df)
# as pointed out by lebatsnok, the general command would be with(df, table(Fruit, Name))
           Name
Fruit       john julius kathy mary
  apple        2      0     0    0
  lemon        0      0     1    0
  orange       0      3     0    0
  pineapple    0      0     0    1

Answer 2

看起来你想要一个简单的双向频率表：

table(Fruit, Name)
#           Name
#Fruit       john julius kathy mary
#  apple        2      0     0    0
#  lemon        0      0     1    0
#  orange       0      3     0    0
#  pineapple    0      0     0    1

Answer 3

> library(gmodels)
> 
> CrossTable(Fruit, Name)


   Cell Contents
|-------------------------|
|                       N |
| Chi-square contribution |
|           N / Row Total |
|           N / Col Total |
|         N / Table Total |
|-------------------------|


Total Observations in Table:  7 


             | Name 
       Fruit |      john |    julius |     kathy |      mary | Row Total | 
-------------|-----------|-----------|-----------|-----------|-----------|
       apple |         2 |         0 |         0 |         0 |         2 | 
             |     3.571 |     0.857 |     0.286 |     0.286 |           | 
             |     1.000 |     0.000 |     0.000 |     0.000 |     0.286 | 
             |     1.000 |     0.000 |     0.000 |     0.000 |           | 
             |     0.286 |     0.000 |     0.000 |     0.000 |           | 
-------------|-----------|-----------|-----------|-----------|-----------|
       lemon |         0 |         0 |         1 |         0 |         1 | 
             |     0.286 |     0.429 |     5.143 |     0.143 |           | 
             |     0.000 |     0.000 |     1.000 |     0.000 |     0.143 | 
             |     0.000 |     0.000 |     1.000 |     0.000 |           | 
             |     0.000 |     0.000 |     0.143 |     0.000 |           | 
-------------|-----------|-----------|-----------|-----------|-----------|
      orange |         0 |         3 |         0 |         0 |         3 | 
             |     0.857 |     2.286 |     0.429 |     0.429 |           | 
             |     0.000 |     1.000 |     0.000 |     0.000 |     0.429 | 
             |     0.000 |     1.000 |     0.000 |     0.000 |           | 
             |     0.000 |     0.429 |     0.000 |     0.000 |           | 
-------------|-----------|-----------|-----------|-----------|-----------|
   pineapple |         0 |         0 |         0 |         1 |         1 | 
             |     0.286 |     0.429 |     0.143 |     5.143 |           | 
             |     0.000 |     0.000 |     0.000 |     1.000 |     0.143 | 
             |     0.000 |     0.000 |     0.000 |     1.000 |           | 
             |     0.000 |     0.000 |     0.000 |     0.143 |           | 
-------------|-----------|-----------|-----------|-----------|-----------|
Column Total |         2 |         3 |         1 |         1 |         7 | 
             |     0.286 |     0.429 |     0.143 |     0.143 |           | 
-------------|-----------|-----------|-----------|-----------|-----------|

基于具有因子的列值的行总和

3 个答案: