25

我正在尝试从特定类型的数据创建列联表。这对于循环等是可行的......但是因为我的最终表格将包含超过 10E5 个单元格，所以我正在寻找一个预先存在的函数。

我的初始数据如下：

PLANT                  ANIMAL                          INTERACTIONS
---------------------- ------------------------------- ------------
Tragopogon_pratensis   Propylea_quatuordecimpunctata         1
Anthriscus_sylvestris  Rhagonycha_nigriventris               3
Anthriscus_sylvestris  Sarcophaga_carnaria                   2
Heracleum_sphondylium  Sarcophaga_carnaria                   1
Anthriscus_sylvestris  Sarcophaga_variegata                  4
Anthriscus_sylvestris  Sphaerophoria_interrupta_Gruppe       3
Cerastium_holosteoides Sphaerophoria_interrupta_Gruppe       1

我想创建一个这样的表：

                       Propylea_quatuordecimpunctata Rhagonycha_nigriventris Sarcophaga_carnaria Sarcophaga_variegata Sphaerophoria_interrupta_Gruppe
---------------------- ----------------------------- ----------------------- ------------------- -------------------- -------------------------------
Tragopogon_pratensis   1                             0                       0                   0                    0
Anthriscus_sylvestris  0                             3                       2                   4                    3
Heracleum_sphondylium  0                             0                       1                   0                    0
Cerastium_holosteoides 0                             0                       0                   0                    1

也就是说，行中的所有植物物种，列中的所有动物物种，有时没有相互作用（而我的初始数据只列出了发生的相互作用）。

4

6 回答 6

34

在基础 R 中，使用tableor xtabs：

with(warpbreaks, table(wool, tension))

    tension
wool L M H
   A 9 9 9
   B 9 9 9

xtabs(~wool+tension, data=warpbreaks)

    tension
wool L M H
   A 9 9 9
   B 9 9 9

这些gmodels软件包有一个功能CrossTable，可以提供类似于 SPSS 或 SAS 用户所期望的输出：

library(gmodels)
with(warpbreaks, CrossTable(wool, tension))


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


Total Observations in Table:  54 


             | tension 
        wool |         L |         M |         H | Row Total | 
-------------|-----------|-----------|-----------|-----------|
           A |         9 |         9 |         9 |        27 | 
             |     0.000 |     0.000 |     0.000 |           | 
             |     0.333 |     0.333 |     0.333 |     0.500 | 
             |     0.500 |     0.500 |     0.500 |           | 
             |     0.167 |     0.167 |     0.167 |           | 
-------------|-----------|-----------|-----------|-----------|
           B |         9 |         9 |         9 |        27 | 
             |     0.000 |     0.000 |     0.000 |           | 
             |     0.333 |     0.333 |     0.333 |     0.500 | 
             |     0.500 |     0.500 |     0.500 |           | 
             |     0.167 |     0.167 |     0.167 |           | 
-------------|-----------|-----------|-----------|-----------|
Column Total |        18 |        18 |        18 |        54 | 
             |     0.333 |     0.333 |     0.333 |           | 
-------------|-----------|-----------|-----------|-----------|
于 2011-09-16T09:06:40.170 回答
10

reshape包应该可以解决问题。

> library(reshape)

> df <- data.frame(PLANT = c("Tragopogon_pratensis","Anthriscus_sylvestris","Anthriscus_sylvestris","Heracleum_sphondylium","Anthriscus_sylvestris","Anthriscus_sylvestris","Cerastium_holosteoides"),
                   ANIMAL= c("Propylea_quatuordecimpunctata","Rhagonycha_nigriventris","Sarcophaga_carnaria","Sarcophaga_carnaria","Sarcophaga_variegata","Sphaerophoria_interrupta_Gruppe","Sphaerophoria_interrupta_Gruppe"),
                   INTERACTIONS = c(1,3,2,1,4,3,1),
                   stringsAsFactors=FALSE)

> df <- melt(df,id.vars=c("PLANT","ANIMAL"))    
> df <- cast(df,formula=PLANT~ANIMAL)
> df <- replace(df,is.na(df),0)

> df
                   PLANT Propylea_quatuordecimpunctata Rhagonycha_nigriventris
1  Anthriscus_sylvestris                             0                       3
2 Cerastium_holosteoides                             0                       0
3  Heracleum_sphondylium                             0                       0
4   Tragopogon_pratensis                             1                       0
  Sarcophaga_carnaria Sarcophaga_variegata Sphaerophoria_interrupta_Gruppe
1                   2                    4                               3
2                   0                    0                               1
3                   1                    0                               0
4                   0                    0                               0

我还在想办法解决这个order问题，有什么建议吗？

于 2011-09-16T09:17:23.363 回答
9

我想指出的是，我们可以在不使用该函数的情况下获得 Andrie 发布的相同结果with：

R 基础包

# 3 options
table(warpbreaks[, 2:3])
table(warpbreaks[, c("wool", "tension")])
table(warpbreaks$wool, warpbreaks$tension, dnn = c("wool", "tension"))

    tension
wool L M H
   A 9 9 9
   B 9 9 9

封装 gmodel：

library(gmodels)
# 2 options    
CrossTable(warpbreaks$wool, warpbreaks$tension)
CrossTable(warpbreaks$wool, warpbreaks$tension, dnn = c("Wool", "Tension"))


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


Total Observations in Table:  54 


                | warpbreaks$tension 
warpbreaks$wool |         L |         M |         H | Row Total | 
----------------|-----------|-----------|-----------|-----------|
              A |         9 |         9 |         9 |        27 | 
                |     0.000 |     0.000 |     0.000 |           | 
                |     0.333 |     0.333 |     0.333 |     0.500 | 
                |     0.500 |     0.500 |     0.500 |           | 
                |     0.167 |     0.167 |     0.167 |           | 
----------------|-----------|-----------|-----------|-----------|
              B |         9 |         9 |         9 |        27 | 
                |     0.000 |     0.000 |     0.000 |           | 
                |     0.333 |     0.333 |     0.333 |     0.500 | 
                |     0.500 |     0.500 |     0.500 |           | 
                |     0.167 |     0.167 |     0.167 |           | 
----------------|-----------|-----------|-----------|-----------|
   Column Total |        18 |        18 |        18 |        54 | 
                |     0.333 |     0.333 |     0.333 |           | 
----------------|-----------|-----------|-----------|-----------|
于 2015-05-18T08:17:18.380 回答
7

基本 R 中的 xtabs 应该可以工作，例如：

dat <- data.frame(PLANT = c("p1", "p2", "p2", "p4", "p5", "p5", "p6"),
                  ANIMAL = c("a1", "a2", "a3", "a3", "a4", "a5", "a5"),
                  INTERACTIONS = c(1,3,2,1,4,3,1),
                  stringsAsFactors = FALSE)

(x2.table <- xtabs(dat$INTERACTIONS ~ dat$PLANT + dat$ANIMAL))

     dat$ANIMAL
dat$PLANT a1 a2 a3 a4 a5
       p1  1  0  0  0  0
       p2  0  3  2  0  0
       p4  0  0  1  0  0
       p5  0  0  0  4  3
       p6  0  0  0  0  1

chisq.test(x2.table, simulate.p.value = TRUE)

我认为这应该很容易做到你正在寻找的东西。我不确定它如何在效率方面扩展到 10E5 列联表，但这可能是统计上的一个单独问题。

于 2013-12-26T05:26:42.400 回答
6

与dplyr / tidyr：

df <- read.table(text='PLANT                  ANIMAL                          INTERACTIONS
                 Tragopogon_pratensis   Propylea_quatuordecimpunctata         1
                 Anthriscus_sylvestris  Rhagonycha_nigriventris               3
                 Anthriscus_sylvestris  Sarcophaga_carnaria                   2
                 Heracleum_sphondylium  Sarcophaga_carnaria                   1
                 Anthriscus_sylvestris  Sarcophaga_variegata                  4
                 Anthriscus_sylvestris  Sphaerophoria_interrupta_Gruppe       3
                 Cerastium_holosteoides Sphaerophoria_interrupta_Gruppe       1', header=TRUE)
library(dplyr)
library(tidyr)
df %>% spread(ANIMAL, INTERACTIONS, fill=0)

#                    PLANT Propylea_quatuordecimpunctata Rhagonycha_nigriventris Sarcophaga_carnaria Sarcophaga_variegata Sphaerophoria_interrupta_Gruppe
# 1  Anthriscus_sylvestris                             0                       3                   2                    4                               3
# 2 Cerastium_holosteoides                             0                       0                   0                    0                               1
# 3  Heracleum_sphondylium                             0                       0                   1                    0                               0
# 4   Tragopogon_pratensis                             1                       0                   0                    0                               0
于 2017-02-10T17:21:48.807 回答
3

只需使用dcast()“ reshape2”包的功能：

ans = dcast( df, PLANT~ ANIMAL,value.var = "INTERACTIONS", fill = 0 )

这里“PLANT”将在左列，“ANIMALS”在顶行，表格的填充将使用“INTERACTIONS”进行，“NULL”值将使用 0 填充。

于 2016-09-30T07:14:58.243 回答