Question

我的数据集很大，需要比较列之间的所有组合。所需的输出将是每个列组合的矩阵。

开始的数据框看起来像Data：

set.seed(1)
Data <- data.frame(
  ID = (1:100),
  A = sample(1:10,10),
  B = sample(1:20,100,replace = T),
  C = sample(1:5,100,replace = T),
  D = sample(1:20,100,replace = T)
  )     
Data

，我想知道两列中相同组合出现的频率。（例如，同一行1中的A和4中的B多久出现一次）对于A到D列之间的所有组合？

我正在使用：

require(dplyr)

X1 <- ddply(Data,.(A,B),transmute, count=length(ID))

并得到一个像这样的对象：

     A  B count
1    1  3     1
2    1  7     1
3    1  9     2
4    1  9     2
5    1 12     1
6    1 13     1
7    1 14     1
8    1 16     1
9    1 18     1
10   1 20     1
11   2  2     1
12   2  6     1
13   2 10     1
14   2 11     1

但是如何获得矩阵格式的count结果呢？

A对B冷的输出如下：

    B1  B2  B3  B4  B5  B6
A1  1   1   2   1   1   ...
A2  1   1   2   1   1   
A3  2   1   1   1   1   
A4  2   1   1   1   1   
A5  1   1   2   1   2   
A6  1   1   2   1   2   
A7  1   3   1   1   1   
A8  1   3   1   1   2   
A9  1   3   2   1   2   
A10 1   1   2   1   1


In the best case the result would be a `list`  containing the objects  `AB`  `AC` ...`CD` as matrix.

Answer 1

您可以这样做：

library(tidyverse)
X2 <-X1 %>% group_by(A,B) %>% 
  summarise(count=max(count)) %>% #use max instead of sum
  ungroup() %>%
  mutate(A=paste0("A",A),B=paste0("B",B)) %>% 
  spread(B,count,fill=0)

X3 <- as.matrix(X2[,2:ncol(X2)])
rownames(X3) <- as.character(X2$A)

Answer 2

使用aggregate和reshape来考虑基数R：

agg <- aggregate(cbind(count=ID) ~ B + A, Data, FUN=length)

rdf <- reshape(agg, timevar = "B", idvar = "A",
               drop = c("ID", "C", "D"),
               direction = "wide")

# CLEAN-UP
rdf <- with(rdf, rdf[order(A), c("A", paste0("count.", 1:20))])  # RE-ORDER ROWS AND COLS
rownames(rdf) <- NULL                                            # RESET ROW NAMES
colnames(rdf) <- gsub("count.", "B", names(rdf))                 # RENAME COL NAMES
rdf[is.na(rdf)] <- 0                                             # CONVERT NAs TO O

rdf

#     A B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 B11 B12 B13 B14 B15 B16 B17 B18 B19 B20
# 1   1  1  1  0  1  0  1  0  0  0   1   0   1   0   1   0   0   0   0   2   1
# 2   2  0  0  0  0  0  0  2  0  1   3   0   0   2   0   1   0   0   0   1   0
# 3   3  1  1  0  0  0  1  0  0  1   0   0   0   0   2   1   1   1   0   0   1
# 4   4  1  0  0  0  1  2  0  0  0   2   1   1   0   1   0   1   0   0   0   0
# 5   5  0  0  0  0  1  2  1  3  0   0   0   1   0   0   0   0   0   0   1   1
# 6   6  1  0  2  0  1  2  1  0  0   0   0   0   1   0   0   0   0   0   0   2
# 7   7  1  0  0  0  1  1  1  0  0   2   1   0   0   0   0   2   1   0   0   0
# 8   8  0  0  2  0  0  0  1  2  1   0   2   0   0   0   0   0   0   1   1   0
# 9   9  1  0  0  0  0  1  1  0  0   0   0   1   1   2   1   0   0   1   1   0
# 10 10  1  1  0  0  0  0  0  0  1   1   0   2   0   1   1   0   1   0   1   0

数据框中的R个汇总/计数组合，并显示为列的新数据框/计数矩阵

2 个答案: