Question

我有一个带杂货的大矩阵。有些值相同但名称不同。例如：

Ketchup  Ketchupwithgarlic  Ketchupspicy Chips Chipsorganic
0               1               0         0      1
1               0               0         0      0
0               0               0         1      0
1               0               0         0      0

我想要做的是将这两个向量组合成一个向量，如果一个名称以完全相同的名称开头，那么输出如下所示：

Ketchup Chips
1        1
1        0
0        1
1        0

我该怎么办？

Answer 1

我相信这可以做你想要的。至少使用您提供的数据集。并且它不依赖于硬编码列名称。

使用@MKR中的代码读取数据：

nms <- names(df)
inx <- which(sapply(seq_along(nms), function(i) any(grepl(paste0("^", nms[i]), nms[-i]))))
result <- sapply(inx, function(i) rowSums(df[, grep(nms[i], nms)]))
colnames(result) <- nms[inx]
result
#     Ketchup Chips
#[1,]       1     1
#[2,]       1     0
#[3,]       0     1
#[4,]       1     0

Answer 2

将矩阵转换为dplyr::coalesce后，可以使用data.frame选项。此外，值为0的单元格应更改为NA以应用coalesce。

library(dplyr)
# First change matrix to data.frame. The same data is created in data.frame 
# so this step can be skipped
df <- as.data.frame(df)

# Replace 0 with NA
df[df==0] <- NA

选项＃1：如果列名较少且已知，则一旦接近

bind_cols(Chips = coalesce(!!!select(df, starts_with("Chips"))), 
          Ketchup = coalesce(!!!select(df, starts_with("Ketchup"))) )
# # A tibble: 4 x 2
#   Chips Ketchup
#   <int>   <int>
# 1     1       1
# 2    NA       1
# 3     1      NA
# 4    NA       1

选项＃2：通用方法可以写成：

overlapName <- names(df)[mapply(function(x)sum(str_detect(names(df),x)), names(df)) >1]
library(stringr)

mapply(function(x)coalesce(!!!select(df, starts_with(x))), overlapName)
#      Ketchup Chips
# [1,]       1     1
# [2,]       1    NA
# [3,]      NA     1
# [4,]       1    NA

数据：

df <- read.table(text = "Ketchup Ketchupwithgarlic Ketchupspicy Chips Chipsorganic 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0", header = TRUE, stringsAsFactors = FALSE)

Answer 3

这是另一个基础R替代品。我认为Rui Barrades的答案可能更好，但看到多种方法会有所帮助。

# save column names
cnms <- colnames(myMat)
# build a matrix that groups on column names using col and grepl
grps <- col(diag(length(cnms))) * sapply(cnms[order(cnms)], grepl, x=cnms)
# run through the groups and perform rowSums to collapse groups into one column
sapply(split(seq_len(ncol(myMat)), 
             colnames(grps)[apply(grps, 1, FUN=function(x) min(x[x != 0]))]),
       function(y) rowSums(myMat[, y]))

返回

     Chips Ketchup
[1,]     1       1
[2,]     0       1
[3,]     1       0
[4,]     0       1

数据

myMat <- structure(c(0L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L), .Dim = 4:5, .Dimnames = list(NULL, c("Ketchup", "Ketchupwithgarlic", "Ketchupspicy", "Chips", "Chipsorganic")))

如果它们具有相同的开头，如何替换矩阵中的值？

3 个答案: