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在 r 编程中检索矩阵中最大值的行名

[英]retrieving the row name of max value in a matrix in r programming

 原文  2020-07-22 14:02:14  3   2    r/ matrix

问题描述

I need to display name of the car which has max acceleration我需要显示具有最大加速度的汽车的名称

# READING THE AUTO FILE
myfile=read.csv("Auto.csv")

#creating the matrix
mpg =c(myfile$mpg)
cylinders=c(myfile$cylinders)
displacement=c(myfile$displacement)
horsepower=c(myfile$horsepower)
weight=c(myfile$weight)
acceleration=c(myfile$acceleration)
year=c(myfile$year)
origin=c(myfile$origin)
name=c(myfile$name)

matrixAuto=matrix(c(mpg,cylinders,displacement,horsepower,weight,acceleration,year,origin),20,8)

rownames(matrixAuto)=c("chevrolet chevelle malibu","buick skylark 320","plymouth satellite","amc rebel sst",
                       "ford torino","ford galaxie 500","chevrolet impala","plymouth fury iii","pontiac catalina",
                       "amc ambassador dpl","dodge challenger se","plymouth 'cuda 340","chevrolet monte carlo",
                       "buick estate wagon (sw)","toyota corona mark ii","plymouth duster","amc hornet",
                       "ford maverick","datsun pl510","volkswagen 1131 deluxe sedan")
matrixAuto
colnames(matrixAuto)=c("mpg","cylinders","displacement","horsepower","weight","accelaration","year","origin")

#calculating max cylinders
maxCylinders=max(cylinders)==cylinders
rownames(matrixAuto[maxCylinders,])

But then I tried the same thing for calculating maximum acceleration and only a single value was returned using max function and when I tried using但是后来我尝试了同样的方法来计算最大加速度，并且使用 max function 并且当我尝试使用时只返回一个值

maxAccelaration=max(acceleration)==acceleration
rownames(matrixAuto[maxAccelaration,])

NULL value was returned. NULL值已返回。 Can you explain me why do I get correct answers for mutliple max values whereas null value for a single max value using my code?你能解释一下为什么我会得到正确的答案多个最大值而 null 值使用我的代码单个最大值？ Also How do i get the row name of the max acceleration?另外我如何获得最大加速度的行名？ I am trying to get row name for max acceleration car which is the volkswagen 1131 deluxe sedan我正在尝试获取最大加速汽车的行名，即大众 1131 豪华轿车

Any help is appreciated.任何帮助表示赞赏。

my file using the dput() function is我使用 dput() function 的文件是

structure(list(mpg = c(18L, 15L, 18L, 16L, 17L, 15L, 14L, 14L, 
14L, 15L, 15L, 14L, 15L, 14L, 24L, 22L, 18L, 21L, 27L, 26L), 
    cylinders = c(8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 
    8L, 8L, 8L, 4L, 6L, 6L, 6L, 4L, 4L), displacement = c(307L, 
    350L, 318L, 304L, 302L, 429L, 454L, 440L, 455L, 390L, 383L, 
    340L, 400L, 455L, 113L, 198L, 199L, 200L, 97L, 97L), horsepower = c(130L, 
    165L, 150L, 150L, 140L, 198L, 220L, 215L, 225L, 190L, 170L, 
    160L, 150L, 225L, 95L, 95L, 97L, 85L, 88L, 46L), weight = c(3504L, 
    3693L, 3436L, 3433L, 3449L, 4341L, 4354L, 4312L, 4425L, 3850L, 
    3563L, 3609L, 3761L, 3086L, 2372L, 2833L, 2774L, 2587L, 2130L, 
    1835L), acceleration = c(12, 11.5, 11, 12, 10.5, 10, 9, 70.5, 
    10, 8.5, 10, 8, 9.5, 10, 15, 15.5, 15.5, 16, 14.5, 70.5), 
    year = c(70L, 70L, 70L, 70L, 70L, 70L, 70L, 70L, 70L, 70L, 
    70L, 70L, 70L, 70L, 70L, 70L, 70L, 70L, 70L, 70L), origin = c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 
    1L, 1L, 3L, 2L), name = c("chevrolet chevelle malibu", "buick skylark 320", 
    "plymouth satellite", "amc rebel sst", "ford torino", "ford galaxie 500", 
    "chevrolet impala", "plymouth fury iii", "pontiac catalina", 
    "amc ambassador dpl", "dodge challenger se", "plymouth 'cuda 340", 
    "chevrolet monte carlo", "buick estate wagon (sw)", "toyota corona mark ii", 
    "plymouth duster", "amc hornet", "ford maverick", "datsun pl510", 
    "volkswagen 1131 deluxe sedan")), row.names = c(NA, 20L), class = "data.frame")

2 个解决方案

解决方案1
 0  2020-07-22 15:16:49

When subsetting a matrix (or dataframe) with only 1 row, it will return a named vector (no rownames).当对只有 1 行的矩阵（或数据框）进行子集化时，它将返回一个命名向量（无行名）。 When you subset more than 1 row, you will get a matrix (or dataframe) back.当您对超过 1 行进行子集化时，您将获得一个矩阵（或数据框）。

Note, that is you use tibble::tibble and subset one row, it will return a tibble .请注意，即您使用tibble::tibble和子集一行，它将返回一个tibble 。

head(mtcars)

matrixMTCARS <- as.matrix(mtcars)

test <- matrixMTCARS[1, ] 
#subsetting a single row will return just a names vector, no row names in a vector

test2 <- matrixMTCARS[1:2, ]
#subsetting multiple rows will return a matrix, will bring the row names with it

str(matrixMTCARS[1, ])

dim(matrixMTCARS[1, ])
#null since it is a vector, there is no dimensions

str(matrixMTCARS[1:2,])

dim(matrixMTCARS[1:2, ])

解决方案2
 0  2020-07-22 16:10:05

Consider calculating max values of all numeric columns then filter accordingly.考虑计算所有数字列的最大值，然后进行相应的过滤。 See even easier way to cast data frame to matrix:查看将数据框转换为矩阵的更简单方法：

Matrix Build矩阵构建

num_cols <- c("mpg", "cylinders", "displacement", "horsepower",
              "weight", "acceleration", "year", "origin")

matrixAuto <- as.matrix(myfile[,num_cols]), 
dimnames(matrixAuto) <- list(myfile$name, num_cols))

# ALTERNATIVELY:
# matrixAuto <- `dimnames<-`(as.matrix(myfile[,num_cols]), list(myfile$name, num_cols))

Matrix Aggregation矩阵聚合

max_vals_matrix <- apply(matrixAuto, 2, max)
max_vals_matrix
#     mpg    cylinders displacement   horsepower    weight acceleration   year   origin 
#    27.0          8.0        455.0        225.0    4425.0         70.5   70.0      3.0

Max Cylinders最大气缸数

mask <- matrixAuto[,"cylinders"] == max_vals_matrix["cylinders"]
matrixAuto[mask,]
#                           mpg cylinders displacement horsepower weight acceleration year origin
# chevrolet chevelle malibu  18         8          307        130   3504         12.0   70      1
# buick skylark 320          15         8          350        165   3693         11.5   70      1
# plymouth satellite         18         8          318        150   3436         11.0   70      1
# amc rebel sst              16         8          304        150   3433         12.0   70      1
# ford torino                17         8          302        140   3449         10.5   70      1
# ford galaxie 500           15         8          429        198   4341         10.0   70      1
# chevrolet impala           14         8          454        220   4354          9.0   70      1
# plymouth fury iii          14         8          440        215   4312         70.5   70      1
# pontiac catalina           14         8          455        225   4425         10.0   70      1
# amc ambassador dpl         15         8          390        190   3850          8.5   70      1
# dodge challenger se        15         8          383        170   3563         10.0   70      1
# plymouth 'cuda 340         14         8          340        160   3609          8.0   70      1
# chevrolet monte carlo      15         8          400        150   3761          9.5   70      1
# buick estate wagon (sw)    14         8          455        225   3086         10.0   70      1

Max Acceleration最大加速度

mask <- matrixAuto[,"acceleration"] == max_vals_matrix["acceleration"]
matrixAuto[mask,]
#                              mpg cylinders displacement horsepower weight acceleration year origin
# plymouth fury iii             14         8          440        215   4312         70.5   70      1
# volkswagen 1131 deluxe sedan  26         4           97         46   1835         70.5   70      2

Max Horsepower最大马力

mask <- matrixAuto[,"horsepower"] == max_vals_matrix["horsepower"]
matrixAuto[mask,]
#                         mpg cylinders displacement horsepower weight acceleration year origin
# pontiac catalina         14         8          455        225   4425           10   70      1
# buick estate wagon (sw)  14         8          455        225   3086           10   70      1

Multiple aggregation多重聚合

agg_vals_matrix <- apply(matrixAuto, 2, function(col) c(max=max(col), min=min(col)))

mask <- matrixAuto[,"horsepower"] == agg_vals_matrix["min", "horsepower"]
matrixAuto[mask, drop=FALSE]   
#                              mpg cylinders displacement horsepower weight acceleration year origin
# volkswagen 1131 deluxe sedan  26         4           97         46   1835         70.5   70      2

row.names(matrixAuto[mask,drop=FALSE])
# [1] "volkswagen 1131 deluxe sedan"

# NOTE: drop PREVENTS ONE-ROW MATRIX CONVERTED INTO VECTOR

In fact, you can handle same operation with original data frame via aggregate :事实上，您可以通过aggregate处理与原始数据帧相同的操作：

Dataframe Aggregation Dataframe 聚合

max_agg_df <- aggregate(cbind(mpg, cylinders, displacement, horsepower, weight,
                              acceleration, year, origin) ~ ., data=myfile[num_cols], FUN=max)
    max_agg_df 
#   mpg cylinders displacement horsepower weight acceleration year origin
# 1  27         8          455        225   4425         70.5   70      3

Max Cylinders最大气缸数

myfile[myfile$cylinder == max_agg_df$cylinder,]
#    mpg cylinders displacement horsepower weight acceleration year origin                      name
# 1   18         8          307        130   3504         12.0   70      1 chevrolet chevelle malibu
# 2   15         8          350        165   3693         11.5   70      1         buick skylark 320
# 3   18         8          318        150   3436         11.0   70      1        plymouth satellite
# 4   16         8          304        150   3433         12.0   70      1             amc rebel sst
# 5   17         8          302        140   3449         10.5   70      1               ford torino
# 6   15         8          429        198   4341         10.0   70      1          ford galaxie 500
# 7   14         8          454        220   4354          9.0   70      1          chevrolet impala
# 8   14         8          440        215   4312         70.5   70      1         plymouth fury iii
# 9   14         8          455        225   4425         10.0   70      1          pontiac catalina
# 10  15         8          390        190   3850          8.5   70      1        amc ambassador dpl
# 11  15         8          383        170   3563         10.0   70      1       dodge challenger se
# 12  14         8          340        160   3609          8.0   70      1        plymouth 'cuda 340
# 13  15         8          400        150   3761          9.5   70      1     chevrolet monte carlo
# 14  14         8          455        225   3086         10.0   70      1   buick estate wagon (sw)

Max Acceleration最大加速度

myfile[myfile$acceleration == max_agg_df$acceleration,]
#    mpg cylinders displacement horsepower weight acceleration year origin                         name
# 8   14         8          440        215   4312         70.5   70      1            plymouth fury iii
# 20  26         4           97         46   1835         70.5   70      2 volkswagen 1131 deluxe sedan

Max Horsepower最大马力

myfile[myfile$horsepower == max_agg_df$horsepower,]
#    mpg cylinders displacement horsepower weight acceleration year origin                    name
# 9   14         8          455        225   4425           10   70      1        pontiac catalina
# 14  14         8          455        225   3086           10   70      1 buick estate wagon (sw)

Multiple Aggregation多重聚合

agg_raw <- aggregate(cbind(mpg, cylinders, displacement, horsepower, weight,
                          acceleration, year, origin) ~ ., data=myfile[num_cols], 
                     FUN=function(col) c(max=max(col), min=min(col)))
agg_df <- do.call(data.frame, agg_raw)
                        
myfile[myfile$horsepower == agg_df$horsepower.min,]
#    mpg cylinders displacement horsepower weight acceleration year origin                         name
# 20  26         4           97         46   1835         70.5   70      2 volkswagen 1131 deluxe sedan
                        
myfile$name[myfile$horsepower == agg_df$horsepower.min]
# [1] "volkswagen 1131 deluxe sedan"

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