我有以下数据框(df),包含5个变量的29次观测:
age height_seca1 height_chad1 height_DL weight_alog1
1 19 1800 1797 180 70
2 19 1682 1670 167 69
3 21 1765 1765 178 80
4 21 1829 1833 181 74
5 21 1706 1705 170 103
6 18 1607 1606 160 76
7 19 1578 1576 156 50
8 19 1577 1575 156 61
9 21 1666 1665 166 52
10 17 1710 1716 172 65
11 28 1616 1619 161 66
12 22 1648 1644 165 58
13 19 1569 1570 155 55
14 19 1779 1777 177 55
15 18 1773 1772 179 70
16 18 1816 1809 181 81
17 19 1766 1765 178 77
18 19 1745 1741 174 76
19 18 1716 1714 170 71
20 21 1785 1783 179 64
21 19 1850 1854 185 71
22 31 1875 1880 188 95
23 26 1877 1877 186 106
24 19 1836 1837 185 100
25 18 1825 1823 182 85
26 19 1755 1754 174 79
27 26 1658 1658 165 69
28 20 1816 1818 183 84
29 18 1755 1755 175 67
我希望获得每个变量的均值,标准差,中位数,最小值,最大值和样本大小,并将输出作为数据帧。我尝试使用下面的代码,但随后我无法使用和使用tapply或聚合似乎超出了我作为新手R程序员。我的任务要求我不要使用任何“额外”R套餐。
apply(df, 2, mean)
apply(df, 2, sd)
apply(df, 2, median)
apply(df, 2, min)
apply(df, 2, max)
apply(df, 2, length)
理想情况下,这是输出数据框的外观,包括每个统计函数的行标题:
age height_seca1 height_chad1 height_DL weight_alog1
mean 20 1737 1736 173 73
sd 3.3 91.9 92.7 9.7 14.5
median 19 1755 1755 175 71
minimum 17 1569 1570 155 50
maximum 31 1877 1880 188 106
sample size 29 29 29 29 29
非常感谢任何帮助。
答案 0 :(得分:14)
或者使用您已经完成的工作,您只需将这些摘要放入列表并使用do.call
df <- psych::read.clipboard()
tmp <- do.call(data.frame,
list(mean = apply(df, 2, mean),
sd = apply(df, 2, sd),
median = apply(df, 2, median),
min = apply(df, 2, min),
max = apply(df, 2, max),
n = apply(df, 2, length)))
tmp
mean sd median min max n
age 20.41379 3.300619 19 17 31 29
height_seca1 1737.24138 91.919474 1755 1569 1877 29
height_chad1 1736.48276 92.682492 1755 1570 1880 29
height_DL 173.37931 9.685828 175 155 188 29
weight_alog1 73.41379 14.541854 71 50 106 29
...或
data.frame(t(tmp))
age height_seca1 height_chad1 height_DL weight_alog1
mean 20.413793 1737.24138 1736.48276 173.379310 73.41379
sd 3.300619 91.91947 92.68249 9.685828 14.54185
median 19.000000 1755.00000 1755.00000 175.000000 71.00000
min 17.000000 1569.00000 1570.00000 155.000000 50.00000
max 31.000000 1877.00000 1880.00000 188.000000 106.00000
n 29.000000 29.00000 29.00000 29.000000 29.00000
答案 1 :(得分:13)
尝试使用fBasics包中的basicStats
> install.packages("fBasics")
> library(fBasics)
> basicStats(df)
age height_seca1 height_chad1 height_DL weight_alog1
nobs 29.000000 29.000000 29.000000 29.000000 29.000000
NAs 0.000000 0.000000 0.000000 0.000000 0.000000
Minimum 17.000000 1569.000000 1570.000000 155.000000 50.000000
Maximum 31.000000 1877.000000 1880.000000 188.000000 106.000000
1. Quartile 19.000000 1666.000000 1665.000000 166.000000 65.000000
3. Quartile 21.000000 1816.000000 1809.000000 181.000000 80.000000
Mean 20.413793 1737.241379 1736.482759 173.379310 73.413793
Median 19.000000 1755.000000 1755.000000 175.000000 71.000000
Sum 592.000000 50380.000000 50358.000000 5028.000000 2129.000000
SE Mean 0.612910 17.069018 17.210707 1.798613 2.700354
LCL Mean 19.158305 1702.277081 1701.228224 169.695018 67.882368
UCL Mean 21.669282 1772.205677 1771.737293 177.063602 78.945219
Variance 10.894089 8449.189655 8590.044335 93.815271 211.465517
Stdev 3.300619 91.919474 92.682492 9.685828 14.541854
Skewness 1.746597 -0.355499 -0.322915 -0.430019 0.560360
Kurtosis 2.290686 -1.077820 -1.086108 -1.040182 -0.311017
您还可以对输出进行子集化以获得所需内容:
> basicStats(df)[c("Mean", "Stdev", "Median", "Minimum", "Maximum", "nobs"),]
age height_seca1 height_chad1 height_DL weight_alog1
Mean 20.413793 1737.24138 1736.48276 173.379310 73.41379
Stdev 3.300619 91.91947 92.68249 9.685828 14.54185
Median 19.000000 1755.00000 1755.00000 175.000000 71.00000
Minimum 17.000000 1569.00000 1570.00000 155.000000 50.00000
Maximum 31.000000 1877.00000 1880.00000 188.000000 106.00000
nobs 29.000000 29.00000 29.00000 29.000000 29.00000
另一种选择是您在this post中定义自己的函数。
(我没有读过“我的任务要求我不要使用任何额外的'R套餐。”部分)
正如我之前所说,您可以使用*apply
族函数定义自己的函数并遍历每一列:
my.summary <- function(x,...){
c(mean=mean(x, ...),
sd=sd(x, ...),
median=median(x, ...),
min=min(x, ...),
max=max(x,...),
n=length(x))
}
# all these calls should give you the same results.
apply(df, 2, my.summary)
sapply(df, my.summary)
do.call(cbind,lapply(df, my.summary))
答案 2 :(得分:10)
您可以使用lapply
遍历每一列和一个匿名函数来执行每项计算:
res <- lapply( mydf , function(x) rbind( mean = mean(x) ,
sd = sd(x) ,
median = median(x) ,
minimum = min(x) ,
maximum = max(x) ,
s.size = length(x) ) )
data.frame( res )
# age height_seca1 height_chad1 height_DL weight_alog1
#mean 20.413793 1737.24138 1736.48276 173.379310 73.41379
#sd 3.300619 91.91947 92.68249 9.685828 14.54185
#median 19.000000 1755.00000 1755.00000 175.000000 71.00000
#minimum 17.000000 1569.00000 1570.00000 155.000000 50.00000
#maximum 31.000000 1877.00000 1880.00000 188.000000 106.00000
#s.size 29.000000 29.00000 29.00000 29.000000 29.00000
答案 3 :(得分:3)
为快速探索性数据分析(EDA)添加了更多选项
1)skimr软件包:
install.packages("skimr")
library(skimr)
skim(df)
2)ExPanDaR
软件包:
install.packages("ExPanDaR")
library(ExPanDaR)
# export data and code to a notebook
ExPanD(df, export_nb_option = TRUE)
# open a shiny app
ExPanD(df)
3)DescTools
软件包:
install.packages("DescTools")
library(DescTools)
Desc(df, plotit = TRUE)
#> ------------------------------------------------------------------------------
#> Describe df (data.frame):
#>
#> data frame: 29 obs. of 5 variables
#> 29 complete cases (100.0%)
#>
#> Nr ColName Class NAs Levels
#> 1 age integer .
#> 2 height_seca1 integer .
#> 3 height_chad1 integer .
#> 4 height_DL integer .
#> 5 weight_alog1 integer .
#>
#>
#> ------------------------------------------------------------------------------
#> 1 - age (integer)
#>
#> length n NAs unique 0s mean meanCI
#> 29 29 0 9 0 20.41 19.16
#> 100.0% 0.0% 0.0% 21.67
#>
#> .05 .10 .25 median .75 .90 .95
#> 18.00 18.00 19.00 19.00 21.00 26.00 27.20
#>
#> range sd vcoef mad IQR skew kurt
#> 14.00 3.30 0.16 1.48 2.00 1.75 2.29
#>
#>
#> level freq perc cumfreq cumperc
#> 1 17 1 3.4% 1 3.4%
#> 2 18 6 20.7% 7 24.1%
#> 3 19 11 37.9% 18 62.1%
#> 4 20 1 3.4% 19 65.5%
#> 5 21 5 17.2% 24 82.8%
#> 6 22 1 3.4% 25 86.2%
#> 7 26 2 6.9% 27 93.1%
#> 8 28 1 3.4% 28 96.6%
#> 9 31 1 3.4% 29 100.0%
#>
#> heap(?): remarkable frequency (37.9%) for the mode(s) (= 19)
来自Desc
的结果可以保存到Microsoft Word docx文件
### RDCOMClient package is needed
install.packages("RDCOMClient", repos = "http://www.omegahat.net/R")
# or
devtools::install_github("omegahat/RDCOMClient")
# create a new word instance and insert title and contents
wrd <- GetNewWrd(header = TRUE)
DescTools::Desc(df, plotit = TRUE, wrd = wrd)
由reprex package(v0.3.0)于2020-01-17创建
答案 4 :(得分:1)
到目前为止,我遇到了同样的问题而且我写了......
h <- function(x, flist){
f <- function(f,...)f(...)
g <- function(x, flist){vapply(flist, f , x, FUN.VALUE = numeric(1))}
df <- as.data.frame(lapply(x, g , flist))
row.names(df) <- names(flist)
df
}
h(cars, flist = list(mean = mean, median = median, std_dev = sd))
只要函数返回单个值,它就可以使用flist中指定的任何函数;即它不适用于范围
请注意,flist的元素应该以其他方式命名,您将获得结果data.frame的奇怪row.names