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社区首页 >问答首页 >如何统计多个列中唯一类别值的出现频率并按年龄分组

如何统计多个列中唯一类别值的出现频率并按年龄分组
EN

Stack Overflow用户
提问于 2019-11-13 14:35:52
回答 3查看 115关注 0票数 2

我有如下数据集:

代码语言:javascript
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Age      Monday Tuesday Wednesday
6-9        a     b        c
6-9        b     a        c
6-9        b     c        a
9-10       c     c        b
9-10       c     a        b

我想要找出a,b,c在不同年龄组中的总频率,使用R如下:

代码语言:javascript
复制
Age    a     b    c
6-9    3     3    3
9-10   1     2    3
EN

回答 3

Stack Overflow用户

发布于 2019-11-13 14:39:16

我们可以获得更长格式的数据,count它们,并再次转换为宽格式。

代码语言:javascript
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library(dplyr)
library(tidyr)

df %>%
  pivot_longer(cols = -Age) %>%
  count(Age, value) %>%
  pivot_wider(names_from = value, values_from = n)

#  Age       a     b     c
#  <fct> <int> <int> <int>
#1 6-9       3     3     3
#2 9-10      1     2     3

data

代码语言:javascript
复制
df <- structure(list(Age = structure(c(1L, 1L, 1L, 2L, 2L), .Label = c("6-9", 
"9-10"), class = "factor"), Monday = structure(c(1L, 2L, 2L, 
3L, 3L), .Label = c("a", "b", "c"), class = "factor"), Tuesday = structure(c(2L, 
1L, 3L, 3L, 1L), .Label = c("a", "b", "c"), class = "factor"), 
Wednesday = structure(c(3L, 3L, 1L, 2L, 2L), .Label = c("a", 
"b", "c"), class = "factor")), class = "data.frame", row.names = c(NA, -5L))
票数 1
EN

Stack Overflow用户

发布于 2019-11-13 17:45:50

给定输入数据df

代码语言:javascript
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df <- structure(list(Age = structure(c(1L, 1L, 1L, 2L, 2L), .Label = c("6-9", 
"9-10"), class = "factor"), Monday = structure(c(1L, 2L, 2L, 
3L, 3L), .Label = c("a", "b", "c"), class = "factor"), Tuesday = structure(c(2L, 
1L, 3L, 3L, 1L), .Label = c("a", "b", "c"), class = "factor"), 
Wednesday = structure(c(3L, 3L, 1L, 2L, 2L), .Label = c("a", 
"b", "c"), class = "factor")), class = "data.frame", row.names = c(NA, -5L))

然后,如果您计划使用base R,以下内容可能会有所帮助

代码语言:javascript
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# make a list of categorized by Age
lst <- split(df,df$Age)
# combine the list of data frame 
zlst <- do.call(rbind,sapply(seq_along(lst), function(k) cbind(data.frame(Age = names(lst)[k]), t(as.data.frame.factor(table(unlist(lst[[k]][,-1]))))),simplify = F))
# rename the row names 
rownames(zlst) <- seq(nrow(zlst))

这最终给出了:

代码语言:javascript
复制
> zlst
   Age a b c
1  6-9 3 3 3
2 9-10 1 2 3
票数 0
EN

Stack Overflow用户

发布于 2019-11-14 01:59:54

我们可以只使用base R中的table

代码语言:javascript
复制
table(rep(df$Age, 3), unlist(df[-1]))
#       a b c
#  6-9  3 3 3
#  9-10 1 2 3

数据

代码语言:javascript
复制
df <- structure(list(Age = structure(c(1L, 1L, 1L, 2L, 2L), .Label = c("6-9", 
"9-10"), class = "factor"), Monday = structure(c(1L, 2L, 2L, 
3L, 3L), .Label = c("a", "b", "c"), class = "factor"), Tuesday = structure(c(2L, 
1L, 3L, 3L, 1L), .Label = c("a", "b", "c"), class = "factor"), 
Wednesday = structure(c(3L, 3L, 1L, 2L, 2L), .Label = c("a", 
"b", "c"), class = "factor")), class = "data.frame", row.names = c(NA, -5L))
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/58831542

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