我正在处理如下所示的数据集。
#Dataframe
df=data.frame(Type=c(1,2,4,5,4,3,3,4,5,1,2,3,2,1,2,3,3,2,1,1,NA),
Q1=c(1,2,6,8,9,10,2,6,7,4,9,9,1,2,NA,4,3,8,7,6,4),
Q2=c(1,2,4,NA,8,2,1,2,10,7,5,5,5,8,2,7,4,8,7,5,1))
上下文
数据框由问卷调查的结果组成。
第一列Type是指回答问卷的员工类型，其中1 ='Worker'，2 ='Factory Lead'，3 ='{{ 1}}'，4 ='Administrative Staff'和5 ='Middle Management'
第二和第三列（Executive和Q1）是问题，评分范围为1 ='Q2'至10（Strongly Agree）。
我要实现的目标
我想根据得分计算每个Strongly Disagree的响应总数。
我为分数创建了垃圾箱，它们是-
1）Type协议-分数从0到4
2）Low协议-得分为5或6
3）Medium协议-得分为7或8
4）High协议-得分为9或10
所以我想计算每个分数箱中每个工人的响应次数。
我的尝试
Very High我的逻辑是我使用library(dplyr)
library(tidyr)
result=df %>%
gather(Item,response,-1) %>%
filter(!is.na(response)) %>%
group_by(Type,Item) %>%
filter(!is.na(Type)) %>%
summarise(Low=sum(response %in% c(0,1,2,3,4)),
Medium=sum(response %in% c(5,6)),
High=sum(response %in% c(7,8)),
VHigh=sum(response %in% c(9,10)) %>%
spread(Type,-Item)
库和第一个tidyr分数来计算总响应。然后展开这些列，这样我就可以得到按工人和分数类别分类的小计。
例如，对于Q1，此列是gather，然后依次是Low-Worker，然后是Medium-Worker，然后是High-Worker，然后是Very High-Worker，然后是Low-Factory Lead ....依此类推，适用于员工和分数类别的所有组合。
很明显，我的代码中有错误之处。
所需的输出
具有两行（Medium-Factory Lead和Q1）和 20列（对于每个员工分数组合）的数据框。
对此将提供任何帮助。
答案 0 :(得分：2)
创建分数数据框
library(tidyr)
library(dplyr)
df <- data_frame(type=c(1,2,4,5,4,3,3,4,5,1,2,3,2,1,2,3,3,2,1,1,NA),
q1=c(1,2,6,8,9,10,2,6,7,4,9,9,1,2,NA,4,3,8,7,6,4),
q2=c(1,2,4,NA,8,2,1,2,10,7,5,5,5,8,2,7,4,8,7,5,1))
scores <- data_frame(score = 0:10,
scorebin = c(rep("Low", 5),
rep("Medium", 2),
rep("High", 2),
rep("Very High", 2)))
以长格式收集数据。加入分数数据框以添加一个scorebin列。按item，type和scorebin分组，并计算每组下的答案数。
df2 <- df %>%
gather(item, score, -type) %>%
left_join(scores, by = "score") %>%
group_by(item, type, scorebin) %>%
summarise(n = n()) %>%
unite(employeescore, type, scorebin)
将employeescore更改为有序水平的因子
这样它们就不会按字母顺序显示（高，低，中）
但顺序正确（低，中，高）。
employeescoreorder <- scores %>%
distinct(scorebin) %>%
merge(distinct(df, type)) %>%
unite(employeescore, type, scorebin)
df2$employeescore <- factor(df2$employeescore,
levels = employeescoreorder$employeescore)
以宽格式扩展数据框以获得20列。
df2 %>%
spread(employeescore, n)
# A tibble: 2 x 20
# Groups: item [2]
item `1_Low` `1_Medium` `1_High` `2_Low` `2_Medium` `2_High` `2_Very High` `4_Low`
* <chr> <int> <int> <int> <int> <int> <int> <int> <int>
1 q1 3 1 1 2 NA 1 1 NA
2 q2 1 1 3 2 2 1 NA 2
# ... with 11 more variables: `4_Medium` <int>, `4_High` <int>, `4_Very High` <int>,
# `5_High` <int>, `5_Very High` <int>, `3_Low` <int>, `3_Medium` <int>, `3_High` <int>,
# `3_Very High` <int>, NA_Low <int>, `<NA>` <int>
答案 1 :(得分：0)
喜欢吗？
df%>%
mutate(Type_real=case_when(
Type==1~"Worker",
Type==2~"Factory Lead",
Type==3~"Administrative Staff",
Type==4~"Middle Management",
Type==5~"Executive"),
Score=case_when(
Q1<5~"Low",
Q1>=5 & Q1<=6~"Medium",
Q1>=7 & Q1<=8~"High",
Q1>8~"Very High"))%>%
na.omit()%>%
group_by(Type_real,Score)%>%
summarise(count=n())
# A tibble: 11 x 3
# Groups: Type_real [?]
Type_real Score count
<chr> <chr> <int>
1 Administrative Staff Low 3
2 Administrative Staff Very High 2
3 Executive High 1
4 Factory Lead High 1
5 Factory Lead Low 2
6 Factory Lead Very High 1
7 Middle Management Medium 2
8 Middle Management Very High 1
9 Worker High 1
10 Worker Low 3
11 Worker Medium 1
答案 2 :(得分：0)
另一种类似于Paul Rougieux的解决方案，但没有加入：
df %>%
mutate(Type = case_when(Type == 1 ~ "Worker",
Type == 2 ~ "Factory Lead",
Type == 3 ~ "Administrative Staff",
Type == 4 ~ "Middle Management",
Type == 5 ~ "Executive")) %>%
mutate_at(c("Q1", "Q2"),
funs(case_when(. %in% 1:4 ~ "Low",
. %in% 5:6 ~ "Medium",
. %in% 7:8 ~ "High",
. %in% 9:10 ~ "Very High"))) %>%
gather(Questions, Score, Q1:Q2) %>%
unite(Type_Score, Type, Score, sep = "_") %>%
count(Questions, Type_Score) %>%
spread(Type_Score, n)
# A tibble: 2 x 21
# Questions `Administrative~ `Administrative~ `Administrative~ `Administrative~ Executive_High Executive_NA `Executive_Very~ `Factory Lead_H~
# <chr> <int> <int> <int> <int> <int> <int> <int> <int>
# 1 Q1 NA 3 NA 2 2 NA NA 1
# 2 Q2 1 3 1 NA NA 1 1 1
# ... with 12 more variables: `Factory Lead_Low` <int>, `Factory Lead_Medium` <int>, `Factory Lead_NA` <int>, `Factory Lead_Very High` <int>,
# `Middle Management_High` <int>, `Middle Management_Low` <int>, `Middle Management_Medium` <int>, `Middle Management_Very High` <int>,
# NA_Low <int>, Worker_High <int>, Worker_Low <int>, Worker_Medium <int>