我有一个与我的 df 结构相关的技术问题。它看起来像这样：
Month District Age Gender Education Disability Religion Occupation JobSeekers GMI
1 2020-01 Dan U17 Male None None Jewish Unprofessional workers 2 0
2 2020-01 Dan U17 Male None None Muslims Sales and costumer service 1 0
3 2020-01 Dan U17 Female None None Other Undefined 1 0
4 2020-01 Dan 18-24 Male None None Jewish Production and construction 1 0
5 2020-01 Dan 18-24 Male None None Jewish Academic degree 1 0
6 2020-01 Dan 18-24 Male None None Jewish Practical engineers and technicians 1 0
ACU NACU NewSeekers NewFiredSeekers
1 0 2 0 0
2 0 1 0 0
3 0 1 0 0
4 0 1 0 0
5 0 1 0 0
6 0 1 1 1
而且我正在寻找一种方法来对两个变量（例如地区和求职者）之间的独立性进行卡方检验，这样我就可以判断北部地区与求职者的关系是否比南部地区更多。据我所知，数据结构有问题（地区是一个字符，求职者是一个整数，表示我有多少基于地区、性别、职业等的求职者）我试图将其子集到地区和求职者喜欢这：
Month District JobSeekers GMI ACU NACU NewSeekers NewFiredSeekers
<chr> <chr> <int> <int> <int> <int> <int> <int>
1 2020-01 Dan 33071 4694 9548 18829 6551 4682
2 2020-01 Jerusalem 21973 7665 3395 10913 3589 2260
3 2020-01 North 47589 22917 4318 20354 6154 3845
4 2020-01 Sharon 25403 6925 4633 13845 4131 2727
5 2020-01 South 37089 18874 2810 15405 4469 2342
6 2020-02 Dan 32660 4554 9615 18491 5529 3689
但这使处理变得更加困难，我当然会接受任何其他可行的测试。
如果您需要更多信息，请帮助并告诉我，
摩西
更新
# t test for district vs new seekers
# sorting
dist.newseek <- Cdata %>%
group_by(Month,District) %>%
summarise(NewSeekers=sum(NewSeekers))
# performing a t test on the mini table we created
t.test(NewSeekers ~ District,data=subset(dist.newseek,District %in% c("Dan","South")))
# results
Welch Two Sample t-test
data: NewSeekers by District
t = 0.68883, df = 4.1617, p-value = 0.5274
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-119952.3 200737.3
sample estimates:
mean in group Dan mean in group South
74608.25 34215.75
#wilcoxon test
# filtering Cdata to New seekers based on month and age
age.newseek <- Cdata %>%
group_by(Month,Age) %>%
summarise(NewSeekers=sum(NewSeekers))
#performing a wilcoxon test on the subset
wilcox.test(NewSeekers ~ Age,data=subset(age.newseek,Age %in% c("25-34","45-54")))
# Results
Wilcoxon rank sum exact test
data: NewSeekers by Age
W = 11, p-value = 0.4857
alternative hypothesis: true location shift is not equal to 0
方差分析测试
# Sorting occupation and month by new seekers
occu.newseek <- Cdata %>%
group_by(Month,Occupation) %>%
summarise(NewSeekers=sum(NewSeekers))
## Make the Occupation as a factor
occu.newseek$District <- as.factor(occu.newseek$Occupation)
## Get the occupation group means and standart deviations
group.mean.sd <- aggregate(
x = occu.newseek$NewSeekers, # Specify data column
by = list(occu.newseek$Occupation), # Specify group indicator
FUN = function(x) c('mean'=mean(x),'sd'= sd(x))
)
## Run one way ANOVA test
anova_one_way <- aov(NewSeekers~ Occupation, data = occu.newseek)
summary(anova_one_way)
## Run the Tukey Test to compare the groups
TukeyHSD(anova_one_way)
## Check the mean differences across the groups
library(ggplot2)
ggplot(occu.newseek, aes(x = Occupation, y = NewSeekers, fill = Occupation)) +
geom_boxplot() +
geom_jitter(shape = 15,
color = "steelblue",
position = position_jitter(0.21)) +
theme_classic()
