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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()

阴谋

4

2 回答 2

1

你不能做卡方,因为JobSeekers它是连续的,所以如果你想知道南北区之间的差异,你可以使用 wilcoxon 或 t.test。这取决于您的数据。wilcoxon 是基于排名的,不需要您的数据呈正态分布。

假设您统计了每个地区和每个月的求职者人数:

df = data.frame(Month=rep(c("2020-01","2020-02","2020-03","2020-04","2020-05","2020-06"),3),
District=rep(c("Dan","North","South"),each=6),JobSeekers=rpois(18,20))

t.test 如下完成,但是如果您的样本是配对的,例如您有 12 个值用于 North,每个月有 12 个值用于南方,那么您需要设置paired=FALSE,请参阅本教程

t.test(JobSeekers ~ District,data=subset(df,District %in% c("North","South")))

    Welch Two Sample t-test

data:  JobSeekers by District
t = 0.27455, df = 9.9435, p-value = 0.7893
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -3.560951  4.560951
sample estimates:
mean in group North mean in group South 
               21.5                21.0 

如果您不确定您的样本是否为正态分布,请使用 wilcoxon:

wilcox.test(JobSeekers ~ District,data=subset(df,District %in% c("North","South")))

    Wilcoxon rank sum test with continuity correction

data:  JobSeekers by District
W = 19.5, p-value = 0.8721
alternative hypothesis: true location shift is not equal to 0
于 2020-06-22T22:56:59.320 回答
0

您可以使用 ANOVA 测试来比较多个组。如果您通过综合 ANOVA 测试发现任何具有统计意义的结果,那么您可以检查哪个地区的差异更好或更差。

您还可以参考 UCLA 的网站,该网站显示了应该使用哪些测试来测试他们的数据。链接在这里

作为一个简单的例子,让我在这里介绍如何运行 ANOVA 测试。

这是您的数据:

head(df)

r$> head(df)
    Month  District   Age Gender Education Disability Religion                          Occupation JobSeekers GMI ACU NACU NewSeekers NewFiredSeekers
1 2020-01       Dan 18-24   Male      None       Hard   Jewish Practical engineers and technicians          1   0   0    1          1               1
2 2020-01     North 18-24   Male      None       Hard   Jewish Practical engineers and technicians          1   0   0    1          1               1
3 2020-01     North 18-24   Male      None       Hard   Jewish Practical engineers and technicians          1   0   0    1          1               1
4 2020-01     South 18-24   Male      None       Hard   Jewish Practical engineers and technicians          1   0   0    1          1               1
5 2020-01       Dan 18-24   Male      None       Hard   Jewish Practical engineers and technicians          1   0   0    1          1               1
6 2020-01 Jerusalem 18-24   Male      None       Hard   Jewish Practical engineers and technicians          1   0   0    1          1               1

因为我需要更多数据点来进行测试,所以我通过引导复制了您的数据。我还增加了南方和北方地区的求职者人数。您无需在数据中执行以下步骤。但这就是我所做的。

# For the sake of this example, I increased the number of observation by bootstrapping the example data

for(i in 1:20) df <- rbind(df[sample(6, 5), ],df)
rownames(df) <- 1:nrow(df)
df$District <- sample(c("Jerusalem", "North", "Sharon", "South", "Dan"), nrow(df),replace = T)
df$JobSeekers[df$District == "North"] <- sample(1:3,length(df$JobSeekers[df$District == "North"]),replace=T,p=c(0.1,0.5,0.4))
df$JobSeekers[df$District == "South"] <- sample(4:6,length(df$JobSeekers[df$District == "South"]),replace=T,p=c(0.1,0.5,0.4))

在分析分类变量时,最好将字符作为一个因素。通过这样做,您可以控制因素的水平。

## Make the District as a factor

df$District <- as.factor(df$District)

接下来,获取组均值和标准差,以查看组间是否存在任何有意义的差异。如您所见,我更改了南区和北区,因此与其他区相比,它们的平均得分最高。

## Get the group means and standart deviations
    group.mean.sd <- aggregate(
        x = df$JobSeekers, # Specify data column
        by = list(df$District), # Specify group indicator
        FUN = function(x) c('mean'=mean(x),'sd'= sd(x))
    )

r$> group.mean.sd
    Group.1    x.mean      x.sd
1       Dan 1.1000000 0.3077935
2 Jerusalem 1.0000000 0.0000000
3     North 2.3225806 0.5992827
4    Sharon 1.1363636 0.3512501
5     South 5.2380952 0.4364358

最后,您可以按如下方式运行 ANOVA 测试和 Tukey 测试。

## Run one way ANOVA test
    anova_one_way <- aov(JobSeekers~ District, data = df)
    summary(anova_one_way)

r$> summary(anova_one_way)
             Df Sum Sq Mean Sq F value Pr(>F)
District      4 260.09   65.02   346.1 <2e-16 ***
Residuals   101  18.97    0.19
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

## Run the Tukey Test to compare the groups 
    TukeyHSD(anova_one_way)

r$> Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = JobSeekers ~ District, data = df)

$District
                        diff        lwr        upr     p adj
Jerusalem-Dan    -0.10000000 -0.5396190  0.3396190 0.9695592
North-Dan         1.22258065  0.8772809  1.5678804 0.0000000
Sharon-Dan        0.03636364 -0.3356042  0.4083315 0.9987878
South-Dan         4.13809524  3.7619337  4.5142567 0.0000000
North-Jerusalem   1.32258065  0.9132542  1.7319071 0.0000000
Sharon-Jerusalem  0.13636364 -0.2956969  0.5684241 0.9048406
South-Jerusalem   4.23809524  3.8024191  4.6737714 0.0000000
Sharon-North     -1.18621701 -1.5218409 -0.8505932 0.0000000
South-North       2.91551459  2.5752488  3.2557803 0.0000000
South-Sharon      4.10173160  3.7344321  4.4690311 0.0000000

最后,您可以用条形图绘制出哪个地区的求职者最多。

## Check the mean differences across the groups 

library(ggplot2)
ggplot(df, aes(x = District, y = JobSeekers, fill = District)) +
    geom_boxplot() +
    geom_jitter(shape = 15,
        color = "steelblue",
        position = position_jitter(0.21)) +
    theme_classic()

跨区求职者

更新

根据您的更新,您可以使用以下语法来缩写 x 标签并更改图例。

library(stringr)
library(ggplot2)
ggplot(occu.newseek, aes(x = Occupation, y = NewSeekers, fill = str_wrap(Occupation,10))) +
    geom_boxplot() +
    geom_jitter(
        shape = 19,
        color = "black",
        position = position_jitter(0.21)
    ) +
     scale_x_discrete(
        labels =
            c(
                "Academic degree" = "Academic",
                "Practical engineers and technicians" = "Engineering",
                'Production and construction'='Production',
                "Sales and costumer service" = "Sales",
                "Unprofessional workers" = "Unprofessional",
                "Undefined" = "Undefined"
            )
    ) +
    labs(fill = "Occupation") +
        theme_classic()+
        theme(
            axis.text.x = element_text(angle = 45, vjust = 1, hjust = 1), legend.key.height=unit(2, "cm")
            #legend.position = "top",
            
        )

你应该得到一个这样的图表。

更新图表

于 2020-06-22T22:55:13.440 回答