I am getting this error while plotting a bar graph and I am not able to get rid of it, I have tried both qplot and ggplot but still the same error.
Following is my code:
library(dplyr)
library(ggplot2)
#Investigate data further to build a machine learning model
data_country = data %>%
group_by(country) %>%
summarise(conversion_rate = mean(converted))
#Ist method
qplot(country, conversion_rate, data = data_country,geom = "bar", stat ="identity", fill = country)
#2nd method
ggplot(data_country)+aes(x=country,y = conversion_rate)+geom_bar()
Error:
stat_count() must not be used with a y aesthetic
Data in data_country:
country conversion_rate
<fctr> <dbl>
1 China 0.001331558
2 Germany 0.062428188
3 UK 0.052612025
4 US 0.037800687
The error is coming in bar chart and not in the dotted chart.
when you want to use your data existing in your data frame as y value, you must add stat = "identity" in mapping parameter. Function geom_bar have default y value. For example,
ggplot(data_country)+
geom_bar(mapping = aes(x = country, y = conversion_rate), stat = "identity")
You can use geom_col() directly. See the differences between geom_bar() and geom_col() in this link https://ggplot2.tidyverse.org/reference/geom_bar.html
geom_bar() makes the height of the bar proportional to the number of cases in each group If you want the heights of the bars to represent values in the data, use geom_col() instead.
ggplot(data_country)+aes(x=country,y = conversion_rate)+geom_col()
I was looking for the same and this may also work
p.Wages.all.A_MEAN <- Wages.all %>%
group_by(`Career Cluster`, Year)%>%
summarize(ANNUAL.MEAN.WAGE = mean(A_MEAN))
names(p.Wages.all.A_MEAN) [1] "Career Cluster" "Year" "ANNUAL.MEAN.WAGE"
p.Wages.all.a.mean <- ggplot(p.Wages.all.A_MEAN, aes(Year, ANNUAL.MEAN.WAGE , color= `Career Cluster`))+
geom_point(aes(col=`Career Cluster` ), pch=15, size=2.75, alpha=1.5/4)+
theme(axis.text.x = element_text(color="#993333", size=10, angle=0)) #face="italic",
p.Wages.all.a.mean
Source: Stackoverflow.com