问题
In R I have 2 datasets: group1 and group2.
For group 1 I have 10 game_id which is the id of a game, and we have number which is the numbers of times this games has been played in group1.
So if we type
group1
we get this output
game_id number
1 758565
2 235289
...
10 87084
For group2 we get
game_id number
1 79310
2 28564
...
10 9048
If I want to test if there is a statistical difference between group1 and group2 for the first 2 game_id I can use Pearson chi-square test.
In R I simply create the matrix
# The first 2 'numbers' in group1
a <- c( group1[1,2] , group1[2,2] )
# The first 2 'numbers' in group2
b <- c( group2[1,2], group2[2,2] )
# Creating it on matrix-form
m <- rbind(a,b)
So m gives us
a 758565 235289
b 79310 28564
Here I can test H: "a is independent from b", meaning that users in group1 play game_id 1 more than 2 compared to group2.
In R we type chisq.test(m) and we get a very low p-value meaning that we can reject H, meaning that a and b is not independent.
How should one find game_id's that are played significantly more in group1 than in group2 ?
回答1:
I created a simpler version of only 3 games. I'm using a chi squared test and a proportions comparison test. Personally, I prefer the second one as it gives you an idea about what percentages you're comparing. Run the script and make sure you understand the process.
# dataset of group 1
dt_group1 = data.frame(game_id = 1:3,
number_games = c(758565,235289,87084))
dt_group1
# game_id number_games
# 1 1 758565
# 2 2 235289
# 3 3 87084
# add extra variables
dt_group1$number_rest_games = sum(dt_group1$number_games) - dt_group1$number_games # needed for chisq.test
dt_group1$number_all_games = sum(dt_group1$number_games) # needed for prop.test
dt_group1$Prc = dt_group1$number_games / dt_group1$number_all_games # just to get an idea about the percentages
dt_group1
# game_id number_games number_rest_games number_all_games Prc
# 1 1 758565 322373 1080938 0.70176550
# 2 2 235289 845649 1080938 0.21767113
# 3 3 87084 993854 1080938 0.08056336
# dataset of group 2
dt_group2 = data.frame(game_id = 1:3,
number_games = c(79310,28564,9048))
# add extra variables
dt_group2$number_rest_games = sum(dt_group2$number_games) - dt_group2$number_games
dt_group2$number_all_games = sum(dt_group2$number_games)
dt_group2$Prc = dt_group2$number_games / dt_group2$number_all_games
# input the game id you want to investigate
input_game_id = 1
# create a table of successes (games played) and failures (games not played)
dt_test = rbind(c(dt_group1$number_games[dt_group1$game_id==input_game_id], dt_group1$number_rest_games[dt_group1$game_id==input_game_id]),
c(dt_group2$number_games[dt_group2$game_id==input_game_id], dt_group2$number_rest_games[dt_group2$game_id==input_game_id]))
# perform chi sq test
chisq.test(dt_test)
# Pearson's Chi-squared test with Yates' continuity correction
#
# data: dt_test
# X-squared = 275.9, df = 1, p-value < 2.2e-16
# create a vector of successes (games played) and vector of total games
x = c(dt_group1$number_games[dt_group1$game_id==input_game_id], dt_group2$number_games[dt_group2$game_id==input_game_id])
y = c(dt_group1$number_all_games[dt_group1$game_id==input_game_id], dt_group2$number_all_games[dt_group2$game_id==input_game_id])
# perform test of proportions
prop.test(x,y)
# 2-sample test for equality of proportions with continuity correction
#
# data: x out of y
# X-squared = 275.9, df = 1, p-value < 2.2e-16
# alternative hypothesis: two.sided
# 95 percent confidence interval:
# 0.02063233 0.02626776
# sample estimates:
# prop 1 prop 2
# 0.7017655 0.6783155
The main thing is that chisq.test is a test that compares counts/proportions, so you need to provide the number of "successes" and "failures" for the groups you compare (contingency table as input). prop.test is another counts/proportions testing command that you need to provide the number of "successes" and "totals".
Now that you're happy with the result and you saw how the process works I'll add a more efficient way to perform those tests.
The first one is using dplyr and broom packages:
library(dplyr)
library(broom)
# dataset of group 1
dt_group1 = data.frame(game_id = 1:3,
number_games = c(758565,235289,87084),
group_id = 1) ## adding the id of the group
# dataset of group 2
dt_group2 = data.frame(game_id = 1:3,
number_games = c(79310,28564,9048),
group_id = 2) ## adding the id of the group
# combine datasets
dt = rbind(dt_group1, dt_group2)
dt %>%
group_by(group_id) %>% # for each group id
mutate(number_all_games = sum(number_games), # create new columns
number_rest_games = number_all_games - number_games,
Prc = number_games / number_all_games) %>%
group_by(game_id) %>% # for each game
do(tidy(prop.test(.$number_games, .$number_all_games))) %>% # perform the test
ungroup()
# game_id estimate1 estimate2 statistic p.value parameter conf.low conf.high
# (int) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
# 1 1 0.70176550 0.67831546 275.89973 5.876772e-62 1 0.020632330 0.026267761
# 2 2 0.21767113 0.24429962 435.44091 1.063385e-96 1 -0.029216006 -0.024040964
# 3 3 0.08056336 0.07738492 14.39768 1.479844e-04 1 0.001558471 0.004798407
The other one is using data.table and broom packages:
library(data.table)
library(broom)
# dataset of group 1
dt_group1 = data.frame(game_id = 1:3,
number_games = c(758565,235289,87084),
group_id = 1) ## adding the id of the group
# dataset of group 2
dt_group2 = data.frame(game_id = 1:3,
number_games = c(79310,28564,9048),
group_id = 2) ## adding the id of the group
# combine datasets
dt = data.table(rbind(dt_group1, dt_group2))
# create new columns for each group
dt[, number_all_games := sum(number_games), by=group_id]
dt[, `:=`(number_rest_games = number_all_games - number_games,
Prc = number_games / number_all_games) , by=group_id]
# for each game id compare percentages
dt[, tidy(prop.test(.SD$number_games, .SD$number_all_games)) , by=game_id]
# game_id estimate1 estimate2 statistic p.value parameter conf.low conf.high
# 1: 1 0.70176550 0.67831546 275.89973 5.876772e-62 1 0.020632330 0.026267761
# 2: 2 0.21767113 0.24429962 435.44091 1.063385e-96 1 -0.029216006 -0.024040964
# 3: 3 0.08056336 0.07738492 14.39768 1.479844e-04 1 0.001558471 0.004798407
You can see that each row represent one game and the comparison is between group 1 and 2. You can get the p values from the corresponding column, but other info of the test/comparison as well.
来源：https://stackoverflow.com/questions/32841453/fishers-and-pearsons-test-for-indepedence