I am trying to fit values in my algorithm so that I could predict a next month's number. I am getting a No data for variable errror when clearly I've defined what the objects are that I am putting into the equation.
I've tried to place them in vectors so that it could use one vector as a training data set to predict the new values. Current script has worked for me for a different dataset but for some reason isn't working here.
The data is small so I was wondering if that has anything to do with it. The data is:
Month io obs Units Sold
12 in 1 114
1 in 2 29
2 in 3 105
3 in 4 30
4 in 5
I'm trying to predict Units Sold with the code below
matt<-TEST1
isdf<-matt[matt$month<=3,]
isdf<-na.omit(isdf)
osdf<-matt[matt$Units.Sold==4,]
lmfit<-lm(Units.Sold~obs+Month,data=isdf,na.action=na.omit)
predict(lmFit,osdf[1,1])
I am expecting to be able to place lmfit in predict and get an output.
In R a dataset data1 that contains game and times. There are 6 games and times simply tells us how many time a game has been played in data1. So head(data1) gives us
game times
1 850
2 621
...
6 210
Similar for data2 we get
game times
1 744
2 989
...
6 711
And sum(data1$times) is a little higher than sum(data2$times). We have about 2000 users in data1 and about 1000 users in data2 but I do not think that information is relevant.
I want to compare the two datasets and see if there is a statistically difference and which game "causes" that difference.
What test should I use two compare these. I don't think Pearson's chisq.test is the right choice in this case, maybe wilcox.test is the right to chose ?
I have two groups (data.frame) in R called good and bad which contain good users and bad users respectively.
The group good contains game_id which is the id for a computergame and number which is how many times this game has been played.
For example good$game_id we get 1 2 3 ... 20. We have 20 games.
Similar good$number we get 45214 1254 23 ... 8914 which is the number the game has been played. For example has game_id==1 been played 45214 times in group good.
Similar for bad.
We also have the same number of users in the two groups.
So for head(good,20) we get
game_id number
1 45214
2 1254
...
20 8914
I want to investigate if there is dependence between the number of times a fixed computergame has been played.
For game_id==1 I would try to use Pearson's Chi test for 'Independence'.
In R I type chisq.test(good[1,2], bad[1,2]) to see if there is indepence between good and bad for game_id==1 but I get an error message: x and y must have same levels.
How can this problem be solved ?
I have a train data set which has 700 records. I prepared the model using c5.0 function with this data.
library(C50)
abc_model <- C5.0(abc_train[-5], abc_train$resultval)
I have test data, which has 5000 records.
I am using predict function to do the prediction on these 5000 recs.
abc_Test <- read.csv("FullData.csv", quote="")
abc_pred <- predict(abc_model, abc_test)
This is giving me the prediction for ONLY 700 recs, not all 5000.
How to make this predict for all 5000?
When I have the train data size larger than test data size, then the result is fine, I get all data, I am able to combine test data with results and get the output into ".CSV". But when train data size is smaller than test data, all records are not getting predicted.
x <- data.frame(abc_test, abc_pred)
Any inputs how to overcome this problem? I am not an expert in R. Any suggestions will help me a lot.
Thanks Richard.
Below is my train data, few recs.
Id Value1 Value2 Country Result
20835 63 1 United States yes
3911156 60 12 Romania no
39321 10 3 United States no
29425 80 9 Australia no
Below is my test data, few recs again.
Id Value1 Value2 Country
3942587 114 12 United States
3968314 25 13 Sweden
3973205 83 10 Russian Federation
17318 159 9 Russian Federation
I am trying to find the Result value and append this to my test data. But, like i described, I am getting the Result only for 700 records, not all 5000
You should try this:
str(abc_train)
str(abc_test)
lapply(abc_train[ names(abc_train) != "Result"] , table)
lapply(abc_train[] , table)
Then you will probably find that some of the levels for some of the variables in abc_test were not in abc_train, so estimates could not be produced. I'm guessing you thought that the numeric values would be handled as though a regression had been done, but that won't happen if those columns are factors in any prediction function and perhaps never depending on the function's behavior. Looking at C50::C5.0.default, it appears there may be no regression option for variables.
Major Edit:
I decided to rewrite this question since my original was poorly put. I will leave the original question below to maintain a record. Basically, I need to do Fisher's Test on tables as big as 4 x 5 with around 200 observations. It turns out that this is often a major computational challenge as explained here (I think, I can't follow it completely). As I use both R and Stata I will frame the question for both with some made-up data.
Stata:
tabi 1 13 3 27 46 \ 25 0 2 5 3 \ 22 2 0 3 0 \ 19 34 3 8 1 , exact(10)
You can increase exact() to 1000 max (but it will take maybe a day before returning an error).
R:
Job <- matrix(c(1,13,3,27,46, 25,0,2,5,3, 22,2,0,3,0, 19,34,3,8,1), 4, 5,
dimnames = list(income = c("< 15k", "15-25k", "25-40k", ">40k"),
satisfaction = c("VeryD", "LittleD", "ModerateS", "VeryS", "exstatic")))
fisher.test(Job)
For me, at least, it errors out on both programs. So the question is how to do this calculation on either Stata or R?
Original Question:
I have Stata and R to play with.
I have a dataset with various categorical variables, some of which have multiple categories.
Therefore I'd like to do Fisher's exact test with more than 2 x 2 categories
i.e. apply Fisher's to a 2 x 6 table or a 4 x 4 table.
Can this be done with either R or Stata ?
Edit: whilst this can be done in Stata - it will not work for my dataset as I have too many categories. Stata goes through endless iterations and even being left for a day or more does not produce a solution.
My question is really - can R do this, and can it do it quickly ?
Have you studied the documentation of R function fisher.test? Quoting from help("fisher.test"):
For 2 by 2 cases, p-values are obtained directly using the (central or
non-central) hypergeometric distribution. Otherwise, computations are
based on a C version of the FORTRAN subroutine FEXACT which implements
the network developed by Mehta and Patel (1986) and improved by
Clarkson, Fan and Joe (1993).
This is an example given in the documentation:
Job <- matrix(c(1,2,1,0, 3,3,6,1, 10,10,14,9, 6,7,12,11), 4, 4,
dimnames = list(income = c("< 15k", "15-25k", "25-40k", "> 40k"),
satisfaction = c("VeryD", "LittleD", "ModerateS", "VeryS")))
fisher.test(Job)
# Fisher's Exact Test for Count Data
#
# data: Job
# p-value = 0.7827
# alternative hypothesis: two.sided
As far as Stata is concerned, your original statement was totally incorrect. search fisher leads quickly to help tabulate twoway and
the help for the exact option explains that it may be applied to r x
c as well as to 2 x 2 tables
the very first example in the same place of Fisher's exact test underlines that Stata is not limited to 2 x 2 tables.
It's a minimal expectation anywhere on this site that you try to read basic documentation. Please!