Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 6 years ago.
Improve this question
I want to run regression for my panel data.
I have a panel data in the following format:
Column 1 has years column 2 has company name and column 3 has Equity variable
Year company name EQUITY
2006 A 12
2007 A 13
2008 A 23
2009 A 24
2010 A 13
2011 A 14
2012 A 12
2013 A 14
2014 A 14
2015 A 15
2006 B 221
2007 B 242
2008 B 262
2009 B 250
2010 B 400
2011 B 411
2012 B 420
2013 B 420
2014 B 422
2015 B 450
I have a data of 10 years for 200 companies. I want to regress the log of equity of each company on number of years(time- 10 years ). I want only slope coefficient.
I want my output like this:
Column 1-years column 2-company name column 3- beta values
Year company name slope(beta) p-value
2006 A beta value (assumed)
2007 A "
2008 A "
2009 A
2010 A
2011 A
2012 A "
2013 A
2014 A "
2015 A "
I mean slope coefficient of each comany.
Can't see what you've tried so far so here's a solution to get you up and running. The final output you sketch out doesn't really make sense since you have a slope for each company - not for each company for each year.
Here's a base R version for running the regressions. by is used to split the data and then lm for the estimation.
res <- by(indata, indata$company, FUN=function(x) { coef(lm(log(EQUITY) ~ Year+0, data=x))} )
This results in the following output of the slopes and the output can be used for plotting or listing
> res
indata$company: A
[1] 0.001344837
-------------------------------------------------------
indata$company: B
[1] 0.002896053
Update
if you want to add the slopes to the dataset for each year you can add
indata$slope <- res[indata$company]
which gives
> indata
Year company EQUITY slope
1 2006 A 12 0.001344837
2 2007 A 13 0.001344837
3 2008 A 23 0.001344837
4 2009 A 24 0.001344837
5 2010 A 13 0.001344837
6 2011 A 14 0.001344837
7 2012 A 12 0.001344837
8 2013 A 14 0.001344837
9 2014 A 14 0.001344837
10 2015 A 15 0.001344837
11 2006 B 221 0.002896053
12 2007 B 242 0.002896053
13 2008 B 262 0.002896053
14 2009 B 250 0.002896053
15 2010 B 400 0.002896053
16 2011 B 411 0.002896053
17 2012 B 420 0.002896053
18 2013 B 420 0.002896053
19 2014 B 422 0.002896053
20 2015 B 450 0.002896053
Related
I have one data.frame with three columns Year, Nominal_Revenue and COEFFICIENT. So I want to forecast with this data like example below
library(dplyr)
TEST<-data.frame(
Year= c(2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013,2014,2015,2016,2017,2018,2019,2020,2021),
Nominal_Revenue=c(8634,5798,6011,6002,6166,6478,6731,7114,6956,6968,7098,7610,7642,8203,9856,10328,11364,12211,13150,NA,NA,NA),
COEFFICIENT=c(NA,1.016,1.026,1.042,1.049,1.106,1.092,1.123,1.121,0.999,1.059,1.066,1.006,1.081,1.055,1.063,1.071,1.04,1.072,1.062,1.07, 1.075))
SIMULATION<-mutate(TEST,
FORECAST=lag(Nominal_Revenue)*COEFFICIENT
)
And results from this code is like picture below, or in other words this code calculate forecasting only for one year or more precisely 2019.
So my intention is get results only for NA in column Nominal_Revenue,like picture below.
So can anybody help me how to fix this code ?
Because each time you need the previously computed value, we can loop for the number of NAs in your variable and apply a dplyr
for (i in 1:length(which(is.na(TEST$Nominal_Revenue)))){
TEST=TEST%>%mutate(Nominal_Revenue=if_else(is.na(Nominal_Revenue),COEFFICIENT*lag(Nominal_Revenue),Nominal_Revenue))
}
> TEST
Year Nominal_Revenue COEFFICIENT
1 2000 8634.00 NA
2 2001 5798.00 1.016
3 2002 6011.00 1.026
4 2003 6002.00 1.042
5 2004 6166.00 1.049
6 2005 6478.00 1.106
7 2006 6731.00 1.092
8 2007 7114.00 1.123
9 2008 6956.00 1.121
10 2009 6968.00 0.999
11 2010 7098.00 1.059
12 2011 7610.00 1.066
13 2012 7642.00 1.006
14 2013 8203.00 1.081
15 2014 9856.00 1.055
16 2015 10328.00 1.063
17 2016 11364.00 1.071
18 2017 12211.00 1.040
19 2018 13150.00 1.072
20 2019 13965.30 1.062
21 2020 14942.87 1.070
22 2021 16063.59 1.075
I have 2 files. One is a time_file which has 3000 rows and the other is userid file which has 2000 rows. I want to merge the two, so that each row (ID) in the userid file is paired with the full data from each row of the time_file.
Rows 1-3000 would show the first userid with each of the dates.
Rows 3001-6000 would show the 2nd userid with each of the dates, and so on.
Thanks in advance!
Time file
mo day year date
11 1 2015 11/1/2015
11 2 2015 11/2/2015
11 3 2015 11/3/2015
11 4 2015 11/4/2015
11 5 2015 11/5/2015
.
.
userid file
userid
154
155
157
158
159
160
.
.
Ideal format(what I want)
mo day year date userid
11 1 2015 11/1/2015 154
11 2 2015 11/2/2015 154
11 3 2015 11/3/2015 154
11 4 2015 11/4/2015 154
11 5 2015 11/5/2015 154
.
.
3 28 2017 3/28/2017 154
3 29 2017 3/29/2017 154
3 30 2017 3/30/2017 154
3 31 2017 3/31/2017 154
11 1 2015 11/1/2015 155
11 2 2015 11/2/2015 155
11 3 2015 11/3/2015 155
11 4 2015 11/4/2015 155
11 5 2015 11/5/2015 155
11 6 2015 11/6/2015 155
Easiest solution in R I can think of, assuming you've gotten your time data in a data frame and your user data in a vector:
final_df <- cbind(date_df, "userid" = rep(user, each = 3000))
This will repeat each user_id 3000 times, then bind the user_id column to the date data frame.
In SPSS you can use the cartesian product function for this:
First this recreates your example data:
data list free/mo day year (3f4) date (a12).
begin data.
11 1 2015 11/1/2015
11 2 2015 11/2/2015
11 3 2015 11/3/2015
11 4 2015 11/4/2015
11 5 2015 11/5/2015
end data.
DATASET NAME time_file.
data list free/ userid.
begin data.
154,155,157,158,159,160
end data.
DATASET NAME userid.
This will now combine the two tables like you requested:
STATS CARTPROD VAR1=userid INPUT2=time_file VAR2=mo day year date
/SAVE OUTFILE="path\your combined data.sav".
My friend is currently working on his assignment about estimation of parameter of a time series model, SARIMAX (Seasonal ARIMA Exogenous), with Maximum Likelihood Estimation (MLE) method. The data used by him is about the monthly rainfall from 2000 - 2012 with Indian Ocean Dipole (IOD) index as the exogenous variable.
Here is data:
MONTH YEAR RAINFALL IOD
1 1 2000 15.3720526 0.0624
2 2 2000 10.3440804 0.1784
3 3 2000 14.6116392 0.3135
4 4 2000 18.6842179 0.3495
5 5 2000 15.2937896 0.3374
6 6 2000 15.0233152 0.1946
7 7 2000 11.1803399 0.3948
8 8 2000 11.0589330 0.4391
9 9 2000 10.1488916 0.3020
10 10 2000 21.1187121 0.2373
11 11 2000 15.3980518 -0.0324
12 12 2000 18.9393770 -0.0148
13 1 2001 19.1075901 -0.2448
14 2 2001 14.9097284 0.1673
15 3 2001 19.2379833 0.1538
16 4 2001 19.6900990 0.3387
17 5 2001 8.0684571 0.3578
18 6 2001 14.0463518 0.3394
19 7 2001 5.9916609 0.1754
20 8 2001 8.4439327 0.0048
21 9 2001 11.8321596 0.1648
22 10 2001 24.3700636 -0.0653
23 11 2001 22.3584436 0.0291
24 12 2001 23.6114379 0.1731
25 1 2002 17.8409641 0.0404
26 2 2002 14.7377067 0.0914
27 3 2002 21.2226294 0.1766
28 4 2002 16.6403125 -0.1512
29 5 2002 10.8074049 -0.1072
30 6 2002 6.3796552 0.0244
31 7 2002 17.0704423 0.0542
32 8 2002 1.7606817 0.0898
33 9 2002 5.3665631 0.6736
34 10 2002 8.3246622 0.7780
35 11 2002 17.8044938 0.3616
36 12 2002 16.7062862 0.0673
37 1 2003 13.5572859 -0.0628
38 2 2003 17.1113997 0.2038
39 3 2003 14.9899967 0.1239
40 4 2003 14.0996454 0.0997
41 5 2003 11.4017542 0.0581
42 6 2003 6.7749539 0.3490
43 7 2003 7.1484264 0.4410
44 8 2003 10.3004854 0.4063
45 9 2003 10.6630202 0.3289
46 10 2003 20.6518764 0.1394
47 11 2003 20.8638443 0.1077
48 12 2003 20.5548048 0.4093
49 1 2004 16.0436903 0.2257
50 2 2004 17.2568827 0.2978
51 3 2004 20.2361063 0.2523
52 4 2004 11.6619038 0.1212
53 5 2004 12.8296532 -0.3395
54 6 2004 8.4202138 -0.1764
55 7 2004 15.5916644 0.0118
56 8 2004 0.9486833 0.1651
57 9 2004 7.2732386 0.2825
58 10 2004 18.0083314 0.3747
59 11 2004 14.4672043 0.1074
60 12 2004 17.3637554 0.0926
61 1 2005 18.9420168 0.0551
62 2 2005 17.0146995 -0.3716
63 3 2005 23.3002146 -0.2641
64 4 2005 17.8689675 0.2829
65 5 2005 17.2365890 0.1883
66 6 2005 14.0178458 0.0347
67 7 2005 12.6925175 -0.0680
68 8 2005 9.3861600 -0.0420
69 9 2005 11.7132404 -0.1425
70 10 2005 18.5768673 -0.0514
71 11 2005 19.6723156 -0.0008
72 12 2005 18.3248465 -0.0659
73 1 2006 18.6252517 0.0560
74 2 2006 18.7002674 -0.1151
75 3 2006 23.4882950 -0.0562
76 4 2006 19.5652754 0.1862
77 5 2006 13.6857590 0.0105
78 6 2006 11.1265448 0.1504
79 7 2006 11.0227038 0.3490
80 8 2006 7.6550637 0.5267
81 9 2006 1.8708287 0.8089
82 10 2006 5.4129474 0.9479
83 11 2006 15.2249795 0.7625
84 12 2006 14.1703917 0.3941
85 1 2007 22.8691932 0.4027
86 2 2007 14.3317829 0.3353
87 3 2007 13.0766968 0.2792
88 4 2007 23.2335964 0.2960
89 5 2007 12.2474487 0.4899
90 6 2007 11.3357840 0.2445
91 7 2007 9.3112835 0.3629
92 8 2007 1.6431677 0.5396
93 9 2007 6.8483575 0.6252
94 10 2007 13.1529464 0.4540
95 11 2007 14.5120639 0.2489
96 12 2007 18.7909553 0.0054
97 1 2008 17.6493626 0.3037
98 2 2008 13.3828248 0.1166
99 3 2008 19.0525589 0.2730
100 4 2008 17.3262806 0.0467
101 5 2008 5.2345009 0.4020
102 6 2008 3.3166248 0.4263
103 7 2008 10.1094016 0.5558
104 8 2008 11.7260394 0.4236
105 9 2008 10.7470926 0.4762
106 10 2008 15.1591557 0.4127
107 11 2008 25.5558213 0.1474
108 12 2008 18.2455474 0.1755
109 1 2009 14.5430396 0.2185
110 2 2009 12.8569048 0.3521
111 3 2009 24.0707291 0.2680
112 4 2009 16.0374562 0.3234
113 5 2009 7.2387844 0.4757
114 6 2009 13.8021737 0.3078
115 7 2009 7.5232972 0.1179
116 8 2009 6.3403470 0.1999
117 9 2009 4.6583259 0.2814
118 10 2009 13.0958008 0.3646
119 11 2009 15.3329710 0.1914
120 12 2009 19.0394328 0.3836
121 1 2010 15.5080624 0.4732
122 2 2010 17.1551742 0.2134
123 3 2010 23.9729014 0.6320
124 4 2010 18.2537667 0.5644
125 5 2010 18.2236111 0.1881
126 6 2010 14.6082169 0.0680
127 7 2010 13.6161669 0.3111
128 8 2010 11.1220502 0.2472
129 9 2010 20.7870152 0.1259
130 10 2010 19.5371441 -0.0529
131 11 2010 24.8837296 -0.2133
132 12 2010 15.5016128 0.0233
133 1 2011 17.3435867 0.3739
134 2 2011 17.6096564 0.4228
135 3 2011 19.0682983 0.5413
136 4 2011 20.4890214 0.3569
137 5 2011 12.0540450 0.1313
138 6 2011 12.5896783 0.2642
139 7 2011 5.0990195 0.5356
140 8 2011 6.5726707 0.6490
141 9 2011 2.5099801 0.5884
142 10 2011 17.6380271 0.7376
143 11 2011 17.5128524 0.6004
144 12 2011 17.2655727 0.0990
145 1 2012 16.6883193 0.2272
146 2 2012 20.8374663 0.1049
147 3 2012 16.7002994 0.1991
148 4 2012 18.7962762 -0.0596
149 5 2012 16.9292646 -0.1165
150 6 2012 11.6490343 0.2207
151 7 2012 6.2529993 0.8586
152 8 2012 5.8991525 0.9473
153 9 2012 7.8485667 0.8419
154 10 2012 12.5817328 0.4928
155 11 2012 24.7770055 0.1684
156 12 2012 23.2486559 0.4899
In doing this, he works with R because it has the package for analysing the SARIMAX model. And so far, he's been doing it good with arimax() function of TSA package with seasonal ARIMA order (1,0,1).
So here I attach his syntax:
#Importing data
data=read.csv("C:/DATA.csv", header=TRUE)
rainfall=data$RAINFALL
exo=data$IOD
#Creating the suitable model of data that is able to be read by R with ts() function
library(forecast)
rainfall_ts=ts(rainfall, start=c(2000, 1), end=c(2012, 12), frequency = 12)
exo_ts=ts(exo, start=c(2000, 1), end=c(2012, 12), frequency = 12)
#Fitting SARIMAX model with seasonal ARIMA order (1,0,1) & estimation method is MLE (or ML)
library(TSA)
model_ts=arimax(log(rainfall_ts), order=c(1,0,1), seasonal=list(order=c(1,0,1), period=12), xreg=exo_ts, method='ML')
Below is the result:
> model_ts
Call:
arimax(x = log(rainfall_ts), order = c(1, 0, 1), seasonal = list(order = c(1,
0, 1), period = 12), xreg = exo_ts, method = "ML")
Coefficients:
ar1 ma1 sar1 sma1 intercept xreg
0.5730 -0.4342 0.9996 -0.9764 2.6757 -0.4894
s.e. 0.2348 0.2545 0.0018 0.0508 0.1334 0.1489
sigma^2 estimated as 0.1521: log likelihood = -86.49, aic = 184.99
Although he claimed the syntax is working, but his lecturer expected more.
Theoretically, because he used MLE, he has proven that the first derivatives of the log-likelihood function give implicit solutions. It means that the estimation process couldn't be done analytically with MLE so we need
to continue our working with the numerical method to get it done.
So this is the expectation of my friend's lecturer. He expected him that he can at least convince him that the estimation is truly need to be done numerically
and if so, he might be able to show him what method that is used by R (the numerical method such as Newton-Raphson, BFGS, BHHH, etc).
But the thing here is the arimax() function doesn't give us the choice on numerical method to choose if the estimation need to be executed numerically like below:
model_ts=arimax(log(rainfall_ts), order=c(1,0,1), seasonal=list(order=c(1,0,1), period=12), xreg=exo_ts, method='ML')
The 'method' above is for the estimation method and the available method are ML, CSS, and CSS-ML. It is clear that the sintax above doesn't consist of the numerical method and this is the matter.
So is there any possible way to know what numerical method used by R? Or my friend just got to construct his own program without depending to arimax() function?
If there are any errors in my code, please let me know. I also apologize for any grammatical or vocabulary mistakes. English is not my native language.
Some suggestions:
Estimate the model with each of the methods: ML, CSS, CSS-ML. Do the parameter estimates agree?
You can view the source code of the arimax() function by typing arimax, View(arimax) or getAnywhere(arimax) in the console.
Or you can do a debug by placing a debug bullet before the line model_ts=arimax(...) and then sourcing or debugSource()-ing your script. You can then step into the arimax function and see/verify yourself which optimization method arimax uses.
I've got a data frame with panel-data, subjects' characteristic through the time. I need create a column with a sequence from 1 to the maximum number of year per every subject. For example, if subject 1 is in the data frame from 2000 to 2005, I need the following sequence: 1,2,3,4,5,6.
Below is a small fraction of my data. The last column (exp) is what I trying to get. Additionally, if you have a look at the first subject (13) you'll see that in 2008 the value of qtty is zero. In this case I need just a NA or a code (0,1, -9999), it doesn't matter which one.
Below the data is what I did to get that vector, but it didn't work.
Any help will be much appreciated.
subject season qtty exp
13 2000 29 1
13 2001 29 2
13 2002 29 3
13 2003 29 4
13 2004 29 5
13 2005 27 6
13 2006 27 7
13 2007 27 8
13 2008 0 NA
28 2000 18 1
28 2001 18 2
28 2002 18 3
28 2003 18 4
28 2004 18 5
28 2005 18 6
28 2006 18 7
28 2007 18 8
28 2008 18 9
28 2009 20 10
28 2010 20 11
28 2011 20 12
28 2012 20 13
35 2000 21 1
35 2001 21 2
35 2002 21 3
35 2003 21 4
35 2004 21 5
35 2005 21 6
35 2006 21 7
35 2007 21 8
35 2008 21 9
35 2009 14 10
35 2010 11 11
35 2011 11 12
35 2012 10 13
My code:
numbY<-aggregate(season ~ subject, data = toCountY,length)
colnames(numbY)<-c("subject","inFish")
toCountY$inFish<-numbY$inFish[match(toCountY$subject,numbY$subject)]
numbYbyFisher<-unique(numbY)
seqY<-aggregate(numbYbyFisher$inFish, by=list(numbYbyFisher$subject), function(x)seq(1,x,1))
I am using ddply and I distinguish 2 cases:
Either you generate a sequence along subjet and you replace by NA where you have qtty is zero
ddply(dat,.(subjet),transform,new.exp=ifelse(qtty==0,NA,seq_along(subjet)))
Or you generate a sequence along qtty different of zero with a jump where you have qtty is zero
ddply(dat,.(subjet),transform,new.exp={
hh <- seq_along(which(qtty !=0))
if(length(which(qtty ==0))>0)
hh <- append(hh,NA,which(qtty==0)-1)
hh
})
EDITED
ind=qtty!=0
exp=numeric(length(subject))
temp=0
for(i in 1:length(unique(subject[ind]))){
temp[i]=list(seq(from=1,to=table(subject[ind])[i]))
}
exp[ind]=unlist(temp)
this will provide what you need
I have managed to aggregate some data into the following:
Month Year Number
1 1 2011 3885
2 2 2011 3713
3 3 2011 6189
4 4 2011 3812
5 5 2011 916
6 6 2011 3813
7 7 2011 1324
8 8 2011 1905
9 9 2011 5078
10 10 2011 1587
11 11 2011 3739
12 12 2011 3560
13 1 2012 1790
14 2 2012 1489
15 3 2012 1907
16 4 2012 1615
I am trying to create a barplot where the bars for the months are next to each other, so for the above example January through April will have two bars (one for 2011 and one for 2012) and the remaining months will only have one bar representing 2011.
I know I have to use beside=T, but I guess I need to create some sort of matrix in order to get the barplot to display properly. I am having an issue figuring out what that step is. I have a feeling it may involve matrix but for some reason I am completely stumped to what seems like a very simple solution.
Also, I have this data: y=c('Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec') which I would like to use in my names.arg. When I try to use it with the above data it tells me undefined columns selected which I am taking to mean that I need 16 variables in y. How can I fix this?
To use barplot you need to rearrange your data:
dat <- read.table(text = " Month Year Number
1 1 2011 3885
2 2 2011 3713
3 3 2011 6189
4 4 2011 3812
5 5 2011 916
6 6 2011 3813
7 7 2011 1324
8 8 2011 1905
9 9 2011 5078
10 10 2011 1587
11 11 2011 3739
12 12 2011 3560
13 1 2012 1790
14 2 2012 1489
15 3 2012 1907
16 4 2012 1615",sep = "",header = TRUE)
y <- c('Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec')
barplot(rbind(dat$Number[1:12],c(dat$Number[13:16],rep(NA,8))),
beside = TRUE,names.arg = y)
Or you can use ggplot2 with the data pretty much as is:
dat$Year <- factor(dat$Year)
dat$Month <- factor(dat$Month)
ggplot(dat,aes(x = Month,y = Number,fill = Year)) +
geom_bar(position = "dodge") +
scale_x_discrete(labels = y)