Re: [R] Adjusting length of series

Mon, 02 Jul 2012 10:50:30 -0700 

Thanks very much A.K. I have to admit that my problem was not clearly stated, 
with the structure of my data provided. Now all is well.
 
Cheers
Lexi
 

________________________________


Cc: R help <r-help@r-project.org> 
Sent: Monday, July 2, 2012 4:40 PM
Subject: Re: [R]  Adjusting length of series
  


Hello,

The class of your data is not dataframe.
Suppose I call your data as ydat1

str(ydat1)

 mts [1:24, 1:7] 68.1 -34.8 90.4 54.6 -172.3 ...
 - attr(*, "dimnames")=List of 2
  ..$ : NULL
  ..$ : chr [1:7] "DCred1" "DCred2" "DCred3" "DBoBC2" ...
 - attr(*, "tsp")= num [1:3] 2001 2003 12
 - attr(*, "class")= chr [1:2] "mts" "ts"

ydat2<-data.frame(ydat1)

str(ydat2)
'data.frame':    24 obs. of  7 variables:
 $ DCred1: num  68.1 -34.8 90.4 54.6 -172.3 ...
 $ DCred2: num  NA -102.9 125.2 -35.8 -226.9 ...
 $ DCred3: num  NA NA 228 -161 -191 ...
 $ DBoBC2: num  NA -164.5 17.1 96 680.2 ...
 $ DBoBC3: num  NA NA 181.5 78.9 584.3 ...
 $ CredL1: num  4937 5005 4970 5061 5115 ...
 $ BoBCL1: num  4188 4296 4240 4201 4258 ...

#Since you wanted only to do lm for these columns, I guess it doesn't really 
matter whether you have month and year in the dataset.
 #With NAs
 regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat2)
> summary(regCred)

Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 + 
    BoBCL1, data = ydat2)

Residuals:
        Min          1Q      Median          
3Q         Max 
-124.988463  -33.133975    7.971083   23.607953   76.813601 

Coefficients:
                 Estimate    Std. Error  t value   
Pr(>|t|)    
(Intercept) -538.61375718  205.91179535 -2.61575   0.020344 *  
DCred2         0.96401908    0.15623660  6.17025 2.4337e-05 ***
DCred3        -0.25720355    0.08983607 -2.86303   0.012524 *  
DBoBC2        -0.11222347    0.07828182 -1.43358   0.173646    
DBoBC3         0.04564621    0.03825169  1.19331   0.252578    
CredL1         0.18499925    0.06565456  2.81777   0.013693 *  
BoBCL1        -0.07682710    0.03406916 -2.25503   0.040666 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ 
’ 1 

Residual standard error: 54.44479 on 14 degrees of freedom
  (3 observations deleted due to missingness)
Multiple R-squared: 0.9324472,    Adjusted R-squared: 0.903496 
F-statistic: 32.20757 on 6 and 14 DF,  p-value: 2.046024e-07 
Without NAs
> ydat3<-na.omit(ydat2)
> regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat3)
> summary(regCred)

Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 + 
    BoBCL1, data = ydat3)

Residuals:
        Min          1Q      Median          
3Q         Max 
-124.988463  -33.133975    7.971083   23.607953   76.813601 

Coefficients:
                 Estimate    Std. Error  t value   
Pr(>|t|)    
(Intercept) -538.61375718  205.91179535 -2.61575   0.020344 *  
DCred2         0.96401908    0.15623660  6.17025 2.4337e-05 ***
DCred3        -0.25720355    0.08983607 -2.86303   0.012524 *  
DBoBC2        -0.11222347    0.07828182 -1.43358   0.173646    
DBoBC3         0.04564621    0.03825169  1.19331   0.252578    
CredL1         0.18499925    0.06565456  2.81777   0.013693 *  
BoBCL1        -0.07682710    0.03406916 -2.25503   0.040666 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ 
’ 1 

Residual standard error: 54.44479 on 14 degrees of freedom
Multiple R-squared: 0.9324472,    Adjusted R-squared: 0.903496 
F-statistic: 32.20757 on 6 and 14 DF,  p-value: 2.046024e-

#Same result
Not sure what you meant by ("This is good, but couldn't I code the process for 
my 15 variable model?")


A.K.

________________________________


Cc: R help <r-help@r-project.org> 
Sent: Monday, July 2, 2012 5:13 AM
Subject: Re: [R]  Adjusting length of series


Hi David and AK,
I have been trying to implement your suggestions since yesterday, but I 
encountered some challenges.

As for David's suggestions, I could only implement it after some 
modifications. Using an abridged version of my data, I dpud my dataset and 
then show my steps below.

> dput(ydata)
structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996, 
54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002, 
-35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1, 
98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8, 
-17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2, -35.7999999999993, 
-226.900000000001, 224.1, 123.2,
-95.1999999999998, -115.500000000001, 
166.200000000001, -13.6999999999998, -184.3, 232, 350.3, -840.900000000001,
424.500000000001, 61.7999999999993, -107, 230.400000000001, -395.200000000001, 
239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999, 
-191.100000000001, 451.000000000001, -100.900000000001, -218.4, 
-20.3000000000011, 281.700000000002, -179.900000000001, -170.6, 
416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999, 
337.400000000001, -625.600000000001, 634.600000000001, -384.500000000001, 
448.700000000001, NA, NA, -164.457840999999, 17.0793539999995, 
95.9767880000009, 680.238166999999, -491.348690999999, -274.694009, 
-256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104, 
757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521, 
-375.277650999999, 293.867029999999, 417.845195, 278.198807, 
-968.592033999999, -314.195986, NA, NA, NA,
181.537194999999, 
78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999, 
18.3611019999998, 725.955867, -616.054851, 105.354689000001, 
-65.8929020000005, 864.658367999999, -2446.902797, 4009.313485, 
-3767.078372, 1963.363941, -891.662171999999, 669.144680999999, 
123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937, 
5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6, 
5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3, 
6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5, 
4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616, 
5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948, 
5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025, 
5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021, 
7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames = list(
    NULL, c("DCred1", "DCred2",
"DCred3", "DBoBC2", "DBoBC3", 
    "CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
), class = c("mts", "ts"))

NB: the NAs in the dataset emanated from lagging or differencing the series

David's suggestion
 df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1, BoBCL1) : 
  arguments imply differing number of rows: 23, 22, 21, 24

So I modified as follows:
length(DCred3)  # finding the minimum length of various series
[1] 21

# Then dataframe construction
dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
+ Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
# Then estimated regression
> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
> summary(regCred)
# Worked well as shown by results below
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL + 
    BoBCL, data = dframe)
Residuals:
    Min      1Q  Median      3Q     Max 
-69.516 -27.695  -8.085  13.851 107.276 
Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 159.32304  157.15209   1.014 0.327873    
Dcre2        -0.75527    0.17262  -4.375 0.000634 ***
Dcre3        -0.21006    0.08656  -2.427 0.029329 *  
Dbobc2        0.05111    0.06565   0.779 0.449197    
Dbobc3        0.03106    0.03510   0.885 0.391108    
CredL        -0.10967    0.04933  -2.223 0.043177 *  
BoBCL         0.09756    0.03097   3.150 0.007087 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ 
’ 1 
Residual standard error: 52.3 on 14 degrees of freedom
Multiple R-squared: 0.9331,     Adjusted R-squared: 0.9044 
F-statistic: 32.55 on 6 and 14 DF,  p-value: 1.911e-07 

This is good, but couldn't I code the process for my 15 variable model?
Perhaps that is where the use of
Dcr<- lapply(..., function(x) ...)
comes in?

AK, if you spare some minutes, please use my dput data to illustrate the 
suggestion you made, I searched the lapply function (using ??lapply) but could 
not get a handle of how to use it in my case. My dput data is as shown below.

         DCred1 DCred2  DCred3      DBoBC2      DBoBC3 
CredL1   BoBCL1
Feb 2001   68.1     NA      NA          
NA          NA 4937.0 4187.500
Mar 2001  -34.8 -102.9      NA  -164.45784          NA 5005.1 
4296.005
Apr 2001   90.4  125.2  
228.1    17.07935   181.53719 4970.3 4240.052
May 2001   54.6  -35.8  -161.0    95.97679    78.89743 5060.7 4201.178
Jun 2001 -172.3 -226.9  -191.1   680.23817   584.26138 5115.3 4258.281
Jul 2001   51.8  224.1   451.0  -491.34869 -1171.58686 4943.0 4995.623
Aug 2001  175.0  123.2  -100.9  -274.69401   216.65468 4994.8 5241.615
Sep 2001   79.8  -95.2  -218.4  -256.33291    18.36110 5169.8 5212.914
Oct 2001  -35.7 -115.5   -20.3   469.62296   725.95587 5249.6 4927.880
Nov 2001  130.5  166.2   281.7  -146.43189  -616.05485 5213.9 5112.468
Dec 2001  116.8  -13.7  -179.9   -41.07720   105.35469 5344.4 5150.625
Jan 2002  -67.5
-184.3  -170.6  -106.97010   -65.89290 5461.2 5147.705
Feb 2002  164.5  232.0   416.3   757.68826   864.65837 5393.7 5037.814
Mar 2002  514.8  350.3   118.3 -1689.21453 -2446.90280 5558.2 5685.612
Apr 2002 -326.1 -840.9 -1191.2  2320.09895  4009.31348 6073.0 4644.195
May 2002   98.4  424.5  1265.4 -1446.97942 -3767.07837 5746.9 5922.877
Jun 2002  160.2   61.8  -362.7   516.38452  1963.36394 5845.3 5754.580
Jul 2002   53.2 -107.0  -168.8  -375.27765  -891.66217 6005.5 6102.667
Aug 2002  283.6  230.4   337.4   293.86703   669.14468 6058.7 6075.477
Sep 2002 -111.6 -395.2  -625.6   417.84519   123.97817 6342.3 6342.153
Oct 2002  127.8  239.4   634.6   278.19881 
-139.64639 6230.7 7026.675
Nov 2002  -17.3 -145.1  -384.5  -968.59203 -1246.79084 6358.5 7989.396
Dec 2002  286.3  303.6   448.7  -314.19599   654.39605 6341.2 7983.524
Jan 2003     NA     NA      NA          
NA          NA 6627.5 7663.457

Thanks kindly. Lexi       
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