There are two matrices:
Matrix with 2 columns: node name and node degree (k1):
Matrix with 1 column: degrees (ms):
I need to split 1st matrix into multiple matrices, where every matrix has nodes of same degree. Then, write matrices to csv-files. But my code is not working. How can i do this correctly?
k1<-read.csv2("VandD.csv", header = FALSE)
fnk1<-as.matrix(k1)
ms<-read.csv2("mas.csv", header = FALSE)
massive<-as.matrix(ms)
wlk<-1
varbl<-1
rtt<-list()
for (wlk in 1:384) {
rtt<-NULL
stepen<-massive[wlk]
for (varbl in 1:2154) {
if(fnk1[varbl,2]==stepen){
kapa<-fnk1[varbl,1]
rtt<-append(rtt,kapa)
}
}
namef<-paste("reslt",stepen,".csv",sep = "")
write.csv2(rtt, file=namef)
}
k1
V1 V2
1 UC7Ucs42FZy3uYzjrqzOIHsw 81
2 UCyWDmyZRjrGHeKF-ofFsT5Q 81
3 UCIZP6nCTyU9VV0zIhY7q1Aw 81
4 UCqk3CdGN_j8IR9z4uBbVPSg 81
5 UCjWzQkWu0l1yAhcBoavokng 81
6 UCRXiA3h1no_PFkb1JCP0yMA 81
7 UC2w9SdXpwq2Uq-MV4W4A8kw 81
8 UCdJqTQJZleoxZFReiyNvn8w 81
9 UC2Qw1dzXDBAZPwS7zm37g8g 81
10 UCTOovOHTf4efJOmGvJBxIQQ 81
ms
V1
1 81
2 82
3 83
4 84
5 85
6 86
7 87
8 88
9 89
10 90
Seems you need split
split(k1,k1$v2)
We can use group_split
library(dplyr)
k1 %>%
group_split(v2)
I have some dataframe. Here is a small expample:
a <- rnorm(100, 5, 2)
b <- rnorm(100, 10, 3)
c <- rnorm(100, 15, 4)
df <- data.frame(a, b, c)
And I have a character variable vect <- "c('a','b')"
When I try to calculate sum of vars using command
df$d <- df[vect]
which must be an equivalent of
df$d <- df[c('a','b')]
But, as a reslut I have got an error
[.data.frame(df, vect) :undefined columns selected
You're assumption that
vect <- "c('a','b')"
df$d <- df[vect]
is equivalent to
df$d <- df[c('a','b')]
is incorrect.
As #Karthik points out, you should remove the quotation marks in the assignment to vect
However, from your question it sounds like you want to then sum the elements specified in vect and then assign to d. To do this you need to slightly change your code
vect <- c('a','b')
df$d <- apply(X = df[vect], MARGIN = 1, FUN = sum)
This does elementwise sum on the columns in df specified by vect. The MARGIN = 1 specifies that we want to apply the sum rowise rather than columnwise.
EDIT:
As #ThomasIsCoding points out below, if for some reason vect has to be a string, you can parse a string to an R expression using str2lang
vect <- "c('a','b')"
parsed_vect <- eval(str2lang(vect))
df$d <- apply(X = df[parsed_vect], MARGIN = 1, FUN = sum)
Perhaps you can try
> df[eval(str2lang(vect))]
a b
1 8.1588519 9.0617818
2 3.9361214 13.2752377
3 5.5370983 8.8739725
4 8.4542050 8.5704234
5 3.9044461 13.2642793
6 5.6679639 12.9529061
7 4.0183808 6.4746806
8 3.6415608 11.0308990
9 4.5237453 7.3255129
10 6.9379168 9.4594150
11 5.1557935 11.6776181
12 2.3829337 3.5170335
13 4.3556430 7.9706624
14 7.3274615 8.1852829
15 -0.5650641 2.8109197
16 7.1742283 6.8161200
17 3.3412044 11.6298940
18 2.5388981 10.1289533
19 3.8845686 14.1517643
20 2.4431608 6.8374837
21 4.8731053 12.7258259
22 6.9534912 6.5069513
23 4.4394807 14.5320225
24 2.0427553 12.1786148
25 7.1563978 11.9671603
26 2.4231207 6.1801862
27 6.5830372 0.9814878
28 2.5443326 9.8774632
29 1.1260322 9.4804636
30 4.0078436 12.9909014
31 9.3599808 12.2178596
32 3.5362245 8.6758910
33 4.6462337 8.6647953
34 2.0698037 7.2750532
35 7.0727970 8.9386798
36 4.8465248 8.0565347
37 5.6084462 7.5676308
38 6.7617479 9.5357666
39 5.2138482 13.6822924
40 3.6259103 13.8659939
41 5.8586547 6.5087016
42 4.3490281 9.5367522
43 7.5130701 8.1699117
44 3.7933813 9.3241308
45 4.9466813 9.4432584
46 -0.3730035 6.4695187
47 2.0646458 10.6511916
48 4.6027309 4.9207746
49 5.9919348 7.1946723
50 6.0148330 13.4702419
51 5.5354452 9.0193366
52 5.2621651 12.8856488
53 6.8580210 6.3526151
54 8.0812166 14.4659778
55 3.6039030 5.9857886
56 9.8548553 15.9081336
57 3.3675037 14.7207681
58 3.9935336 14.3186175
59 3.4308085 10.6024579
60 3.9609624 6.6595521
61 4.2358603 10.6600581
62 5.1791856 9.3241118
63 4.6976289 13.2833055
64 5.1868906 7.1323826
65 3.1810915 12.8402472
66 6.0258287 9.3805249
67 5.3768112 6.3805096
68 5.7072092 7.1130150
69 6.5789349 8.0092541
70 5.3175820 17.3377234
71 9.7706112 10.8648956
72 5.2332127 12.3418373
73 4.7626124 13.8816910
74 3.9395911 6.5270785
75 6.4394724 10.6344965
76 2.6803695 10.4501753
77 3.5577834 8.2323369
78 5.8431140 7.7932460
79 2.8596818 8.9581837
80 2.7365174 10.2902512
81 4.7560973 6.4555758
82 4.6519084 8.9786777
83 4.9467471 11.2818536
84 5.6167284 5.2641380
85 9.4700525 2.9904731
86 4.7392906 11.3572521
87 3.1221908 6.3881556
88 5.6949432 7.4518023
89 5.1435241 10.8912283
90 2.1628966 10.5080671
91 3.6380837 15.0594135
92 5.3434709 7.4034042
93 -0.1298439 0.4832707
94 7.8759390 2.7411723
95 2.0898649 9.7687250
96 4.2131549 9.3175228
97 5.0648105 11.3943350
98 7.7225193 11.4180456
99 3.1018895 12.8890257
100 4.4166832 10.4901303
I am converting SAS programs that demonstrate temporal data analysis into R. I would like to reproduce SAS PROC SPECTRA output using R.
So, my question is, can the spectral density values produced by the spectrum() function in R be converted into the values for spectral density produced by SAS PROC SPECTRA?
DATA AR1_09;
INPUT t U;
OUTPUT;
CARDS;
1 -5.19859
2 4.91364
3 -3.86515
4 4.02932
5 -4.12263
6 3.46548
7 -3.01139
8 3.13753
9 -2.34875
10 2.1531
11 -2.01086
12 1.88911
13 -2.22766
14 1.94077
15 0.1786
16 0.84228
17 -1.51301
18 2.62644
19 -3.44148
20 3.13813
21 -2.34959
22 2.70754
23 -2.54789
24 2.04427
25 -2.34041
26 1.13443
27 -0.11853
28 0.74645
29 0.02448
30 0.57811
31 -1.54715
32 1.05646
33 -0.56458
34 0.6863
35 -0.53347
36 0.60813
37 -1.22044
38 0.13136
39 -0.45568
40 0.13459
41 -0.10892
42 0.46324
43 1.01367
44 -2.44015
45 1.62849
46 1.54928
47 -2.7146
48 2.20448
49 -1.58668
50 1.06419
51 -1.41402
52 1.30755
53 -1.55331
54 1.58191
55 -2.38216
56 1.45702
57 0.79562
58 -0.91078
59 -0.59827
60 1.44958
61 -1.81996
62 -0.05101
63 -0.13188
64 1.34861
65 -1.81912
66 0.73641
67 -0.32049
68 -0.37179
69 2.26288
70 -2.2773
71 0.95193
72 -1.24679
73 0.67123
74 -0.40868
75 1.46308
76 -0.71945
77 1.07481
78 -2.25127
79 1.87573
80 -1.52811
81 1.27772
82 -2.96657
83 3.58684
84 -1.7656
85 2.92004
86 -2.36525
87 2.17087
88 -1.65458
89 0.86588
90 0.19505
91 -2.34264
92 3.51124
93 -3.33501
94 3.13522
95 -1.8957
96 0.93527
97 -0.96551
98 0.08307
99 -0.14018
100 0.48641
;
PROC SPECTRA DATA=AR1_09 OUT=AR1_09PSPEC1 P S WHITETEST;
VAR U;
WEIGHTS 1 2 1;
RUN;
PROC SPECTRA DATA=AR1_09 OUT=AR1_09PSPEC2 S WHITETEST;
VAR U;
WEIGHTS 1 2 3 4 5 4 3 2 1;
RUN;
DATA AR1_09PSPEC12;
SET AR1_09PSPEC1;
n=100;
fre=0.5*n*FREQ/(4*ATAN(1));
P_01=P_01/(16*ATAN(1));
KEEP fre P_01 S_01;
RUN;
DATA AR1_09PSPEC22;
SET AR1_09PSPEC2;
n=100;
fre=0.5*n*FREQ/(4*ATAN(1));
S_02=S_01;
KEEP fre S_02;
DATA AR1_09TRUESPEC;
SET AR1_09PSPEC1;
n=100;
rho=-0.9;
theoreticalS=1.0/(8*ATAN(1)*(1-2*rho*cos(FREQ)+rho*rho));
fre=0.5*n*FREQ/(4*ATAN(1));
KEEP fre theoreticalS;
DATA AR1_09PSPEC;
MERGE AR1_09PSPEC12 AR1_09PSPEC22 AR1_09TRUESPEC;
PROC PRINT DATA=AR1_09PSPEC;
VAR fre P_01 S_01 S_02 theoreticalS;
RUN;
and so far, after entering the data using read.xlsx() in R, this is what I've got:
AR1.09 <- as.ts(AR1.09[, 2])
install.packages("forecast")
library(forecast)
Using ma for the moving average smoother of order 2.
MA2AR1.09 <- ma(AR1.09, order = 2)
AR1.09PSPEC1 <- spectrum(na.omit(MA2AR1.09))
Tests for White Noise for Variable U
Box.test (MA2AR1.09)
Box.test (MA2AR1.09, type = "Ljung")
Periodogram of Fourier analysis. The periodogram values produced by SAS look to be approximately P (below) times 4. This isn't surprising, I have been told that some software produces periodogram values divided by 4pi.
n <- length(AR1.09)
FF <- abs(fft(AR1.09) / sqrt(n))^2
P <- (4 / n) * FF[1:((n / 2) + 1)]
f <- (0:(n/2)) / n
plot(f, P, type = "h")
So, the $spec values that are the spectral density values produced by R's spectrum() function are not the same as those produced by SAS PROC SPECTRA. Can I transform my R values into the SAS values?
That's all. Thank you for your time.
i need to calculate the mean value for each row (mean of interval). Here is a basic example (maybe anyone has even better idea to do it):
M_1_mb <- (15 : -15)#creating a vector value --> small
M_31 <- cut(M_31_mb,128)# getting 128 groups from the small vector
#M_1_mb <- (1500 : -1500)#creating a vector value
#M_1 <- cut(M_1_mb,128)# getting 128 groups from the vector
I do need to get the mean value for each row/group out of 128 intervals created in M_1 (actually i do not need even those intervals, i just need the mean of them) and i cannot figure out how to do it...
I had a look at the cut2 function from Hmisc library but unfortunatelly there is no option to set up number of intervals into which vector is to be cut (-> but there is an option to get the mean value of created intervals: levels.mean...)
I would appreciate any help! Thanks!
Additional Info:
cut2 function is working well for bigger vectors (M_1_mb), however when my vector is small (M_31_mb), then i am getting a Warning message:
Warning message:
In min(xx[xx > upper]) : no non-missing arguments to min; returning Inf
and only 31 groups are created:
M_31_mb <- (15 : -15) # smaller vector
M_31 <- table(cut2(M_31_mb,g=128,levels.mean = TRUE))
whereas
g = number of quantile groups
like this?
aggregate(M_1_mb,by=list(M_1),mean)
EDIT: Result
Group.1 x
1 (-1.5e+03,-1.48e+03] -1488.5
2 (-1.48e+03,-1.45e+03] -1465.0
3 (-1.45e+03,-1.43e+03] -1441.5
4 (-1.43e+03,-1.41e+03] -1418.0
5 (-1.41e+03,-1.38e+03] -1394.5
6 (-1.38e+03,-1.36e+03] -1371.0
7 (-1.36e+03,-1.34e+03] -1347.5
8 (-1.34e+03,-1.31e+03] -1324.0
9 (-1.31e+03,-1.29e+03] -1301.0
10 (-1.29e+03,-1.27e+03] -1277.5
11 (-1.27e+03,-1.24e+03] -1254.0
12 (-1.24e+03,-1.22e+03] -1230.5
13 (-1.22e+03,-1.2e+03] -1207.0
14 (-1.2e+03,-1.17e+03] -1183.5
15 (-1.17e+03,-1.15e+03] -1160.0
16 (-1.15e+03,-1.12e+03] -1136.5
17 (-1.12e+03,-1.1e+03] -1113.0
18 (-1.1e+03,-1.08e+03] -1090.0
19 (-1.08e+03,-1.05e+03] -1066.5
20 (-1.05e+03,-1.03e+03] -1043.0
21 (-1.03e+03,-1.01e+03] -1019.5
22 (-1.01e+03,-984] -996.0
23 (-984,-961] -972.5
24 (-961,-938] -949.0
25 (-938,-914] -926.0
26 (-914,-891] -902.5
27 (-891,-867] -879.0
28 (-867,-844] -855.5
29 (-844,-820] -832.0
30 (-820,-797] -808.5
31 (-797,-773] -785.0
32 (-773,-750] -761.5
33 (-750,-727] -738.0
34 (-727,-703] -715.0
35 (-703,-680] -691.5
36 (-680,-656] -668.0
37 (-656,-633] -644.5
38 (-633,-609] -621.0
39 (-609,-586] -597.5
40 (-586,-562] -574.0
41 (-562,-539] -551.0
42 (-539,-516] -527.5
43 (-516,-492] -504.0
44 (-492,-469] -480.5
45 (-469,-445] -457.0
46 (-445,-422] -433.5
47 (-422,-398] -410.0
48 (-398,-375] -386.5
49 (-375,-352] -363.0
50 (-352,-328] -340.0
51 (-328,-305] -316.5
52 (-305,-281] -293.0
53 (-281,-258] -269.5
54 (-258,-234] -246.0
55 (-234,-211] -222.5
56 (-211,-188] -199.0
57 (-188,-164] -176.0
58 (-164,-141] -152.5
59 (-141,-117] -129.0
60 (-117,-93.8] -105.5
61 (-93.8,-70.3] -82.0
62 (-70.3,-46.9] -58.5
63 (-46.9,-23.4] -35.0
64 (-23.4,0] -11.5
65 (0,23.4] 12.0
66 (23.4,46.9] 35.0
67 (46.9,70.3] 58.5
68 (70.3,93.8] 82.0
69 (93.8,117] 105.5
70 (117,141] 129.0
71 (141,164] 152.5
72 (164,188] 176.0
73 (188,211] 199.0
74 (211,234] 222.5
75 (234,258] 246.0
76 (258,281] 269.5
77 (281,305] 293.0
78 (305,328] 316.5
79 (328,352] 340.0
80 (352,375] 363.5
81 (375,398] 387.0
82 (398,422] 410.0
83 (422,445] 433.5
84 (445,469] 457.0
85 (469,492] 480.5
86 (492,516] 504.0
87 (516,539] 527.5
88 (539,562] 551.0
89 (562,586] 574.0
90 (586,609] 597.5
91 (609,633] 621.0
92 (633,656] 644.5
93 (656,680] 668.0
94 (680,703] 691.5
95 (703,727] 715.0
96 (727,750] 738.5
97 (750,773] 762.0
98 (773,797] 785.0
99 (797,820] 808.5
100 (820,844] 832.0
101 (844,867] 855.5
102 (867,891] 879.0
103 (891,914] 902.5
104 (914,938] 926.0
105 (938,961] 949.0
106 (961,984] 972.5
107 (984,1.01e+03] 996.0
108 (1.01e+03,1.03e+03] 1019.5
109 (1.03e+03,1.05e+03] 1043.0
110 (1.05e+03,1.08e+03] 1066.5
111 (1.08e+03,1.1e+03] 1090.0
112 (1.1e+03,1.12e+03] 1113.5
113 (1.12e+03,1.15e+03] 1137.0
114 (1.15e+03,1.17e+03] 1160.0
115 (1.17e+03,1.2e+03] 1183.5
116 (1.2e+03,1.22e+03] 1207.0
117 (1.22e+03,1.24e+03] 1230.5
118 (1.24e+03,1.27e+03] 1254.0
119 (1.27e+03,1.29e+03] 1277.5
120 (1.29e+03,1.31e+03] 1301.0
121 (1.31e+03,1.34e+03] 1324.0
122 (1.34e+03,1.36e+03] 1347.5
123 (1.36e+03,1.38e+03] 1371.0
124 (1.38e+03,1.41e+03] 1394.5
125 (1.41e+03,1.43e+03] 1418.0
126 (1.43e+03,1.45e+03] 1441.5
127 (1.45e+03,1.48e+03] 1465.0
128 (1.48e+03,1.5e+03] 1488.5
As a begginer in R i have a, probably, simple question.
I have a linear regression with this specification:
X1 = X1_t-h + X2_t-h
h for is equal to 1,2,3,4,5:
For example, when h=1 i run this code:
Modelo11 <- dynlm(X1 ~ L(X1,1) + L(X2, 1)-1, data = GDP)
Its a simple regression.
I want to implement a function that gives me the five linear regressions (h=1,2,3,4 and 5) with and without HAC heteroscedasticity estimation:
I did this, and didnt work:
for(h in 1:5){
Modelo1[h] <- dynlm(GDPTrimestralemT ~ L(SpreademT,h) + L(GDPTrimestralemT, h)-1, data = MatrizDadosUS)
coeftest(Modelo1[h], df = Inf, vcov = parzenHAC)
return(list(summary(Modelo1[h])))
}
One of the error message is:
number of items to replace is not a multiple of replacement length
This is my data.frame:
GDP <- data.frame(data )
GDP
X1 X2
1 0.542952690 0.226341364
2 0.102328393 0.743360185
3 0.166345969 0.186533485
4 1.406733422 1.392420181
5 -0.469811005 -0.114609464
6 -0.509268267 0.687555461
7 1.470439930 0.298655018
8 1.046456428 -1.056387597
9 -0.492462197 -0.530284962
10 -0.516065519 0.645957530
11 0.624638996 1.044731264
12 0.213616470 -1.652979785
13 0.669747432 1.398602289
14 0.552089131 -0.821013792
15 0.452715216 1.420094663
16 -0.892063248 -1.436600779
17 1.429284965 0.559738610
18 0.853740565 -0.898976767
19 0.741864168 1.352012831
20 0.171494650 1.704764705
21 0.422326351 -0.267064235
22 -1.261643503 -2.090694608
23 -1.321086283 -0.273954212
24 0.365226000 1.965167113
25 -0.080888690 -0.594498893
26 -0.183293801 -0.483053404
27 -1.033792032 0.586491772
28 0.718322432 1.776210145
29 -2.822693790 -0.731509917
30 -1.251740437 -1.918124078
31 1.184256949 -0.016548037
32 2.255202675 0.303438286
33 -0.930446147 0.803126180
34 -1.691383225 -0.157839283
35 -1.081643279 -0.006652717
36 1.034162006 -1.970063305
37 -0.716827488 0.306792930
38 0.098471514 0.338333164
39 0.343536547 0.389775011
40 1.442117465 -0.668885360
41 0.095131066 -0.298356861
42 0.222524607 0.291485267
43 -0.499969717 1.308312472
44 0.588162304 0.026539575
45 0.581215173 0.167710855
46 0.629343124 -0.052835206
47 0.811618963 0.716913172
48 1.463610069 -0.356369304
49 -2.000576321 1.226446201
50 1.278233553 0.313606888
51 -0.700373666 0.770273988
52 -1.206455648 0.344628878
53 0.024602262 1.001621886
54 0.858933385 -0.865771777
55 -1.592291995 -0.384908852
56 -0.833758365 -1.184682199
57 -0.281305858 2.070391729
58 -0.122848757 -0.308397782
59 -0.661013984 1.590741535
60 1.887869805 -1.240283364
61 -0.313677463 -1.393252994
62 1.142864110 -1.150916732
63 -0.633380499 -0.223923970
64 -0.158729527 -1.245647224
65 0.928619010 -1.050636078
66 0.424317087 0.593892028
67 1.108704956 -1.792833100
68 -1.338231248 1.138684394
69 -0.647492569 0.181495183
70 0.295906675 -0.101823172
71 -0.079827607 0.825158278
72 0.050353111 -0.448453121
73 0.129068772 0.205619797
74 -0.221450137 0.051349511
75 -1.300967949 1.639063824
76 -0.861963677 1.273104220
77 -1.691001610 0.746514122
78 0.365888734 -0.055308006
79 1.297349754 1.146102001
80 -0.652382297 -1.095031447
81 0.165682952 -0.012926971
82 0.127996446 0.510673745
83 0.338743162 -3.141650682
84 -0.266916587 -2.483389321
85 0.148135154 -1.239997153
86 1.256591385 0.051984536
87 -0.646281986 0.468210275
88 0.180472423 0.393014848
89 0.231892902 -0.545305005
90 -0.709986273 0.104969765
91 1.231712844 -1.703489840
92 0.435378714 0.876505107
93 -1.880394798 -0.885893722
94 1.083580732 0.117560662
95 -0.499072654 -1.039222894
96 1.850756855 -1.308752222
97 1.653952857 0.440405804
98 -1.057618294 -1.611779530
99 -0.021821282 -0.807071503
100 0.682923562 -2.358596342
101 -1.132293845 -1.488806929
102 0.319237353 0.706203968
103 -2.393105781 -1.562111727
104 0.188653972 -0.637073832
105 0.667003685 0.047694037
106 -0.534018861 1.366826933
107 -2.240330371 -0.071797320
108 -0.220633546 1.612879694
109 -0.022442941 1.172582601
110 -1.542418139 0.635161458
111 -0.684128812 -0.334973482
112 0.688849615 0.056557966
113 0.848602803 0.785297518
114 -0.874157558 -0.434518305
115 -0.404999060 -0.078893114
116 0.735896917 1.637873669
117 -0.174398836 0.542952690
118 0.222418628 0.102328393
119 0.419461884 0.166345969
120 -0.042602368 1.406733422
121 2.135670836 -0.469811005
122 1.197644287 -0.509268267
123 0.395951293 1.470439930
124 0.141327444 1.046456428
125 0.691575897 -0.492462197
126 -0.490708151 -0.516065519
127 -0.358903359 0.624638996
128 -0.227550909 0.213616470
129 -0.766692832 0.669747432
130 -0.001690915 0.552089131
131 -1.786701123 0.452715216
132 -1.251495762 -0.892063248
133 1.123462446 1.429284965
134 0.237862653 0.853740565
Thanks.
Your variable Modelo1 is a vector which cannot store lm objects. When Modelo1 is a list it should work.
library(dynlm)
df<-data.frame(rnorm(50),rnorm(50))
names(df)<-c("a","b")
c<-list()
for(h in 1:5){
c[[h]] <- dynlm(a ~ L(a,h) + L(b, h)-1, data = df)
}
To get the summary you have to access the single list elements. For example:
summary(c[[1]])
*edit in response to Richard Scriven comment
The most efficent way to to get all summaries would be:
lapply(c, summary)
This applies the summary function to each element of the list and returns a list with the results.