Hello I'm looking for a way to turn user inputted strings into matrices for example: 28 = SPACE, 27 = ?, 26 = 0, 25 = A, 24=B 23=C 22=D 21=E 20=F 19=G 18=H
17=I 16=J 15=K 14=L 13=M 12=N 11=O 10=P 9=Q 8=R 7=S 6=T 5=U 4=V 3=W 2=X 1=Y 0=Z
"HI HOW ARE YOU?" -> "[18 17 28 18 11][3 28 25 8 21][28 1 11 5 27]"
wherein each letter/symbol of the string is converted to a numerical value (special attention to spacebar I really don't know how to turn space into numbers). I'll be using these matrices to make a cryptograph
You could use utf8ToInt
x <- "HI HOW ARE YOU?"
We need pmin to get your condition 28 = SPACE right.
pmin(abs(utf8ToInt("HI HOW ARE YOU?") - utf8ToInt("Z")), 28)
# [1] 18 17 28 18 11 3 28 25 8 21 28 1 11 5 27
From ?utf8ToInt :
Conversion of UTF-8 encoded character vectors to and from integer vectors representing a UTF-32 encoding.
First step is
utf8ToInt("HI HOW ARE YOU?")
[1] 72 73 32 72 79 87 32 65 82 69 32 89 79 85 63
from which we substract utf8ToInt("Z"), i.e. 90 because you wrote 0=Z.
Call abs on the result to get positive numbers.
abs(utf8ToInt("HI HOW ARE YOU?") - utf8ToInt("Z"))
# [1] 18 17 58 18 11 3 58 25 8 21 58 1 11 5 27
The last piece is your condition 28 = SPACE, which is where pmin helps you out.
I need to find an index for all the values that present Q2 or Q4.
So basically skipping the first element that is a title. I need to save: 3,5,7,9..... till the end of the object.
How can I do this analyzing a string?
This is the code:
line = c("ISSUE_CUR", "1993-Q1", "1993-Q2", "1993-Q3", "1993-Q4", "1994-Q1",
"1994-Q2", "1994-Q3", "1994-Q4", "1995-Q1", "1995-Q2", "1995-Q3",
"1995-Q4", "1996-Q1", "1996-Q2", "1996-Q3", "1996-Q4", "1997-Q1",
"1997-Q2", "1997-Q3", "1997-Q4", "1998-Q1", "1998-Q2", "1998-Q3",
"1998-Q4", "1999-Q1", "1999-Q2", "1999-Q3", "1999-Q4", "2000-Q1",
"2000-Q2", "2000-Q3", "2000-Q4", "2001-Q1", "2001-Q2", "2001-Q3",
"2001-Q4", "2002-Q1", "2002-Q2", "2002-Q3", "2002-Q4", "2003-Q1",
"2003-Q2", "2003-Q3", "2003-Q4", "2004-Q1", "2004-Q2", "2004-Q3",
"2004-Q4", "2005-Q1", "2005-Q2", "2005-Q3", "2005-Q4", "2006-Q1",
"2006-Q2", "2006-Q3", "2006-Q4", "2007-Q1", "2007-Q2", "2007-Q3",
"2007-Q4", "2008-Q1", "2008-Q2", "2008-Q3", "2008-Q4", "2009-Q1",
"2009-Q2", "2009-Q3", "2009-Q4", "2010-Q1", "2010-Q2", "2010-Q3",
"2010-Q4", "2011-Q1", "2011-Q2", "2011-Q3", "2011-Q4", "2012-Q1",
"2012-Q2", "2012-Q3", "2012-Q4", "2013-Q1", "2013-Q2", "2013-Q3",
"2013-Q4", "2014-Q1", "2014-Q2", "2014-Q3", "2014-Q4", "2015-Q1",
"2015-Q2", "2015-Q3", "2015-Q4", "2016-Q1", "2016-Q2", "2016-Q3",
"2016-Q4", "2017-Q1", "2017-Q2", "2017-Q3", "2017-Q4", "2018-Q1",
"2018-Q2", "2018-Q3", "2018-Q4", "2019-Q1", "2019-Q2", "2019-Q3",
"2019-Q4")
We can use grep to return the index by matching 'Q2' or (|) 'Q4' at the end ($) of the string
grep("(Q2|Q4)$", line)
#[1] 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61
#[31] 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109
Or the 'Q' can be placed outside as it is common
grep("Q(2|4)$", line)
Or another option is endsWith
which(endsWith(line, "Q2")|endsWith(line, "Q4"))
You can use the grep() function.
grep("[24]$", line, value = T)
The "[24]$" is a regular expression pattern. For a good tutorial on it, you may check here.
https://www.regular-expressions.info/rlanguage.html
envelope of the K funcition (and its derivative such as L) is very useful for validating a fitted spatial points process model. for instance, I fit a poisson model for a data J1a2, which is as following:
J1a2.points:
# X.1 X Y
1 1 118.544 1638.445
2 2 325.995 1761.223
3 3 681.625 1553.771
4 4 677.392 1816.261
5 5 986.451 1685.016
6 6 1469.093 1354.787
7 7 1608.805 1625.744
8 8 1994.071 1782.391
9 9 1968.669 1375.955
10 10 2362.403 1337.852
11 11 2701.099 1773.924
12 12 2900.083 1820.495
13 13 2963.588 1668.081
14 14 3412.360 1676.549
15 15 3378.490 1456.396
16 16 3721.420 1464.863
17 17 3823.028 1701.951
18 18 4072.817 1790.859
19 19 4089.751 1388.656
20 20 97.375 715.497
21 21 376.799 1033.025
22 22 563.082 1126.166
23 23 935.647 1206.607
24 24 512.277 486.876
25 25 935.647 757.834
26 26 1409.821 410.670
27 27 1435.223 639.290
28 28 1706.180 1045.726
29 29 1968.669 876.378
30 30 2307.365 711.263
31 31 2624.892 897.546
32 32 2654.528 1236.243
33 33 2857.746 423.371
34 34 3039.795 639.290
35 35 3298.050 707.029
36 36 3111.767 1011.856
37 37 3361.555 1227.775
38 38 4047.414 1185.438
39 39 3569.007 508.045
40 40 4250.632 469.942
41 41 4386.110 872.144
42 42 93.141 237.088
43 43 554.614 186.283
44 44 757.832 148.180
45 45 965.283 220.153
46 46 1723.115 296.360
47 47 1744.283 423.371
48 48 1913.631 203.218
49 49 2167.653 292.126
50 50 2629.126 211.685
51 51 3217.610 283.658
52 52 3827.262 325.996
and:
J1a2.Win<-owin(c(0, 4500.42),c(0, 1917.87))
if you draw evelope for the data with Lest:
library(spatstat)
env.data<-envelope(J1a2, Lest,correction="border",
nsim=19, global=TRUE)
plot(env.data,.-r~r, shade=NULL, legend=FALSE,
xlab=expression(paste("r(",mu,"m)")),ylab="L(r)-r", main = "")
the Lest() curve goes out of the envelope. however, if you use Linhom instead of Lest, you will find the Linhom() are all inside of the envelope.
it seems that this suggest a inhomogenous density kernel of the data. so I use y as covariate in fitting:
poisson.J1a2<-ppm(J1a2~1,Poisson(),correction="border")
y1.J1a2<-ppm(J1a2~y,correction="border")
anova(poisson.J1a2,y.J1a2,test="LR") #p=0.6484
I don't find any evidence of a spatial trend of density along y, or x, or their combinations.
then why the Linhom() outperform the Lest() in this case?
furthermore, when should one decide to use Linhom() instead of Lest?
You should first decide whether or not the intensity can be assumed to be constant. To help you with this you can look at kernel density estimates or do formal tests such as a quadrat test etc. If you decide that the intensity can be assumed to be constant you use Lest() if this is not the case you use Linhom().
I have a matrix of data (see below) and I am trying to turn it into a density contour map (Can1 and Can2 variables), maybe with ks or sm packages.
My question is how do I select those points in the dataset which lie above (say) 80% contour of the density map?
Thanks
ID Can1 Can2
4 -12.3235137 -1.0788867664
1 -12.2949912 -0.9321009837
5 -12.2835123 -1.0164225574
2 -12.2571822 -0.7094457036
3 -12.2713779 -0.9908419863
10 -12.9870438 -1.0936405526
6 -12.7167605 -1.4620772026
7 -12.8193776 -1.0911349785
8 -12.9781963 -1.1762698594
9 -12.7983478 -1.3453369581
13 -14.0389948 0.2855210115
11 -14.0015922 0.1467552738
15 -14.0723604 0.0244576488
14 -14.0743560 0.1417245145
12 -13.9898266 0.0005437008
20 -6.5881994 0.5124980991
17 -6.1812321 0.6789584579
16 -6.4704200 0.5942317307
18 -6.6960456 0.5720874622
19 -6.1159788 0.5960966790
22 -2.4794887 2.5493267897
24 -2.4918040 2.7823374576
21 -2.5145044 2.5877290160
23 -2.5048371 2.4916280770
25 -2.5018765 2.8536302559
29 -0.1781852 2.0805229401
26 -0.1581308 2.0151355747
28 -0.2118605 1.9658284615
27 -0.4184119 2.0540218901
30 -0.2994573 2.0205573385
35 2.6254869 1.3858705991
31 2.3146430 1.3510499304
33 2.5346138 1.2524229847
34 2.3741699 1.3842499455
32 2.6008389 1.3446707509
37 3.0920503 1.5807032840
38 3.1559727 1.4924092104
36 3.1593556 1.5803284343
39 3.0801444 1.6031732981
40 3.2562384 1.5810975265
43 4.8414364 2.1539254215
41 4.7938193 2.1613978258
44 4.7919209 2.2151527426
42 4.9830802 2.2374622446
45 4.7629268 2.4217335005
46 5.5631728 0.9986762598
50 5.5250403 1.0549399894
48 5.5833619 1.1368625963
47 5.5660312 1.1881215490
49 5.6224256 1.1634998303
53 5.5536366 0.2513665533
54 5.5276808 0.2685455911
51 5.7103045 0.2193839293
52 5.6014729 0.2353172964
55 5.5959034 0.2447836618
56 5.1542133 0.6070006863
59 5.0043394 0.4518710615
58 5.2314146 0.5656457888
60 5.1318728 0.4771275341
57 5.3599822 0.4918185651
61 7.0235173 -0.2669136870
63 7.0216315 -0.0097862523
64 7.0521253 -0.2457722410
62 7.0150637 -0.1456269078
65 7.0729018 -0.3573952321
69 5.8115406 -1.4652084167
67 5.7624475 -1.4147564126
68 5.8692888 -1.4695783153
70 5.9088094 -1.4927034632
66 5.8400205 -1.4817447808
71 4.8586107 -1.3111515744
73 4.7198564 -1.2891991780
72 4.9153659 -1.4499710448
74 4.7653488 -1.2839433419
75 4.7754971 -1.4655359108
77 3.8955675 -7.0922887151
78 3.8338151 -7.1595858283
80 3.7255063 -7.2147373050
79 3.7367055 -7.3468877516
76 4.0166957 -7.1952570639
Calculate the 80% point. One way: y<- x[x > 0.8 * max(x)] (I'm assuming you wanted 80% of the max level, not the 80th percentile) .
Then plot y .
After a bit of searching I think it can be achieved using the kde2d function from the MASS package.
I'm previously a SAS user - since I don't have SAS anymore I need to learn to use R for work.
The dataset has the following column:
market date sitename impression clicks
I want to transpose it into:
market date sitename-impression sitename-clicks
I think in SAS I used to do:
Proc Transpose
by market date;
id sitename;
var impression clicks;
run;
I do have a book on R and googled a lot, but couldn't find the solution that works...
Would really appreciate if anyone can help.
Thanks in advance!!!
Let me start by saying welcome to stackoverflow. Glad to have anew user. When you ask a question it's helpful and encouraged for you to provide the code you're using and a reproducible data set that looks like the original. This is called a minimal reproducible example. To get a data set into here you can use several options, here are two: use dput() around the object name and cut and paste what is displayed in the console or just post the dataframe directly. For the code provide all the code necessary to replicate your problem. I hope you find this helpful for future questions you'll ask.
I may not fully understand but I think you want to transform, not transpose, the data.
dat <- data.frame(market=rnorm(10), date=rnorm(10), #let's create a data set
sitename=rnorm(10), impression=rnorm(10), clicks=rnorm(10))
dat #look at it (I pasted it below)
# > dat
# market date sitename impression clicks
# 1 -0.9593797 -0.08411994 1.6079129 -0.5204772 -0.31633966
# 2 -0.5088689 1.78799500 -0.2469315 1.3476964 -0.04344779
# 3 -0.1527465 0.81673996 1.7824969 -1.5531260 -1.28304384
# 4 -0.7026194 0.52072913 -0.1174356 0.5722210 -1.20474443
# 5 -0.4537490 -0.69139062 1.1124277 -0.2452974 -0.33025320
# 6 0.7466588 0.36318337 -0.4623319 -0.9036768 -0.65754302
# 7 0.8007612 2.59588554 0.1820732 0.4318629 -0.36308748
# 8 1.0781715 -1.01512734 0.2297475 0.9219439 -1.15687902
# 9 0.3731450 -0.19004572 0.5190749 -1.4020371 -0.97370295
# 10 0.7724259 1.76528303 0.5781786 -0.5490849 -0.83819036
#now to create the new columns (I think this is what you want)
#the easiest way is to use transform. ?tranform for more
dat.new <- transform(dat, sitename.clicks=sitename-clicks,
impression.clicks=impression-clicks)
dat.new #here's the new data set. Notice it has the new and old columns.
#To get rid of the old columns you can use indexing and specify the columns you want.
dat.new[, c(1:2, 6:7)]
#We could have also done:
dat.new[, c(1,2,6,7)]
#or said the columns not wanted with negative indexing:
dat.new[, -c(3:5)]
EDIT In looking at Brian's comments and the variables I would think that a long to wide transformation is what the poster desires. I would likely approach it using Wickham's reshape2 package as well, as this method is easier for me to work with and I imagine it would be easier for an R beginner as well. However, here is a base way to do the long to wide format using the same data set Brian provided:
wide <- reshape(DF, v.names=c("impression", "clicks"), idvar=c("market", "date"),
timevar="sitename", direction="wide")
reshape(wide)
The reshape function is very flexible but takes some getting used to to use appropriately. I'm leaving my previous response up as well to keep the history of this post though I now believe this is not the posters intent. It serves as a reminder that a reproducible example is very helpful in providing clarity to your query.
Example data, as Tyler said, is important. I interpreted your question differently because I thought your data was different. I didn't take the - as a literal subtraction of numerics, but a combination of variables.
DF <- expand.grid(market = LETTERS[1:5],
date = Sys.Date()+(0:5),
sitename = letters[1:2])
n <- nrow(DF)
DF$impression <- sample(100, n, replace=TRUE)
DF$clicks <- sample(100, n, replace=TRUE)
I find the reshape2 package useful for these sort of transpositions/transformations/rearrangements.
library("reshape2")
dcast(melt(DF, id.vars=c("market","date","sitename")),
market+date~sitename+variable)
gives
market date a_impression a_clicks b_impression b_clicks
1 A 2012-02-28 74 97 11 71
2 A 2012-02-29 34 30 88 35
3 A 2012-03-01 40 85 40 49
4 A 2012-03-02 46 12 99 20
5 A 2012-03-03 6 95 85 56
6 A 2012-03-04 61 61 42 64
7 B 2012-02-28 4 53 74 9
8 B 2012-02-29 43 27 92 59
9 B 2012-03-01 34 26 86 43
10 B 2012-03-02 81 47 84 35
11 B 2012-03-03 3 5 91 48
12 B 2012-03-04 19 26 99 21
13 C 2012-02-28 22 31 100 53
14 C 2012-02-29 40 83 95 27
15 C 2012-03-01 78 89 81 29
16 C 2012-03-02 57 55 79 87
17 C 2012-03-03 37 61 3 97
18 C 2012-03-04 83 61 41 77
19 D 2012-02-28 81 18 47 3
20 D 2012-02-29 90 100 17 83
21 D 2012-03-01 12 40 35 93
22 D 2012-03-02 85 14 63 67
23 D 2012-03-03 63 53 29 58
24 D 2012-03-04 40 79 56 70
25 E 2012-02-28 97 62 68 31
26 E 2012-02-29 24 84 17 63
27 E 2012-03-01 94 93 32 2
28 E 2012-03-02 6 26 86 26
29 E 2012-03-03 100 34 37 80
30 E 2012-03-04 89 87 72 11
The column names have a _ between them rather than a -, but you can change that if you want. I wouldn't recommend it, though, because then you will have problems later referencing the column since the - will be taken as subtraction (you would need to quote the name).