Draw this loop in Postscript - adobe

Any idea how to draw this in Postscript using for loop and ifelse conditional?
My idea was to make a large red circle, then a smaller white circle and a smaller red circle again...
Also we can see that the color is getting darker so it should also be saved as a variable that gets darker.
50 50 translate
/coordinate_system {0.5 0.3 0 0 setcmykcolor
gsave
2 setlinewidth 500 0 moveto 0 0 lineto 0 500 lineto stroke
grestore
gsave
0.3 setlinewidth
9 { 30 100 moveto 500 100 lineto stroke 0 50 translate } repeat
grestore
gsave
0.3 setlinewidth
10 { 100 20 moveto 100 500 lineto stroke 50 0 translate } repeat
grestore
gsave
/tekst 3 string def /Helvetica findfont 10 scalefont setfont
100 100 500 { /y exch def 5 y 2 sub moveto y tekst cvs show } for
90 100 500 { /x exch def x 5 moveto x 10 add tekst cvs show } for
grestore
0 setgray } bind def
/s { mark pstack pop } def
coordinate_system
And this is the code so far...
100 100 translate
%100 -3 0 {{1 0 0 setrgbcolor exch 0 exch 0 360 arc stroke}{0 0 0 setrgbcolor exch 0 exch 0 360 arc stroke} ifelse} for
3 4 lt {1 0 0 setrgbcolor 0 0 50 0 360 arc stroke}{0 0 0 setrgbcolor 0 0 100 0 360 arc stroke} ifelse

The following code loops for i = 1, 2, ..., 10. I am using i to control the radius and color of the circle.
/i 1 def
{
i 0.1 mul 0 0 setrgbcolor % RGB (i*0.1, 0, 0)
i 10 gt { exit } if % exit the loop if i > 10
300 300 % center at 300 300
20 i mul % radius 20*i
drawcircle
/i i 1 add def % i = i + 1
} loop
drawcircle code:
/drawcircle % XO YO R
{
newpath
0 360 arc
closepath
stroke
} bind def
3 setlinewidth
My output (cropped a bit) is:

Related

QML how to use a dynamic topmargin correctly

I want to enlarge or reduce a button depending on the presence of an icon. The buttons are in a rectangle and I would like to realize it with the top margin, because the buttons also have rounded corners and only the top corners should be visible.
The goal is a representation like this
If I set the topmargin fix to 10 it looks good
If I calculate the value depending on whether an icon is present, the buttons without icon are correctly displayed deeper, but the buttons with icon sit too deep
Like the picture before the topmargin for the icon buttons is 10 but they are moved lower.
Any idea - the code for display.qml is reduced and does not show all the properties. if they are needed i will add them.
FooterButton.qml
Rectangle {
id: button
property string p_identity
property string p_icon
property string p_source
property string p_backgroundColor
property int p_topmargin: 10
height: 70
width: 80
Layout.leftMargin: 25
Layout.topMargin: p_topmargin
color: p_backgroundColor
radius: 10
border.color: "black"
border.width: 0
Connections {
target: m_screen;
onScreenChanged: {
p_icon = m_screen.getButtonIcon(p_identity)
p_source = (!p_icon || p_icon.length === 0) ? "" : "image://iconprovider/" + p_icon)
// p_topmargin = 10
p_topmargin = (!p_icon || p_icon.length === 0) ? 45 : 10
}
}
Footer.qml
Item {
id: footer
property string p_footerBackgroundColor: "yellow" //m_config.getColor(Colors.FooterBackground)
property string p_buttonBackgroundColor: m_config.getColor(Colors.ButtonBackground)
Rectangle { anchors.fill: parent; x: footer.x; y: footer.y; width: footer.width; height: footer.height; color: p_footerBackgroundColor
RowLayout{ anchors.fill: parent
FooterButton{ p_identity: "FB1"; p_backgroundColor: p_buttonBackgroundColor }
FooterButton{ p_identity: "FB2"; p_backgroundColor: p_buttonBackgroundColor }
FooterButton{ p_identity: "FB3"; p_backgroundColor: p_buttonBackgroundColor }
FooterButton{ p_identity: "FB4"; p_backgroundColor: p_buttonBackgroundColor }
FooterButton{ p_identity: "FB5"; p_backgroundColor: p_buttonBackgroundColor }
}
}
Display.qml
Item {
id: display
Header { x: 0; y: 0; width: display.width; height: p_headerHeight; visible: p_headerVisible; color: p_backgroundColor; p_buttonColor: p_buttonBackgroundColor }
Left { x: 0; y: p_headerHeight; width: p_borderWidth; height: p_contentHeight; color: "blue" }
Right { x: display.width - p_encoderWidth; y: p_headerHeight; width: p_encoderWidth; height: p_contentHeight; p_color: "magenta" }
Footer { x: 0; y: display.height - p_footerHeight; width: display.width; height: p_footerHeight; visible: p_footerVisible }
just making sure you're not over-engineering your problem, but, I want to point out that the standard Button has both icon support and the ability to change the background to a rounded Rectangle:
import QtQuick
import QtQuick.Controls
import QtQuick.Layouts
Page {
footer: Frame {
background: Rectangle {
color: "yellow"
}
RowLayout {
anchors.horizontalCenter: parent.horizontalCenter
spacing: 20
AppButton {
//icon.source: "hammer.svg"
}
AppButton {
//icon.source: "hammer.svg"
}
AppButton {
icon.source: "hammer.svg"
}
AppButton {
//icon.source: "hammer.svg"
}
AppButton {
icon.source: "check.svg"
}
}
}
}
// AppButton.qml
import QtQuick
import QtQuick.Controls
Button {
width: 100
height: 100
background: Rectangle {
color: pressed ? palette.mid : palette.button
radius: 20
}
icon.source: "blank.svg"
icon.width: 64
icon.height: 64
}
// blank.svg
<svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 32 32">
</svg>
// hammer.svg
<svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 32 32"><path d="M29.64 25.462c-1.186-1.62-3.535-4.176-6.254-7.136-2.657-2.893-5.671-6.173-8.066-9.11a3.883 3.883 0 0 1-1.044-1.531 6.899 6.899 0 0 0-.215-1.271 3.427 3.427 0 0 1-.08-.348 7.985 7.985 0 0 1 3.153-1.61 25.43 25.43 0 0 1 4.095-.527l1.08-.043-1.14-1.239-.148-.035a24.293 24.293 0 0 0-5.123-.606A13.096 13.096 0 0 0 7.53 4.82c-.225.2-1.433 1.478-1.338 2.334.078.73-.212.949-.792 1.383a8.35 8.35 0 0 0-.558.444c-1.468-.125-1.92.252-3.014 1.16l-.39.32-.095.105a1.472 1.472 0 0 0-.277 1.24 7.214 7.214 0 0 0 2.294 3.029 2.25 2.25 0 0 0 2.404-.483 18.003 18.003 0 0 0 1.577-2.018 2.67 2.67 0 0 1 1.633-1.26 12.815 12.815 0 0 1 2.588.88c.11.046.2.077.277.104.05.018.111.032.116.036 4.108 5.004 6.896 8.936 8.93 11.807 1.401 1.976 2.413 3.404 3.3 4.412l.912 1.038a1.935 1.935 0 0 0 1.362.651l.078.001a1.939 1.939 0 0 0 1.334-.534l1.548-1.486a1.927 1.927 0 0 0 .22-2.52zM12.059 11.028l-.029.034c-.03-.012-.052-.018-.088-.033a10.285 10.285 0 0 0-3-.954 3.577 3.577 0 0 0-2.454 1.738 21.031 21.031 0 0 1-1.375 1.786c-.605.434-.936.519-1.313.338a6.931 6.931 0 0 1-1.792-2.446.85.85 0 0 1 .125-.305l.334-.275c1.045-.867 1.228-1.021 2.299-.933a1.02 1.02 0 0 0 .738-.247A7.72 7.72 0 0 1 6 9.337a2.27 2.27 0 0 0 1.186-2.288A3.785 3.785 0 0 1 8.19 5.571a12.232 12.232 0 0 1 7.706-2.565 20.9 20.9 0 0 1 2.624.178c-.523.076-1.076.173-1.614.298A9.024 9.024 0 0 0 13.34 5.3a1.176 1.176 0 0 0-.25 1.356 5.831 5.831 0 0 1 .19 1.1 3.345 3.345 0 0 0 .842 1.625 9.48 9.48 0 0 0-.994.683 7.036 7.036 0 0 0-1.068.964zm16.668 16.234l-1.547 1.485a.945.945 0 0 1-.678.256.924.924 0 0 1-.652-.312l-.912-1.038c-.853-.97-1.905-2.452-3.236-4.33-2.018-2.848-4.78-6.742-8.838-11.696a6.433 6.433 0 0 1 .875-.772 8.145 8.145 0 0 1 .964-.66l.09-.05C17.14 13 20.06 16.182 22.65 19.001c2.7 2.939 5.032 5.477 6.184 7.051a.923.923 0 0 1-.106 1.209z"/><path fill="none" d="M0 0h32v32H0z"/></svg>
// check.svg
<svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 32 32"><path d="M13.5 22.142L7.59 16.42l.636-.636L13.5 20.87 26.721 7.8l.637.637z"/><path fill="none" d="M0 0h32v32H0z"/></svg>
You can Try it Online!

r if else for loop with logical and boollean operators

I'm trying to use an if else statement to create a new column of binary data in my data frame, but what I get is all zeros...
command:
for(i in 1:nrow(asort)){
if(asort$recip==0 && asort$dist<.74){
asort$temp[i]<-0
} else{
asort$temp[i]<-1
}
}
#temp ends up being all 0's
In addition, I would actually like to ask something along the lines of this:
# if the data in the recip column = 0, and the distances is < 0.74, OR if the #data is greater than 1.85 give me a zero, else 1
for(i in 1:nrow(asort)){
if(asort$recip==0 && asort$dist<.74 || asort$dist>1.85){
asort$temp[i]<-0
} else{
asort$temp[i]<-1
}
}
> head(asort)
coordinates CLASS_ID Flight UFID dist nnid nnid2 observed recip temp
157 (285293.3, 4426017) 0 F4_ F4_156 0.3857936 158 F4_157 0 0 0
158 (285293.2, 4426017) 0 F4_ F4_157 0.3857936 157 F4_156 0 0 0
259 (285255, 4426014) 0 F4_ F4_258 0.5226039 261 F4_260 1 0 0
261 (285255, 4426014) 0 F4_ F4_260 0.5226039 259 F4_258 1 0 0
402 (285275.3, 4426004) 0 F4_ F4_401 0.5598427 403 F4_402 1 0 0
403 (285275.6, 4426004) 0 F4_ F4_402 0.5598427 402 F4_401 1 0 0
Using df data.frame
dist <- runif(10, 0.3, 2)
recip<- c(0,1,1,0,1,0,1,0,0,1)
df <- data.frame(dist, recip)
and ifelse
df$temp<-ifelse(df$dist < 0.74 & df$recip == 0 , 0,
ifelse(df$dist > 1.85 & df$recip == 0, 0, 1))
> head(df)
# dist recip temp
#1 1.1878002 0 1
#2 0.4147835 1 1
#3 1.3707311 1 1
#4 0.9008034 0 1
#5 1.0220149 1 1
#6 1.9384069 0 0

How can I plot square wave data in R?

Consider the following data, where the left column represents a bit (1 or 0), and the right column represents the number of microseconds that we observe the bit.
0 664
1 63
0 404
1 544
0 651
1 686
0 507
1 1155
0 664
1 271
0 456
1 2763
0 664
1 115
0 456
1 4010
0 664
1 63
0 351
1 3855
I would like to plot this data such that there is a horizontal line at 0 with a width of 664, followed by a rise to a horizontal line at 1 with a width of 63, followed by a fall to a horizontal line at 0 with a width of 404, and so on.
Is there an efficient and direct way to plot this in R that does not involve manual comparison against bounds?
Here is my current code for doing this which is extremely inefficient and naive, so I hope there is a better way.
args <- commandArgs(trailingOnly = TRUE)
data = read.table(args[1])
current = 1
sumA = 0
pf = function(x) {
if (x < sumA) {
return(data[current,1])
}
for (i in current: length(data[,1])) {
sumA <<- sumA + data[i,2]
if (x < sumA) {
current <<- i + 1
return(data[i,1])
}
}
return("OUT OF BOUNDS")
}
cumSum = colSums(data)[[2]]
print(cumSum - 1);
h = Vectorize(pf)
plot(h, 1, cumSum-1, n=cumSum-1, lwd=0.001, xlim=c(0,cumSum-1))
As mentioned in my comment, plot command with type flag set to s should do the trick.
E.g., for you first 10 samples:
x <- c(0,664,63,404,544,651,686,507,1155,664,271)
xC <- cumsum(x)
y <- c(0,1,0,1,0,1,0,1,0,1,0)
plot(xC,y,type='s')

How do I make big.mark apply to more than just the first column?

Here's what I'm getting:
> panderOptions('big.mark', ',')
> foo <- rbind(cancer, cancer); for(i in 1:8) foo <- rbind(foo, foo)
> pander(table(foo$ph.karno, foo$pat.karno))
-----------------------------------------------------
30 40 50 60 70 80 90 100
--------- ----- ---- ---- ---- ---- ----- ----- -----
**50** 0 512 512 512 512 512 0 0
**60** 0 512 512 2560 4608 1536 0 0
**70** 1,024 0 1024 4608 3072 3072 1536 1536
**80** 0 0 0 6144 6656 6656 10240 4608
**90** 0 0 0 1536 5120 10752 12288 7680
**100** 0 0 0 0 512 3584 6656 4096
-----------------------------------------------------
I would like the comma delimiter to show up in the other columns too. How do I do that?
Best result I got (with t being your table call result):
> pander(format(t,big.mark=','))
----------------------------------------------------------
30 40 50 60 70 80 90 100
--------- ----- ---- ----- ----- ----- ------ ------ -----
**50** 0 512 512 512 512 512 0 0
**60** 0 512 512 2,560 4,608 1,536 0 0
**70** 1,024 0 1,024 4,608 3,072 3,072 1,536 1,536
**80** 0 0 0 6,144 6,656 6,656 10,240 4,608
**90** 0 0 0 1,536 5,120 10,752 12,288 7,680
**100** 0 0 0 0 512 3,584 6,656 4,096
----------------------------------------------------------
I assume this is a bug in pander.table.return but I did not dig enough to get the root cause.
Edit: I found the reason, line 283 of the function there's a for loop calling sapply on each col, but once the first column has been processed, the full array of the table is converted to char as we bring in chars from the output of format.
Then the subsequent calls to format can't format the number as they've been coerced to char.
Code from pander.table.return giving this behavior:
for (j in 1:ncol(t)) {
temp.t[, j] <- sapply(temp.t[, j], format, trim = TRUE,
digits = digits[j], decimal.mark = decimal.mark,
big.mark = big.mark)
}

Efficient (repeat) looping

I am trying to evaluate if a price, price(k), in a given row,(k), is equal to the one above, price(k-1). If it is I want to sum the volume from the prior and the price in question, volume(k)+volume(k+1), and then remove the row with the duplicate price, row k.
I have the following repeat loop which I am applying to a large dataset looking to delete repeated values.
k <- 1
repeat{
if( Prices$Price[ k + 1 ] == Prices$Price[ k ] ){
Prices$CumVolume[ k + 1 ] <- Prices$CumVolume[ k + 1 ] + Prices$CumVolume[ k ]
Prices <- Prices[ -k , ]
k <- k + 1
if( k > nrow( Prices ) ) break
}
}
The loop is very slow and I was wondering if there are ways to speed it up. Unfortunately I am relatively new to R and am having difficulty working out the best way to go about this.
Also is there a way in R to observe the iteration the loop is currently up too? i.e. have it displayed in the workspace on each iteration?
Example data:
Date Time Price CumVolume Ret MeanRet VolRet
26 01-JAN-2009 21:30:01.783 96.660 537 0 0 0
31 01-JAN-2009 21:30:58.041 96.650 78 0 0 0
33 01-JAN-2009 21:34:09.589 96.640 60 0 0 0
35 01-JAN-2009 21:34:10.879 96.640 40 0 0 0
37 01-JAN-2009 21:35:55.001 96.635 50 0 0 0
It appears you want something like this:
DF <- read.table(text=" Date Time Price CumVolume Ret MeanRet VolRet
26 01-JAN-2009 21:30:01.783 96.660 537 0 0 0
31 01-JAN-2009 21:30:58.041 96.650 78 0 0 0
33 01-JAN-2009 21:34:09.589 96.640 60 0 0 0
35 01-JAN-2009 21:34:10.879 96.640 40 0 0 0
37 01-JAN-2009 21:35:55.001 96.635 50 0 0 0", header=TRUE)
#create a run id
DF$runs <- cumsum(c(TRUE, diff(DF$Price) != 0))
#sum per each price run
DF$CCVolume <- with(DF, ave(CumVolume, runs, FUN=sum))
#remove duplicated prices
DF[!duplicated(DF$Price), ]
# Date Time Price CumVolume Ret MeanRet VolRet runs CCVolume
#26 01-JAN-2009 21:30:01.783 96.660 537 0 0 0 1 537
#31 01-JAN-2009 21:30:58.041 96.650 78 0 0 0 2 78
#33 01-JAN-2009 21:34:09.589 96.640 60 0 0 0 3 100
#37 01-JAN-2009 21:35:55.001 96.635 50 0 0 0 4 50
I think your code is going in infinite loop because of your increment index.K=k+1 and Break is always within the condition,I hope you want this
k=1
z=unique(Prices$Price)
for(i in 1:length(z))
{
dupindex=which(z[i]==Prices$Price)
Prices$CumVolume[tail(dupindex,n=1)]=sum(Prices$CumVolume[dupindex])
Prices=Prices[-(dupindex[1:length(dupindex)-1]),]
}
I hope it help,thanks.

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