Updating a vector using a while loop in R - r

I got the following dataframe called nodes_df:
x y node_demand
1 2 62 3
2 80 25 14
3 36 88 1
4 57 23 14
5 33 17 19
6 76 43 2
7 77 85 14
8 94 6 6
10 59 72 6
. . . .
. . . .
. . . .
. . . .
45 60 84 8
46 35 100 5
47 38 2 1
48 9 9 7
50 1 58 2
I have to split this dataframe between hubs and clients.
hubs <- nodes_df[keep <- sample(1:total_nodes, requested_hubs, replace = FALSE),]
client_nodes <- nodes_df[-keep, ]
I need to randomly select 1 row at a time from clients_nodes and calculate the total node_demand, I need to keep adding rows until random_clients$node_demand exceedes 120.
random_clients <- client_nodes[sample(nrow(client_nodes), size = 1, replace = FALSE),]
I created the following variables and while loop
node_demand <- c(0)
cumulative_demand <- cumsum(node_demand)
client_nodes <- nodes_df[-keep, ]
last_node <- cumsum(cumulative_demand) >= max_supply_capacity
condition = TRUE
while(condition){
random_clients <- client_nodes[sample(nrow(client_nodes), size = 1, replace = FALSE),]
node_demand <- c(node_demand,random_clients$node_demand)
cumulative_demand <- cumsum(node_demand)
if(cumulative_demand <= max_supply_capacity){
condition == FALSE
}
}
The loop doesn't stop and I get the following return value:
cumulative_demand
[1] 0 14 20 26 27 35 49 50 68 79 97 100 101 104 109 118
[17] 119 137 150 164 178 185 188 191 208 209 219 222 227 246 252 272 (it carries on and on)
I am not sure why the loop doesn't stop despite the condition cumulative_demand <= max_supply_capacity being met.
Anybody could show me how to fix it?
I managed to fix it :).
I had to use ifelse() so R could evaluate the condition of a vector. The normal if() statement wouldn't work in this case
while(TRUE){
random_clients <- client_nodes[sample(nrow(client_nodes), size = 1, replace = FALSE),]
node_demand <- c(node_demand,random_clients$node_demand)
cumulative_demand <- cumsum(node_demand)
last_node <- (cumulative_demand <= max_supply_capacity)
ifelse(last_node == FALSE,break,next)
}

I had to use and ifelse() instead of an if() statement as shown in the problem description.

Related

Splitting a matrix into multiple matrices

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)

Function with a for loop to create a column with values 1:n conditioned by intervals matched by another column

I have a data frame like the following
my_df=data.frame(x=runif(100, min = 0,max = 60),
y=runif(100, min = 0,max = 60)) #x and y in cm
With this I need a new column with values from 1 to 36 that match x and y every 10 cm. For example, if 0<=x<=10 & 0<=y<=10, put 1, then if 10<=x<=20 & 0<=y<=10, put 2 and so on up to 6, then 0<=x<=10 & 10<=y<=20 starting with 7 up to 12, etc. I tried to make a function with an if repeating the interval for x 6 times, and increasing by 10 the interval for y every iteration. Here is the function
#my miscarried function 'zones'
>zones= function(x,y) {
i=vector(length = 6)
n=vector(length = 6)
z=vector(length = 36)
i[1]=0
z[1]=0
n[1]=1
for (t in 1:6) {
if (0<=x & x<10 & i[t]<=y & y<i[t]+10) { z[t] = n[t]} else
if (10<=x & x<20 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+1} else
if (20<=x & x<30 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+2} else
if (30<=x & x<40 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+3} else
if (40<=x & x<50 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+4}else
if (50<=x & x<=60 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+5}
else {i[t+1]=i[t]+10
n[t+1]=n[t]+6}
}
return(z)
}
>xy$z=zones(x=xy$x,y=xy$y)
and I got
There were 31 warnings (use warnings() to see them)
>xy$z
[1] 0 0 0 0 25 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Please,help me before I die alone!
I think think this does the trick.
a <- cut(my_df$x, (0:6) * 10)
b <- cut(my_df$y, (0:6) * 10)
z <- interaction(a, b)
levels(z)
[1] "(0,10].(0,10]" "(10,20].(0,10]" "(20,30].(0,10]" "(30,40].(0,10]"
[5] "(40,50].(0,10]" "(50,60].(0,10]" "(0,10].(10,20]" "(10,20].(10,20]"
[9] "(20,30].(10,20]" "(30,40].(10,20]" "(40,50].(10,20]" "(50,60].(10,20]"
[13] "(0,10].(20,30]" "(10,20].(20,30]" "(20,30].(20,30]" "(30,40].(20,30]"
[17] "(40,50].(20,30]" "(50,60].(20,30]" "(0,10].(30,40]" "(10,20].(30,40]"
[21] "(20,30].(30,40]" "(30,40].(30,40]" "(40,50].(30,40]" "(50,60].(30,40]"
[25] "(0,10].(40,50]" "(10,20].(40,50]" "(20,30].(40,50]" "(30,40].(40,50]"
[29] "(40,50].(40,50]" "(50,60].(40,50]" "(0,10].(50,60]" "(10,20].(50,60]"
[33] "(20,30].(50,60]" "(30,40].(50,60]" "(40,50].(50,60]" "(50,60].(50,60]"
If this types of levels aren't for your taste, then change as below:
levels(z) <- 1:36
Is this what you're after? The resulting numbers are in column res:
# Get bin index for x values and y values
my_df$bin1 <- as.numeric(cut(my_df$x, breaks = seq(0, max(my_df$x) + 10, by = 10)));
my_df$bin2 <- as.numeric(cut(my_df$y, breaks = seq(0, max(my_df$x) + 10, by = 10)));
# Multiply bin indices
my_df$res <- my_df$bin1 * my_df$bin2;
> head(my_df)
x y bin1 bin2 res
1 49.887499 47.302849 5 5 25
2 43.169773 50.931357 5 6 30
3 10.626466 43.673533 2 5 10
4 43.401454 3.397009 5 1 5
5 7.080386 22.870539 1 3 3
6 39.094724 24.672907 4 3 12
I've broken down the steps for illustration purposes; you probably don't want to keep the intermediate columns bin1 and bin2.
We probably need a table showing the relationship between x, y, and z. After that, we can define a function to do the join.
The solution is related and inspired by this post (R dplyr join by range or virtual column). You may also find other solutions are useful.
# Set seed for reproducibility
set.seed(1)
# Create example data frame
my_df <- data.frame(x=runif(100, min = 0,max = 60),
y=runif(100, min = 0,max = 60))
# Load the dplyr package
library(dplyr)
# Create a table to show the relationship between x, y, and z
r <- expand.grid(x_from = seq(0, 50, 10), y_from = seq(0, 50, 10)) %>%
mutate(x_to = x_from + 10, y_to = y_from + 10, z = 1:n())
# Define a function for dynamic join
dynamic_join <- function(d, r){
if (!("z" %in% colnames(d))){
d[["z"]] <- NA_integer_
}
d <- d %>%
mutate(z = ifelse(x >= r$x_from & x < r$x_to & y >= r$y_from & y < r$y_to,
r$z, z))
return(d)
}
re_dynamic_join <- function(d, r){
r_list <- split(r, r$z)
for (i in 1:length(r_list)){
d <- dynamic_join(d, r_list[[i]])
}
return(d)
}
# Apply the function
re_dynamic_join(my_df, r)
x y z
1 15.930520 39.2834357 20
2 22.327434 21.1918363 15
3 34.371202 16.2156088 10
4 54.492467 59.5610437 36
5 12.100916 38.0095959 20
6 53.903381 12.7924881 12
7 56.680516 7.7623409 6
8 39.647868 28.6870821 16
9 37.746843 55.4444682 34
10 3.707176 35.9256580 19
11 12.358474 58.5702417 32
12 10.593405 43.9075507 26
13 41.221371 21.4036147 17
14 23.046223 25.8884214 15
15 46.190485 8.8926936 5
16 29.861955 0.7846545 3
17 43.057110 42.9339640 29
18 59.514366 6.1910541 6
19 22.802111 26.7770609 15
20 46.646713 38.4060627 23
21 56.082314 59.5103172 36
22 12.728551 29.7356147 14
23 39.100426 29.0609715 16
24 7.533306 10.4065401 7
25 16.033240 45.2892567 26
26 23.166846 27.2337294 15
27 0.803420 30.6701870 19
28 22.943277 12.4527068 9
29 52.181451 13.7194886 12
30 20.420940 35.7427198 21
31 28.924807 34.4923319 21
32 35.973950 4.6238628 4
33 29.612478 2.1324348 3
34 11.173056 38.5677295 20
35 49.642399 55.7169120 35
36 40.108004 35.8855453 23
37 47.654392 33.6540449 23
38 6.476618 31.5616634 19
39 43.422657 59.1057134 35
40 24.676466 30.4585093 21
41 49.256778 40.9672847 29
42 38.823612 36.0924731 22
43 46.975966 14.3321207 11
44 33.182179 15.4899556 10
45 31.783175 43.7585774 28
46 47.361374 27.1542499 17
47 1.399872 10.5076061 7
48 28.633804 44.8018962 27
49 43.938824 6.2992584 5
50 41.563893 51.8726969 35
51 28.657177 36.8786983 21
52 51.672569 33.4295723 24
53 26.285826 19.7266391 9
54 14.687837 27.1878867 14
55 4.240743 30.0264584 19
56 5.967970 10.8519817 7
57 18.976302 31.7778362 20
58 31.118056 4.5165447 4
59 39.720305 16.6653560 10
60 24.409811 12.7619712 9
61 54.772555 17.0874289 12
62 17.616202 53.7056462 32
63 27.543944 26.7741194 15
64 19.943680 46.7990934 26
65 39.052228 52.8371421 34
66 15.481007 24.7874526 14
67 28.712715 3.8285088 3
68 45.978640 20.1292495 17
69 5.054815 43.4235568 25
70 52.519280 20.2569200 18
71 20.344376 37.8248473 21
72 50.366421 50.4368732 36
73 20.801009 51.3678999 33
74 20.026496 23.4815569 15
75 28.581075 22.8296331 15
76 53.531900 53.7267256 36
77 51.860368 38.6589458 24
78 23.399373 44.4647189 27
79 46.639242 36.3182068 23
80 57.637080 54.1848967 36
81 26.079569 17.6238093 9
82 42.750881 11.4756066 11
83 23.999662 53.1870566 33
84 19.521129 30.2003691 20
85 45.425229 52.6234526 35
86 12.161535 11.3516173 8
87 42.667273 45.4861831 29
88 7.301515 43.4699336 25
89 14.729311 56.6234891 32
90 8.598263 32.8587952 19
91 14.377765 42.7046321 26
92 3.536063 23.3343060 13
93 38.537296 6.0523876 4
94 52.576153 55.6381253 36
95 46.734881 16.9939500 11
96 47.838530 35.4343895 23
97 27.316467 6.6216363 3
98 24.605045 50.4304219 33
99 48.652215 19.0778211 11
100 36.295997 46.9710802 28

Flip Every Nth Coin in R [duplicate]

This question already has answers here:
R: How to use ifelse statement for a vector of characters
(2 answers)
Closed 6 years ago.
My friend gave me a brain teaser that I wanted to try on R.
Imagine 100 coins in a row, with heads facing up for all coins. Now every 2nd coin is flipped (thus becoming tails). Then every 3rd coin is flipped. How many coins are now showing heads?
To create the vector, I started with:
flips <- rep('h', 100)
levels(flips) <- c("h", "t")
Not sure how to proceed from here. Any help would be appreciated.
Try this:
coins <- rep(1, 100) # 1 = Head, 0 = Tail
n = 3 # run till the time when you flip every 3rd coin
invisible(sapply(2:n function(i) {indices <- seq(i, 100, i); coins[indices] <<- (coins[indices] + 1) %% 2}) )
which(coins == 1)
# [1] 1 5 6 7 11 12 13 17 18 19 23 24 25 29 30 31 35 36 37 41 42 43 47 48 49 53 54 55 59 60 61 65 66 67 71 72 73 77 78 79 83 84 85 89 90 91 95 96 97
sum(coins==1)
#[1] 49
If you run till n = 100, only the coins at the positions which are perfect squares will be showing heads.
coins <- rep(1, 100) # 1 = Head, 0 = Tail
n <- 100
invisible(sapply(2:n, function(i) {indices <- seq(i, 100, i); coins[indices] <<- (coins[indices] + 1) %% 2}) )
which(coins == 1)
# [1] 1 4 9 16 25 36 49 64 81 100
sum(coins==1)
# [1] 10

Filter rows based on values of multiple columns in R

Here is the data set, say name is DS.
Abc Def Ghi
1 41 190 67
2 36 118 72
3 12 149 74
4 18 313 62
5 NA NA 56
6 28 NA 66
7 23 299 65
8 19 99 59
9 8 19 61
10 NA 194 69
How to get a new dataset DSS where value of column Abc is greater than 25, and value of column Def is greater than 100.It should also ignore any row if value of atleast one column in NA.
I have tried few options but wasn't successful. Your help is appreciated.
There are multiple ways of doing it. I have given 5 methods, and the first 4 methods are faster than the subset function.
R Code:
# Method 1:
DS_Filtered <- na.omit(DS[(DS$Abc > 20 & DS$Def > 100), ])
# Method 2: which function also ignores NA
DS_Filtered <- DS[ which( DS$Abc > 20 & DS$Def > 100) , ]
# Method 3:
DS_Filtered <- na.omit(DS[(DS$Abc > 20) & (DS$Def >100), ])
# Method 4: using dplyr package
DS_Filtered <- filter(DS, DS$Abc > 20, DS$Def >100)
DS_Filtered <- DS %>% filter(DS$Abc > 20 & DS$Def >100)
# Method 5: Subset function by default ignores NA
DS_Filtered <- subset(DS, DS$Abc >20 & DS$Def > 100)

Custom sorting of a dataframe in R

I have a binomail dataset that looks like this:
df <- data.frame(replicate(4,sample(1:200,1000,rep=TRUE)))
addme <- data.frame(replicate(1,sample(0:1,1000,rep=TRUE)))
df <- cbind(df,addme)
df <-df[order(df$replicate.1..sample.0.1..1000..rep...TRUE..),]
The data is currently soreted in a way to show the instances belonging to 0 group then the ones belonging to the 1 group. Is there a way I can sort the data in a 0-1-0-1-0... fashion? I mean to show a row that belongs to the 0 group, the row after belonging to the 1 group then the zero group and so on...
All I can think about is complex functions. I hope there's a simple way around it.
Thank you,
Here's an attempt, which will add any extra 1's at the end:
First make some example data:
set.seed(2)
df <- data.frame(replicate(4,sample(1:200,10,rep=TRUE)),
addme=sample(0:1,10,rep=TRUE))
Then order:
with(df, df[unique(as.vector(rbind(which(addme==0),which(addme==1)))),])
# X1 X2 X3 X4 addme
#2 141 48 78 33 0
#1 37 111 133 3 1
#3 115 153 168 163 0
#5 189 82 70 103 1
#4 34 37 31 174 0
#6 189 171 98 126 1
#8 167 46 72 57 0
#7 26 196 30 169 1
#9 94 89 193 134 1
#10 110 15 27 31 1
#Warning message:
#In rbind(which(addme == 0), which(addme == 1)) :
# number of columns of result is not a multiple of vector length (arg 1)
Here's another way using dplyr, which would make it suitable for within-group ordering. It's also probably pretty quick. If there's unbalanced numbers of 0's and 1's, it will leave them at the end.
library(dplyr)
df %>%
arrange(addme) %>%
mutate(n0 = sum(addme == 0),
orderme = seq_along(addme) - (n0 * addme) + (0.5 * addme)) %>%
arrange(orderme) %>%
select(-n0, -orderme)

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