I have the code below which plots two time-series. I'd like to add a vertical line every say 10 units on the x-axis to basically divide the plot up into like 5 squares. Any tips are very much appreciated.
Code:
## Plot Forecast & Actual
ts.plot(ts(CompareDf$stuff1),ts(CompareDf$stuff2),col=1:2,xlab="Hour",ylab="Minu tes",main='testVar')
legend("topleft", legend = c("Actual","Forecast"), col = 1:2, lty = 1)
Data:
dput(CompareDf)
structure(list(stuff1 = c(6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,
32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47,
48, 49, 50, 51, 52, 53, 54, 55), stuff2 = c(8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44,
45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57)), .Names = c("stuff1",
"stuff2"), row.names = c(NA, -50L), class = "data.frame")
After plotting timeseries data, use abline to draw vertical lines.
abline(v = seq(10, 50, 10))
Related
I want to perform an operation over a vector without using a loop. The operation is the following:
This is how I am coding in R
meanx <- mean(rankx)
Numerador <- (rankx[] - meanx)*(rankx[+1] - meanx)
This is the input:
> dput(rankx)
c(15, 11, 12, 30, 58, 14, 41, 10, 57, 32, 28, 52, 61, 18, 54,
37, 19, 7, 29, 66, 5, 47, 25, 6, 50, 65, 62, 23, 40, 63, 42,
64, 38, 56, 45, 17, 8, 59, 55, 67, 24, 60, 2, 35, 44, 20, 3,
39, 4, 31, 26, 51, 21, 22, 53, 33, 46, 9, 16, 36, 13, 27, 34,
48, 1, 49, 43)
For example for the first case it will be: (15 - mean(rankx))(11 - mean(rankx))
For the next: (11 - mean(rankx))(12 - mean(rankx))
I am not sure how to refer to the second element and my error is in rankx[+1]
Any idea in how to solve this operation without using a loop?
You can use dplyr::lead
rankx[+1] is equivalent to rankx[1], which is 15.
If you want a copy of rankx that's displaced by one unit, use dplyr::lead(rankx) - like this:
rankx <- c(15, 11, 12, 30, 58, 14, 41)
dplyr::lead(rankx)
#> [1] 11 12 30 58 14 41 NA
meanx <- mean(rankx)
Numerador <- (rankx - meanx)*(dplyr::lead(rankx) - meanx)
Numerador
#> [1] 161.30612 205.87755 -57.40816 133.16327 -381.12245 -179.55102 NA
Created on 2021-04-20 by the reprex package (v1.0.0)
I have created a simple minimum spanning tree and now have a data frame with columns 'from', 'to' and 'distance'.
Based on this, I found communities using the Louvain method, which I plotted. As far as I understand it, for clustering and plotting I need only the columns from and to, and the distance is not used.
How can I keep the communities I found, ideally each in a different color, but remove the box around the communities?
library(igraph)
from <- c(14, 25, 18, 19, 29, 23, 24, 36, 5, 22, 21, 29, 18, 26, 2, 45, 8, 7, 36, 42, 3, 23, 13, 13, 20, 15, 13, 7, 28, 9, 6, 37, 8, 4, 15, 27, 10, 2, 39, 1, 43, 21, 14, 4, 14, 8, 9, 40, 31, 1)
to <- c(16, 26, 27, 20, 32, 34, 35, 39, 6, 32, 35, 30, 22, 28, 45, 46, 48, 12, 38, 43, 42, 24, 27, 25, 30, 20, 50, 29, 34, 49, 40, 39, 11, 41, 46, 47, 50, 16, 46, 40, 44, 31, 17, 40, 44, 23, 33, 42, 33, 1)
distance <- c(0.3177487, 0.3908324, 0.4804059, 0.4914682, 0.5610357, 0.6061082, 0.6357532, 0.6638961, 0.7269725, 0.8136463, 0.8605391, 0.8665838, 0.8755252, 0.8908454, 0.9411793, 0.9850834, 1.0641603, 1.0721154, 1.0790506, 1.1410964, 1.1925349, 1.2115428, 1.2165045, 1.2359032, 1.2580204, 1.2725243, 1.2843610, 1.2906908, 1.3070725, 1.3397053, 1.3598817, 1.3690732, 1.3744088, 1.3972220, 1.4472312, 1.4574936, 1.4654772, 1.4689660, 1.5999424, 1.6014316, 1.6305410, 1.6450413, 1.6929959, 1.7597620, 1.8113320, 2.0380866, 3.0789517, 4.0105981, 5.1212614, 0.0000000)
mst <- cbind.data.frame(from, to, distance)
g <- graph.data.frame(mst[, 1:2], directed = FALSE)
lou <- cluster_louvain(g)
set.seed(1)
plot(lou, g, vertex.label = NA, vertex.size=5)
The blobs around the groups can be turned off like this:
plot(lou, g, vertex.label = NA, vertex.size=5, mark.groups = NULL)
Do you want this?
plot(lou, g, vertex.label = NA, vertex.size = 5, mark.border = NA)
While there are functions for saving data as a separate CSV file (write.table) or as an R-data file (save, saveRDS), I have not found a way to store or print a data frame as R code that recreates this data frame.
Background of my question is that I want to include data with a script (instead of storing it in a separate file), and am thus looking for a way to generate the specific code provided the data frame already exists. I could hack on with sed or other external tools, but I wonder whether someone knows of a built-in method in R.
Try with "dput" like so:
dput(cars)
# Returns:
structure(list(speed = c(4, 4, 7, 7, 8, 9, 10, 10, 10, 11, 11,
12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16,
16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 20,
22, 23, 24, 24, 24, 24, 25), dist = c(2, 10, 4, 22, 16, 10, 18,
26, 34, 17, 28, 14, 20, 24, 28, 26, 34, 34, 46, 26, 36, 60, 80,
20, 26, 54, 32, 40, 32, 40, 50, 42, 56, 76, 84, 36, 46, 68, 32,
48, 52, 56, 64, 66, 54, 70, 92, 93, 120, 85)), class = "data.frame",
row.names = c(NA, -50L))
I'm having difficulty using the Tidyr function "complete()" to fill in columns for absent weeks. While the complete() function does work, it loops through the entire year 35 times and fills in 4,375 entries rather than just 125.
In short, when I try to use the complete function, it does not just complete the dataframe but duplicates all columns 35 times.
I have tried several different approaches including with and w/o the full_seq function.
1st approach:
Df %>%
group_by(week = week(`Local Start Time`)) %>%
complete(week = 1:52)
Second approach:
Df %>%
group_by(week = week(`Local Start Time`)) %>%
complete(week = full_seq(week <- c(1:52), 1L))
I expected the dataframe to stop at row 125 but instead the complete() loops over the entire yearly data (35 times!) and continues until column 4375.
Any advice is appreciated, thanks!
The data I used is here...
structure(list(`Local Start Time` = structure(c(1483846399, 1483846519,
1483851979, 1484734742, 1485017522, 1485190862, 1486236902, 1486238462,
1486347422, 1486448822, 1487221742, 1487392502, 1487449502, 1487678750,
1487679111, 1487679411, 1487683370, 1488321651, 1488745130, 1489353950,
1489710710, 1491043550, 1492036467, 1492105535, 1492150284, 1492180823,
1492772358, 1493428578, 1493440398, 1493465717, 1493476518, 1493484558,
1493495837, 1493622558, 1493639598, 1493718078, 1493718858, 1493720778,
1495021772, 1495598357, 1495599017, 1496175677, 1496428517, 1496439678,
1496494637, 1496632757, 1496887457, 1496887757, 1496888117, 1497031577,
1497207557, 1497318797, 1497368657, 1497491178, 1497558017, 1497857478,
1498220117, 1498245977, 1498246577, 1498255277, 1498257797, 1499203517,
1499470577, 1500752057, 1500899837, 1502036477, 1502392277, 1502410817,
1502428157, 1502429957, 1503492618, 1503500417, 1503507318, 1503672677,
1503674057, 1503674370, 1503675077, 1503923478, 1503928037, 1503932777,
1503943037, 1503972019, 1503989537, 1504383497, 1504421837, 1504639337,
1504656977, 1504672937, 1504682418, 1504722677, 1506766878, 1507180518,
1507184597, 1507228877, 1507229657, 1507370717, 1508326217, 1508343977,
1508357297, 1508374397, 1508492838, 1508555177, 1508560158, 1508868737,
1509231244, 1509252184, 1509845644, 1510709818), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), week = c(2, 2, 2, 3, 3, 4, 5, 5, 6,
6, 7, 7, 7, 8, 8, 8, 8, 9, 10, 11, 11, 13, 15, 15, 15, 15, 16,
17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 20, 21, 21, 22, 22,
22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 25, 25, 25, 25,
25, 25, 27, 27, 29, 30, 32, 32, 32, 32, 32, 34, 34, 34, 34, 34,
34, 34, 35, 35, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 36, 39,
40, 40, 40, 40, 40, 42, 42, 42, 42, 42, 42, 42, 43, 43, 44, 45,
46)), class = "data.frame", row.names = c(NA, -108L), .Names = c("Local Start Time",
"week"))
My friend suggested me to try to solve this problem before interview, but I have no idea on how to approach it.
I need to write a code to shuffle a deck of 52 cards without using a built-in standard random function.
Update
Thanks to Yifei Wu, his answer was very helpful.
Here is a link for my github project where I executed the given algorithm
https://github.com/Dantsj16/Shuffle-Without-Random.git
Your question does not say it must be a random shuffle of 52 cards. There is such a thing as a perfect shuffle, where a riffle shuffle is done with the top card remaining on the top and every other card comes from the other half of the deck. Many magicians and card sharks can do this shuffle as desired. It is well known that eight perfect shuffles in a row of a standard 52-card deck returns the cards to their original order, if the top card remains on top for each shuffle.
Here are 8 perfect shuffles in python Note that this shuffle is done differently than an actual manual shuffle would be done, to simplify the code.
In [1]: d0=[x for x in range(1,53)] # the card deck
In [2]: print(d0)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52]
In [3]: d1=d0[::2]+d0[1::2] # a perfect shuffle
In [4]: print(d1)
[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, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52]
In [5]: d2=d1[::2]+d1[1::2]
In [6]: d3=d2[::2]+d2[1::2]
In [7]: d4=d3[::2]+d3[1::2]
In [8]: d5=d4[::2]+d4[1::2]
In [9]: d6=d5[::2]+d5[1::2]
In [10]: d7=d6[::2]+d6[1::2]
In [11]: d8=d7[::2]+d7[1::2]
In [12]: print(d8)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52]
In [13]: print(d0 == d8)
True
If you want the perfect shuffle as done by hand, use
d1=[None]*52
d1[::2]=d0[:26]
d1[1::2]=d0[26:]
This gives, for d1,
[1, 27, 2, 28, 3, 29, 4, 30, 5, 31, 6, 32, 7, 33, 8, 34, 9, 35, 10, 36, 11, 37, 12, 38, 13, 39, 14, 40, 15, 41, 16, 42, 17, 43, 18, 44, 19, 45, 20, 46, 21, 47, 22, 48, 23, 49, 24, 50, 25, 51, 26, 52]
Let me know if you really need a random shuffle. I can adapt my Borland Delphi code into python if you need it.