I have a data.frame P1 (5000rows x 4cols) and would like to save the subset of data in columns 2,3 and 4 when the time-stamp in column 1 falls into a set range determined by a vector TimeStamp (in seconds).
E.g. put all values in columns 2, 3, and 4 into a new data.frame and call each section of data: Condition.1.P1, Condition.2.P1, etc.
The reason I'd like to label separately as I have 35 versions of P1 (P2, P3, P33, etc) and need to be able to melt them together to plot them.
dput(TimeStamp)
c(18, 138, 438, 678, 798, 1278, 1578, 1878, 2178)
dput(head(P1))
structure(list(Time = c(0, 5, 100, 200, 500, 1200), SkinTemp = c(27.781,
27.78, 27.779, 27.779, 27.778, 27.777), HeartRate = c(70, 70,
70, 70, 70, 70), RespirationRate = c(10, 10, 10, 10, 10, 10)), .Names = c("Time",
"SkinTemp", "HeartRate", "RespirationRate"), row.names = c(NA,
6L), class = "data.frame")
Do you want to seperate the data by the timestamp range and put it in a list? Than this might be what you are looking for:
TimeStamp <- c(18, 138, 438, 678, 798, 1278, 1578, 1878, 2178)
dat <- structure(list(Time = c(0, 5, 100, 200, 500, 1200), SkinTemp =(27.781,
27.78, 27.779, 27.779, 27.778, 27.777), HeartRate = c(70, 70,
70, 70, 70, 70), RespirationRate = c(10, 10, 10, 10, 10, 10)), .Names = c ("Time",
"SkinTemp", "HeartRate", "RespirationRate"), row.names = c(NA,
6L), class = "data.frame")
dat$Segment <- cut(dat$Time,c(-Inf,TimeStamp))
split(dat,dat$Segment)
P2 = data.frame(NA, NA, NA, NA) # Create empty data.frame
for (i in 1:length(ts)){
P3 = data.frame() # Create empty changing data.frame
if (i == 1) {ts1 = 0} else {ts1 = ts[i-1]} #First time stamp starts at 0
ts2 = ts[i]
P3 = subset(P1, P1$Time > ts1 & P1$Time < ts2)[,c(2,3,4)] #Subset the columns and assign to P3
if (nrow(P3) == 0){P3 = data.frame(NA, NA, NA)} #If the subset is empty, assign NA
P3$TimeStamp = paste(ts1,ts2,sep="-") # Append TimeStamp to the P3
colnames(P3) = colnames(P2) #Make sure column names are same to allow rbind
P2 = rbind(P2,P3) #Append P3 to P2
}
P2 = P2[c(2:nrow(P2)),] #Remove the first row (that has NA)
colnames(P2) = c("SkinTemp", "HeartRate", "RespirationRate", "TimeStamp") #Provide column names)
rm(P3); rm(i); rm(ts1); rm(ts2) #Cleanup
Related
Using a dataframe with missing values:
structure(list(id = c("id1", "test", "rew", "ewt"), total_frq_1 = c(54, 87, 10, 36), total_frq_2 = c(45, 24, 202, 43), total_frq_3 = c(24, NA, 25, 8), total_frq_4 = c(36, NA, 104, NA)), row.names = c(NA, 4L), class = "data.frame")
How is is possible to create a bar plot with the mean for every column, excluding the id column, but without filling the missing values with 0 but leaving out the row with missing values example for total_frq_3 24+25+8 = 57/3 = 19
You can use colMeans function and pass it the appropriate argument to ignore NA.
library(ggplot2)
xy <- structure(list(id = c("id1", "test", "rew", "ewt"),
total_frq_1 = c(54, 87, 10, 36), total_frq_2 = c(45, 24, 202, 43), total_frq_3 = c(24, NA, 25, 8),
total_frq_4 = c(36, NA, 104, NA)),
row.names = c(NA, 4L),
class = "data.frame")
xy.means <- colMeans(x = xy[, 2:ncol(xy)], na.rm = TRUE)
xy.means <- as.data.frame(xy.means)
xy.means$total <- rownames(xy.means)
ggplot(xy.means, aes(x = total, y = xy.means)) +
theme_bw() +
geom_col()
Or just use base image graphic
barplot(height = colMeans(x = xy[, 2:ncol(xy)], na.rm = TRUE))
I came up with the idea to represent stats on a chart like this. Example of the plot. And made it like this.
df_n <- df_normalized %>%
transmute(
Height_x = round(Height*cos_my(45), 2),
Height_y = round(Height*sin_my(45), 2),
Weight_x = round(Weight*cos_my(45*2), 2),
Weight_y = round(Weight*sin_my(45*2), 2),
Reach_x = round(Reach*cos_my(45*3), 2),
Reach_y = round(Reach*sin_my(45*3), 2),
SLpM_x = round(SLpM*cos_my(45*4), 2),
SLpM_y = round(SLpM*sin_my(45*4), 2),
Str_Def_x = round(`Str_Def %`*cos_my(45*5), 2),
Str_Def_y = round(`Str_Def %`*sin_my(45*5), 2),
TD_Avg_x = round(TD_Avg*cos_my(45*6), 2),
TD_Avg_y = round(TD_Avg*sin_my(45*6), 2),
TD_Acc_x = round(`TD_Acc %`*cos_my(45*7), 2),
TD_Acc_y = round(`TD_Acc %`*sin_my(45*7), 2),
Sub_Avg_x = round(Sub_Avg*cos_my(45*8), 2),
Sub_Avg_y = round(Sub_Avg*sin_my(45*8), 2))
Now I want to do this smart way, so I created a data frame with same number of rows empty_df, and later in for loop I try to mutate and array, with every iteration. So for example I want to multiply 1st column by cos(30), 2nd by cos(30*2), and so on
But...
It mutate only last column because all columns during iteration have the same name 'column'.
I want to name each column by the variable column, made with paste0().
reprex_df <- structure(list(Height = c(190, 180, 183, 196, 185),
Weight = c(120, 77, 93, 120, 84),
Reach = c(193, 180, 188, 203, 193),
SLpM = c(2.45, 3.8, 2.05, 7.09, 3.17),
`Str_Def %` = c(58, 56, 55, 34, 44),
TD_Avg = c(1.23, 0.33, 0.64, 0.91, 0),
`TD_Acc %` = c(24, 50, 20, 66, 0),
Sub_Avg = c(0.2, 0, 0, 0, 0)), row.names = c(NA, -5L),
class = c("tbl_df", "tbl", "data.frame"))
temp <- apply(reprex_df[,1], function(x) x*cos(60), MARGIN = 2)
temp
empty_df <- data.frame(first_column = replicate(length(temp),1))
for (x in 1:8) {
temp <- apply(df[,x], function(x) round(x*cos((360/8)*x),2), MARGIN = 2)
column <- paste0("Column_",x)
empty_df <- mutate(empty_df, column = temp)
}
Later I want to make it a function where I can pass data frame and receive data frame with X, and Y coordinates.
So, how should I make it?
Perhaps this helps
library(purrr)
library(stringr)
nm1 <- names(reprex_df)
nm_cos <- str_c(names(reprex_df), "_x")
nm_sin <- str_c(names(reprex_df), "_y")
reprex_df[nm_cos] <- map2(reprex_df, seq_along(nm1),
~ round(.x * cos(45 *.y ), 2))
reprex_df[nm_sin] <- map2(reprex_df[nm1], seq_along(nm1),
~ round(.x * sin(45 *.y ), 2))
I'm trying to conditionally replace values in multiple columns based on a string match in a different column but I'd like to be able to do so in a single line of code using the across() function but I keep getting errors that don't quite make sense to me. I feel like this is probably a simple solution so if anyone could point me in the right direction, that would be fantastic!
df <- data.frame("type" = c("Park", "Neighborhood", "Airport", "Park", "Neighborhood", "Neighborhood"),
"total" = c(34, 56, 75, 89, 21, 56),
"group_a" = c(30, 26, 45, 60, 3, 46),
"group_b" = c(4, 30, 30, 29, 18, 10))
# working but not concise
df %>%
mutate(total = ifelse(str_detect(type, "Park"), NA, total),
group_a = ifelse(str_detect(type, "Park"), NA, group_a),
group_b = ifelse(str_detect(type, "Park"), NA, group_b))
# concise but not working
df %>% mutate(across(total, group_a, group_b), ifelse(str_detect(type, "Park"), NA, .))
Update
We got a solution that works with my dummy dataset but is not working with my real data, so I am going to share a small snippet of my real data frame with the numbers changed and organization names hidden. When I run this line of code (df %>% mutate(across(c(Attempts, Canvasses, Completes)), ~ifelse(str_detect(long_name, "park-cemetery"), NA, .))) on these data, I get the following error message:
Error: Problem with mutate() input ..2. x Input ..2 must be a
vector, not a formula object. i Input ..2 is
~ifelse(str_detect(long_name, "park-cemetery"), NA, .).
This a small sample of the data that produces this error:
df <- structure(list(Org = c("OrgName", "OrgName", "OrgName", "OrgName",
"OrgName", "OrgName", "OrgName", "OrgName", "OrgName", "OrgName"
), nCode = c("M34", "R36", "R46", "X29", "M31", "K39", "Q12",
"Q39", "X41", "K27"), Attempts = c(100, 100, 100, 100, 100, 100,
100, 100, 100, 100), Canvasses = c(80, 80, 80, 80, 80, 80, 80,
80, 80, 80), Completes = c(50, 50, 50, 50, 50, 50, 50, 50, 50,
50), van_nocc_id = c(999, 999, 999, 999, 999, 999, 999, 999,
999, 999), van_name = c("M-Upper West Side", "SI-Rosebank", "SI-Tottenville",
"BX-park-cemetery-etc-Bronx", "M-Stuyvesant Town-Cooper Village",
"BK-Kensington", "Q-Broad Channel", "Q-Lindenwood", "BX-Wakefield",
"BK-East New York"), boro_short = c("M", "SI", "SI", "BX", "M",
"BK", "Q", "Q", "BX", "BK"), long_name = c("Upper West Side",
"Rosebank", "Tottenville", "park-cemetery-etc-Bronx", "Stuyvesant Town-Cooper Village",
"Kensington", "Broad Channel", "Lindenwood", "Wakefield", "East New York"
)), row.names = c(NA, -10L), class = "data.frame")
Final update
The curse of the misplaced closing bracket! Thanks to everyone for your help... the correct solution was df %>% mutate(across(c(Attempts, Canvasses, Completes), ~ifelse(str_detect(long_name, "park-cemetery"), NA, .)))
If you use the newly introduced function across (which is the correct way to approach this task), you have to specify inside across itself the function you want to apply. In this case the function ifelse(...) has to be a purrr-style lambda (so starting with ~). Check out across documentation and look for the arguments .cols and .fns.
df %>%
mutate(across(c(total, group_a, group_b), ~ifelse(str_detect(type, "Park"), NA, .)))
Output
# type total group_a group_b
# 1 Park NA NA NA
# 2 Neighborhood 56 26 30
# 3 Airport 75 45 30
# 4 Park NA NA NA
# 5 Neighborhood 21 3 18
# 6 Neighborhood 56 46 10
Here a data.table solution.
require(data.table)
df <- data.frame("type" = c("Park", "Neighborhood", "Airport", "Park", "Neighborhood", "Neighborhood"),
"total" = c(34, 56, 75, 89, 21, 56),
"group_a" = c(30, 26, 45, 60, 3, 46),
"group_b" = c(4, 30, 30, 29, 18, 10))
setDT(df)
df[type == "Park", c("total", "group_a", "group_b") := NA]
Update: that didn't take long to figure out! Just needed to place the columns in a vector:
# concise AND working!
df %>% mutate(across(c(total, group_a, group_b)), ifelse(str_detect(type, "Park"), NA, .))
I had tried this initially but placed the columns in quotes... don't do that :)
I have the following code for creating a boxplot in ggplot2:
throughput <- c(1, 2, 3, 4, 5)
response_time_min <- c(9, 19, 29, 39, 49)
response_time_10 <- c(50, 55, 60, 60, 61)
response_time_med <- c(100, 100, 100, 100, 100)
response_time_90 <- c(201, 201, 250, 200, 230)
response_time_max <- c(401, 414, 309, 402, 311)
df <- data.frame(throughput, response_time_min, response_time_10, response_time_med,response_time_90, response_time_max)
df
library(ggplot2)
g <- ggplot(df) +
geom_boxplot(aes(x=factor(throughput),ymax = response_time_max,upper = response_time_90,
y = response_time_med,
middle = response_time_med,
lower = response_time_10,
ymin = response_time_min), stat = "identity")
g
But now when I want to apply ggplotly(g) the graph does not render correctly. What can I do to make this work?
I don't think 90th percentile and 10th percentile can be done. Assuming they are q3 and q1, respectively, the code below
bp <- plot_ly(color=c("orange")) %>%
add_trace(lowerfence = response_time_min, q1 = response_time_10,
median = response_time_med, q3 = response_time_90,
upperfence = response_time_max, type = "box") %>%
layout(xaxis=list(title="throughput"),
yaxis=list(title="response_time"))
bp
gives the following output:
I have a dataframe df
dput(df)
structure(list(x = c(49, 50, 51, 52, 53, 54, 55, 56, 1, 2, 3,
4, 5, 14, 15, 16, 17, 2, 3, 4, 5, 6, 10, 11, 3, 30, 64, 66, 67,
68, 69, 34, 35, 37, 39, 2, 17, 18, 99, 100, 102, 103, 67, 70,
72), y = c(2268.14043972082, 2147.62290922552, 2269.1387550775,
2247.31983098201, 1903.39138268307, 2174.78291538358, 2359.51909126411,
2488.39004804939, 212.851575751527, 461.398994384333, 567.150629704352,
781.775113821961, 918.303706148872, 1107.37695799186, 1160.80594193377,
1412.61328924168, 1689.48879626486, 260.737164468854, 306.72700499362,
283.410379620422, 366.813913489692, 387.570173754128, 388.602676983443,
477.858510450125, 128.198042456082, 535.519377609133, 1028.8780498564,
1098.54431357711, 1265.26965941035, 1129.58344809909, 820.922447928053,
749.343583476846, 779.678206156474, 646.575242339517, 733.953282899613,
461.156280127354, 906.813018662913, 798.186995701282, 831.365377249207,
764.519073183124, 672.076289062505, 669.879217186302, 1341.47673353751,
1401.44881976186, 1640.27575962036)), .Names = c("x", "y"), row.names = c(NA,
-45L), class = "data.frame")
I have created two non-linear regression (nls1 and nls2) based on my dataset.
library(stats)
nls1 <- nls(y~A*(x^B)*(exp(k*x)),
data = df,
start = list(A = 1000, B = 0.170, k = -0.00295))
nls2<-nls(y~A*x^3+B*x^2+C*x+D, data=df,
start = list(A=0.02, B=-0.6, C= 50, D=200))
I then computed bootstrap objects for these two functions to get multiple sets of parameters (A,B and k for nls1 and A, B, C and D for nls2).
library(nlstools)
Boo1 <- nlsBoot(nls1, niter = 200)
Boo2 <- nlsBoot(nls2, niter = 200)
Based on this bootstrap objects, I would like to compute r-squared of each combination of parameters and save the min, max and median of my r-squared values for each bootstrap object into one new dataframe. The dataframe could look like new.df.
structure(list(Median = c(NA, NA), Max = c(NA, NA), Min = c(NA,
NA)), .Names = c("Median", "Max", "Min"), row.names = c("nls1",
"nls2"), class = "data.frame")
The idea is then to do some box plots with the median, min and max values for each non-linear model based on bootstrapping to compare them. Can someone help me out with that? Thanks in advance.
Answer from #bunk
stat <- function(dat, inds) { fit <- try(nls(y~A*(x^B)*(exp(k*x)), data = dat[inds,], start = list(A = 1000, B = 0.170, k = -0.00295)), silent=TRUE); f1 <- if (inherits(fit, "nls")) AIC(fit) else NA; fit2 <- try(nls(y~A*x^3+B*x^2+C*x+D, data = dat[inds,], start = list(A=0.02, B=-0.6, C= 50, D=200)), silent=TRUE); f2 <- if (inherits(fit2, "nls")) AIC(fit2) else NA; c(f1, f2) }; res <- boot(df, stat, R=200). Then, to get medians for example, apply(res$t, 2, median, na.rm=TRUE)