I have this kind of data:
> data_example
date A B C D E F
1 2020-09-22 1.3 0.0 1.3 0.3 0.9 0.0
2 2020-09-23 0.7 0.0 0.7 0.0 0.7 0.0
3 2020-09-24 0.4 0.0 0.4 0.0 0.4 0.0
4 2020-09-25 0.2 0.2 0.5 0.0 0.2 0.0
5 2020-09-26 1.0 0.0 1.0 0.0 1.0 0.0
6 2020-09-27 0.2 0.2 0.5 0.1 0.1 0.0
7 2020-09-28 0.6 0.1 0.7 0.0 0.6 0.0
8 2020-09-29 0.4 0.1 0.5 0.1 0.2 0.0
9 2020-09-30 0.4 0.1 0.6 0.0 0.4 0.0
10 2020-10-01 1.0 0.1 1.1 0.8 0.1 0.0
11 2020-10-02 0.6 0.1 0.8 0.2 0.4 0.0
I would like to plot more than one of the columns (A, B, C...) in the same time series plot BUT without using the add_trace. The reason is I am building a Shiny app where dynamically the user can choose, using the selectize input, which variables want to plot, so to do it dynamically it's a must to not to be in an add_trace way.
Is there another way to achieve that?
Thanks.
Edit:
Output of the dput(data_example)
data_example <- structure(list(date = c("2020-09-22", "2020-09-23", "2020-09-24",
"2020-09-25", "2020-09-26", "2020-09-27", "2020-09-28", "2020-09-29",
"2020-09-30", "2020-10-01", "2020-10-02"), A = c(1.3, 0.7, 0.4,
0.2, 1, 0.2, 0.6, 0.4, 0.4, 1, 0.6), B = c(0, 0, 0, 0.2, 0, 0.2,
0.1, 0.1, 0.1, 0.1, 0.1), C = c(1.3, 0.7, 0.4, 0.5, 1, 0.5, 0.7,
0.5, 0.6, 1.1, 0.8), D = c(0.3, 0, 0, 0, 0, 0.1, 0, 0.1, 0, 0.8,
0.2), E = c(0.9, 0.7, 0.4, 0.2, 1, 0.1, 0.6, 0.2, 0.4, 0.1, 0.4
), F = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), class = "data.frame", row.names = c(NA,
-11L))
You should reshape your data.frame to long format.
I prefer library(data.table) for this - see the melt call. After that you may use split or color to generate the traces:
library(data.table)
library(plotly)
DF <- data.frame(
date = c("2020-09-22","2020-09-23","2020-09-24",
"2020-09-25","2020-09-26","2020-09-27","2020-09-28",
"2020-09-29","2020-09-30","2020-10-01","2020-10-02"),
A = c(1.3, 0.7, 0.4, 0.2, 1, 0.2, 0.6, 0.4, 0.4, 1, 0.6),
B = c(0, 0, 0, 0.2, 0, 0.2, 0.1, 0.1, 0.1, 0.1, 0.1),
C = c(1.3, 0.7, 0.4, 0.5, 1, 0.5, 0.7, 0.5, 0.6, 1.1, 0.8),
D = c(0.3, 0, 0, 0, 0, 0.1, 0, 0.1, 0, 0.8, 0.2),
E = c(0.9, 0.7, 0.4, 0.2, 1, 0.1, 0.6, 0.2, 0.4, 0.1, 0.4),
F = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
)
setDT(DF)
longDF <- melt(DF, id.vars = "date")
plot_ly(longDF, type = "scatter", mode = "lines+markers", x = ~date, y = ~value, split = ~variable)
Related
I randomly sample values from the first row of matrix A:
#1 2 3 4 5 6 7 8 9
A <- matrix(data=c( 0, 0, 0.33, 0.37, 0, 0, 0, 0.3, 0, #1
0, 0, 0, 0, 0.1, 0, 0, 0.9, 0, #2
0.2, 0, 0.1, 0, 0.4, 0, 0, 0.3, 0, #3
0.5, 0, 0,0, 0, 0, 0, 0, 0.5, #4
0, 0.4, 0,0, 0, 0.5, 0, 0, 0.1, #5
0, 0, 0,0, 1, 0, 0, 0, 0, #6
0, 0.2, 0, 0.8, 0, 0, 0, 0,0, #7
1, 0, 0, 0, 0, 0, 0, 0,0, #8
0.1, 0.1, 0.1, 0.1, 0.2, 0.1, 0.1, 0.1, 0.1 ),#9
nrow=9, ncol=9)
colnames(A) <- c(1:9)
rownames(A) <- c(1:9)
x <- sample(x=A[,1], size=2, prob=A[,1])
The result is a numeric object that looks, for example, like this:
> x
3 8
0.33 0.30
The 3 and 8 represent important information that I need to store and use for downstream calculations. I have no idea how to extract them - these row numbers seem to be stored as metadata when I View(x):
How can I reshape the results of sample() so that the numeric object that gets output is a vector containing the row numbers (i.e., in this example, I want x to equal c(3, 8))?
How about:
A <- matrix(data=c( 0, 0, 0.33, 0.37, 0, 0, 0, 0.3, 0, #1
0, 0, 0, 0, 0.1, 0, 0, 0.9, 0, #2
0.2, 0, 0.1, 0, 0.4, 0, 0, 0.3, 0, #3
0.5, 0, 0,0, 0, 0, 0, 0, 0.5, #4
0, 0.4, 0,0, 0, 0.5, 0, 0, 0.1, #5
0, 0, 0,0, 1, 0, 0, 0, 0, #6
0, 0.2, 0, 0.8, 0, 0, 0, 0,0, #7
1, 0, 0, 0, 0, 0, 0, 0,0, #8
0.1, 0.1, 0.1, 0.1, 0.2, 0.1, 0.1, 0.1, 0.1 ),#9
nrow=9, ncol=9)
colnames(A) <- c(1:9)
rownames(A) <- c(1:9)
x <- names(sample(x=A[,1], size=2, prob=A[,1]))
x
#> [1] "3" "8"
Created on 2022-02-01 by the reprex package (v2.0.1)
I am writing a function that uses a dataframe as filtering criteria for a big dataframe containing model outputs. These are the filtering criteria (as a df):
parameter value
1 alpha 0.1
2 beta 0.1
3 eta 0.1
4 zeta 0.1
5 lambda 0.5
6 phi 5.0
7 kappa 1.0
dput(values)
structure(list(parameter = structure(c(1L, 2L, 3L, 7L, 5L, 6L,
4L), .Label = c("alpha", "beta", "eta", "kappa", "lambda", "phi",
"zeta"), class = "factor"), value = c(0.1, 0.1, 0.1, 0.1, 0.5,
5, 1)), class = "data.frame", row.names = c(NA, -7L))
And this is how the 'outputs' df looks like:
time w x y z alpha beta eta zeta lambda phi kappa
1 0.0 10.00000 10.00000 10.000000 10.000000 0.1 0.1 0.1 0.1 0.95 5 1
1.1 0.1 10.00572 11.04680 9.896057 9.054394 0.1 0.1 0.1 0.1 0.95 5 1
1.2 0.2 10.01983 12.17827 9.592536 8.215338 0.1 0.1 0.1 0.1 0.95 5 1
1.3 0.3 10.04010 13.37290 9.112223 7.483799 0.1 0.1 0.1 0.1 0.95 5 1
1.4 0.4 10.06377 14.60353 8.489174 6.855626 0.1 0.1 0.1 0.1 0.95 5 1
1.5 0.5 10.08778 15.83982 7.764470 6.323152 0.1 0.1 0.1 0.1 0.95 5 1
dput(outputs)
structure(list(time = c(0, 0.1, 0.2, 0.3, 0.4, 0.5, 276.5, 276.6,
276.7, 276.8, 276.9, 276.961144437566), w = c(10, 10.0057192322758,
10.0198266325956, 10.040096099625, 10.0637654242843, 10.087779652849,
-1.71585943177118, -2.04004317987084, -2.56315700921588, -3.56775247519687,
-6.37643561014456, -13.828470036737), x = c(10, 11.0467963604334,
12.1782709261765, 13.3728962503142, 14.6035317074526, 15.8398164069251,
27.2774474452024, 26.3099862348669, 24.8705756934881, 22.3379071188018,
15.8960461541267, 3.62452931346518e-144), y = c(10, 9.89605687874935,
9.59253574727296, 9.11222320249057, 8.48917353431654, 7.76447036695841,
-0.604572230605542, -0.878231815857628, -1.46586965791714, -3.20623046085508,
-14.9365932475767, -3.30552834129368e+146), z = c(10, 9.05439359565339,
8.21533762023494, 7.48379901688836, 6.85562632179817, 6.3231517466183,
42.3149654949179, 43.8836626616462, 46.4372543252026, 51.7183454733949,
72.7027555440752, 3.30552834129368e+146), alpha = c(0.1, 0.1,
0.1, 0.1, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5), beta = c(0.1,
0.1, 0.1, 0.1, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5), eta = c(0.1,
0.1, 0.1, 0.1, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5), zeta = c(0.1,
0.1, 0.1, 0.1, 0.1, 0.1, 0.9, 0.9, 0.9, 0.9, 0.9, 0.9), lambda = c(0.9,
0.9, 0.5, 0.5, 0.9, 0.9, 0.5, 0.9, 0.5, 0.9, 0.5, 0.5
), phi = c(5, 5, 5, 5, 5, 5, 20, 20, 20, 20, 20, 20), kappa = c(1,
1, 1, 1, 1, 1, 10, 10, 10, 10, 10, 10), ode_outputs..iteration.. = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c("1",
"1.1", "1.2", "1.3", "1.4", "1.5", "2916.2765", "2916.2766",
"2916.2767", "2916.2768", "2916.2769", "2916.2770"), class = "data.frame")
So it should be something like:
filtered_outputs <- outputs %>% filter(all rows in column 1 == all values in column 2)
The names under the 'parameter' column correspond to column names in the 'outputs' df. I'd like this to be not hard-coded, so that I can feed in any filtering criteria as a df and the function will filter 'outputs'. I'd like to use dplyr or baseR preferably.
So you want to select all the rows in outputs dataframe which matches the values in values dataframe?
Here is a base R approach using sweep and rowSums.
result <- outputs[rowSums(sweep(outputs[as.character(values$parameter)], 2,
values$value, `!=`)) == 0, ]
result
# time w x y z alpha beta eta zeta lambda phi kappa
#1.2 0.2 10.01983 12.17827 9.592536 8.215338 0.1 0.1 0.1 0.1 0.5 5 1
#1.3 0.3 10.04010 13.37290 9.112223 7.483799 0.1 0.1 0.1 0.1 0.5 5 1
# ode_outputs..iteration..
#1.2 NA
#1.3 NA
A possible dplyr and tidyr solution:
Create a helper data frame by turning the values data frame into wide format, and apply a semi-join to filter by the required conditions.
You could easily wrap this up in one continuous workflow but I think it's easier to understand in separate steps.
library(dplyr)
library(tidyr)
conditions <-
values %>%
pivot_wider(names_from = parameter, values_from = value)
outputs %>%
semi_join(conditions)
#> Joining, by = c("alpha", "beta", "eta", "zeta", "lambda", "phi", "kappa")
#> time w x y z alpha beta eta zeta lambda phi
#> 1.2 0.2 10.01983 12.17827 9.592536 8.215338 0.1 0.1 0.1 0.1 0.5 5
#> 1.3 0.3 10.04010 13.37290 9.112223 7.483799 0.1 0.1 0.1 0.1 0.5 5
#> kappa ode_outputs..iteration..
#> 1.2 1 NA
#> 1.3 1 NA
Created on 2021-07-08 by the reprex package (v2.0.0)
I often find these kind of things are easier when the data is in long-form format - although this is just preference:
outputs %>%
tidyr::pivot_longer(
cols = -c(time, w, x, y, z, ode_outputs..iteration..),
names_to="parameter", values_to="value_truth"
) %>%
dplyr::left_join(filter_df) %>%
dplyr::group_by(time) %>%
dplyr::filter(all(value == value_truth)) %>%
dplyr::select(-value) %>%
tidyr::pivot_wider(
names_from="parameter",
values_from="value_truth"
)
Output:
# A tibble: 2 x 13
# Groups: time [2]
time w x y z ode_outputs..iteration.. alpha beta eta zeta lambda phi kappa
<dbl> <dbl> <dbl> <dbl> <dbl> <lgl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0.2 10.0 12.2 9.59 8.22 NA 0.1 0.1 0.1 0.1 0.5 5 1
2 0.3 10.0 13.4 9.11 7.48 NA 0.1 0.1 0.1 0.1 0.5 5 1
Data:
outputs = structure(list(time = c(0, 0.1, 0.2, 0.3, 0.4, 0.5, 276.5, 276.6,
276.7, 276.8, 276.9, 276.961144437566), w = c(10, 10.0057192322758,
10.0198266325956, 10.040096099625, 10.0637654242843, 10.087779652849,
-1.71585943177118, -2.04004317987084, -2.56315700921588, -3.56775247519687,
-6.37643561014456, -13.828470036737), x = c(10, 11.0467963604334,
12.1782709261765, 13.3728962503142, 14.6035317074526, 15.8398164069251,
27.2774474452024, 26.3099862348669, 24.8705756934881, 22.3379071188018,
15.8960461541267, 3.62452931346518e-144), y = c(10, 9.89605687874935,
9.59253574727296, 9.11222320249057, 8.48917353431654, 7.76447036695841,
-0.604572230605542, -0.878231815857628, -1.46586965791714, -3.20623046085508,
-14.9365932475767, -3.30552834129368e+146), z = c(10, 9.05439359565339,
8.21533762023494, 7.48379901688836, 6.85562632179817, 6.3231517466183,
42.3149654949179, 43.8836626616462, 46.4372543252026, 51.7183454733949,
72.7027555440752, 3.30552834129368e+146), alpha = c(0.1, 0.1,
0.1, 0.1, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5), beta = c(0.1,
0.1, 0.1, 0.1, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5), eta = c(0.1,
0.1, 0.1, 0.1, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5), zeta = c(0.1,
0.1, 0.1, 0.1, 0.1, 0.1, 0.9, 0.9, 0.9, 0.9, 0.9, 0.9), lambda = c(0.9,
0.9, 0.5, 0.5, 0.9, 0.9, 0.5, 0.9, 0.5, 0.9, 0.5, 0.5
), phi = c(5, 5, 5, 5, 5, 5, 20, 20, 20, 20, 20, 20), kappa = c(1,
1, 1, 1, 1, 1, 10, 10, 10, 10, 10, 10), ode_outputs..iteration.. = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c("1",
"1.1", "1.2", "1.3", "1.4", "1.5", "2916.2765", "2916.2766",
"2916.2767", "2916.2768", "2916.2769", "2916.2770"), class = "data.frame")
filter_df = fread(' parameter value
1 alpha 0.1
2 beta 0.1
3 eta 0.1
4 zeta 0.1
5 lambda 0.5
6 phi 5.0
7 kappa 1.0') %>% dplyr::select(-V1)
Context: I am trying to plot crop rotations from transition matrix describing the changes from one crop to another. For those wondering, using transition matrix to represent crop rotations has been published but the reference is not about making the plots; see:
Castellazzi, M.S., Wood, G.A., Burgess, P.J., Morris, J., Conrad,
K.F., Perry, J.N., 2008. A systematic representation of crop
rotations. Agricultural Systems 97, 26–33.
https://doi.org/10.1016/j.agsy.2007.10.006
By stochastic matrix I mean a square matrix representing the transition probability from Ci to Cj in one step. The sum of each row is equal to one.
This comes down to use a stochastic transition matrix as an adjacency matrix to build a weighted and directed graph. This can be done using diagram::plotmat() and igraph::graph_from_data_frame() but I just can't find out how to order vertices and make the edges look "nice".
Expected plot:
The matrix I would like to use looks like this:
> transmat
C1 C2 C3 C4 C5
C1 0 1 0.0 0.0 0
C2 0 0 0.5 0.5 0
C3 0 0 0.0 0.0 1
C4 0 0 0.0 0.0 1
C5 1 0 0.0 0.0 0
> dput(transmat)
structure(list(C1 = c(0, 0, 0, 0, 1), C2 = c(1, 0, 0, 0, 0),
C3 = c(0, 0.5, 0, 0, 0), C4 = c(0, 0.5, 0, 0, 0), C5 = c(0,
0, 1, 1, 0)), class = "data.frame", row.names = c("C1", "C2",
"C3", "C4", "C5"))
Attempt with diagram::plotmat():
## With diagram::plotmat ----------------------------------------------------
plot.new()
plotmat(t(transmat),
pos = c(1,1,2,1), # non-0 count in each line of transmat
curve = 0.3,
absent = 0, # don't connect crops linked by 0
arr.type = "triangle",
arr.pos = 0.6,
box.type = "rect",
box.prop=0.3,
box.lwd=2,
shadow.size = 0,
cex.txt=0.8,
endhead = FALSE)
Output:
Another example with igraph::graph_from_data_frame() and layout = layout_as_tree():
Expected plot:
The associated matrix:
> complex_ex
C1 C2 C3 C4 C5
C1 0.00 1.00 0.0 0.0 0.00
C2 0.00 0.00 0.5 0.5 0.00
C3 0.25 0.25 0.0 0.0 0.50
C4 0.00 0.00 0.0 0.0 1.00
C5 0.75 0.00 0.0 0.0 0.25
> dput(complex_ex)
structure(list(C1 = c(0, 0, 0.25, 0, 0.75), C2 = c(1, 0, 0.25,
0, 0), C3 = c(0, 0.5, 0, 0, 0), C4 = c(0, 0.5, 0, 0, 0), C5 = c(0,
0, 0.5, 1, 0.25)), class = "data.frame", row.names = c("C1",
"C2", "C3", "C4", "C5"))
The code I used:
# Some data transformation to make it work with igraph::graph_from_data_frame()
df_transmat <- as.data.frame(complex_ex)
df_transmat$from <- rownames(df_transmat)
df_transmat <- reshape(df_transmat,
idvar = "from",
varying = colnames(df_transmat)[1:(ncol(df_transmat)-1)],
times = colnames(df_transmat)[1:(ncol(df_transmat)-1)],
timevar = "to",
v.names = "change",
direction = "long")
rownames(df_transmat) <- NULL
rotation <- subset(df_transmat, subset = df_transmat$change != 0)
# The plot
g1 <- graph_from_data_frame(rotation, directed = TRUE)
plot(g1,
layout = layout_as_tree(g1, root = "C1"),
edge.arrow.mode = 2,
edge.arrow.size = 0.5,
edge.curved = 0,
edge.width = 0.5,
vertex.label.cex = 1,
vertex.label.font = 2,
vertex.shape = "rectangle",
vertex.color = "white",
vertex.size = 50,
vertex.size2 = 30,
vertex.label.dist = 0,
vertex.label.color = "black")
The output:
I have data that looks like this:
time sucrose fructose glucose galactose molasses water
1 5 0.0 0.00 0.0 0.0 0.3 0
2 10 0.3 0.10 0.1 0.0 1.0 0
3 15 0.8 0.20 0.2 0.2 1.4 0
4 20 1.3 0.35 0.7 0.4 2.5 0
5 25 2.2 0.80 1.6 0.5 3.5 0
6 30 3.1 1.00 2.3 0.6 4.5 0
7 35 3.6 1.60 3.1 0.7 5.7 0
8 40 5.1 2.80 4.3 0.7 6.7 0
How can i make a time series plot that uses the time column? They are all increasing values.
I saw this post multiple-time-series-in-one-plot which uses ts.plot to achieve something similar to what i want to show, which is this:
Input data for the table above:
structure(list(time = c(5, 10, 15, 20, 25, 30, 35, 40), sucrose = c(0,
0.3, 0.8, 1.3, 2.2, 3.1, 3.6, 5.1), fructose = c(0, 0.1, 0.2,
0.35, 0.8, 1, 1.6, 2.8), glucose = c(0, 0.1, 0.2, 0.7, 1.6, 2.3,
3.1, 4.3), galactose = c(0, 0, 0.2, 0.4, 0.5, 0.6, 0.7, 0.7),
molasses = c(0.3, 1, 1.4, 2.5, 3.5, 4.5, 5.7, 6.7), water = c(0,
0, 0, 0, 0, 0, 0, 0)), .Names = c("time", "sucrose", "fructose",
"glucose", "galactose", "molasses", "water"), row.names = c(NA,
-8L), class = "data.frame")
It doesn't seem like a ts plot is necessary. Here's how you could do it in base-R:
with(df, plot(time, sucrose, type="n", ylab="contents"))
var <- names(df)[-1]
for(i in var) lines(df$time, df[,i])
The more elegant solution would however be using the 'dplyrandggplot2` package:
df <- df %>%
gather(content, val, -time)
ggplot(df, aes(time, val, col=content)) + geom_line()
I have a dataframe with measurements stored as a list by row.
Subject Measurements
1 s1 -0.4, -0.9, -1.1, -0.1, 0.1
2 s2 -1.4, -1.7, -1.7, -0.6, -1.7
3 s3 -1.0, -0.1, -0.6, -0.5, -0.1
4 s4 -0.2, -0.5, -0.2, 0.1, -0.7
5 s5 0.7, 0.2, 0.4, 0.7, 0.2
6 s6 -0.3, -0.1, 0.1, -0.2, -0.1
How do I average/find standard deviation/other list manipulations and add the output to a new column in data frame (e.g "mean")
Edit
Here's the data structure I'm working with:
structure(list(Subject = structure(1:6, .Label = c("s1", "s2",
"s3", "s4", "s5", "s6"), class = "factor"), Measurements = list(
c(-0.4, -0.9, -1.1, -0.1, 0.1), c(-1.4, -1.7, -1.7, -0.6,
-1.7), c(-1, -0.1, -0.6, -0.5, -0.1), c(-0.2, -0.5, -0.2,
0.1, -0.7), c(0.7, 0.2, 0.4, 0.7, 0.2), c(-0.3, -0.1, 0.1,
-0.2, -0.1))), .Names = c("Subject", "Measurements"), row.names = c(NA,
6L), class = "data.frame")
If you store your data more efficiently, this becomes much easier:
dat<- structure(list(Subject = structure(1:6, .Label = c("s1", "s2",
"s3", "s4", "s5", "s6"), class = "factor"), Measurements = list(
c(-0.4, -0.9, -1.1, -0.1, 0.1), c(-1.4, -1.7, -1.7, -0.6,
-1.7), c(-1, -0.1, -0.6, -0.5, -0.1), c(-0.2, -0.5, -0.2,
0.1, -0.7), c(0.7, 0.2, 0.4, 0.7, 0.2), c(-0.3, -0.1, 0.1,
-0.2, -0.1))), .Names = c("Subject", "Measurements"), row.names = c(NA,
6L), class = "data.frame")
> dat <- data.frame(subject = dat$Subject,do.call(rbind,dat$Meas))
> dat$means <- apply(dat[,-1],1,mean)
> dat
subject X1 X2 X3 X4 X5 means
1 s1 -0.4 -0.9 -1.1 -0.1 0.1 -0.48
2 s2 -1.4 -1.7 -1.7 -0.6 -1.7 -1.42
3 s3 -1.0 -0.1 -0.6 -0.5 -0.1 -0.46
4 s4 -0.2 -0.5 -0.2 0.1 -0.7 -0.30
5 s5 0.7 0.2 0.4 0.7 0.2 0.44
6 s6 -0.3 -0.1 0.1 -0.2 -0.1 -0.12
Once you have each measurement in its own column, you can simply use apply (or rowMeans) os some similar function.
It looks like Measurements is a matrix within your data.frame (df).
df$means <- rowMeans(df$Measurements)
For a more general solution you can use apply with Margin = 1 for a given function.
df$SDs <- apply(df$Measurements, 1, sd)
If Measurements were actually a genuine list you'd use
df$SDs <- lapply(df$Measurements, sd)
That gives maximum performance but now your SDs column is a list so to make it a vector I'd go with...
df$SDs <- sapply(df$Measurements, sd)
(when I made a data.frame with a list included it didn't look like that so I didn't think it was really a list at first).