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All combinations of a list of lists
...eed itertools.product:
>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (...
Difference between numpy.array shape (R, 1) and (R,)
... how to interpret the data buffer.
For example, if we create an array of 12 integers:
>>> a = numpy.arange(12)
>>> a
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
Then a consists of a data buffer, arranged something like this:
┌────┬────┬──...
Does assignment with a comma work?
Why does aaa = 1,2,3 work and set the value of aaa to 1 ?
4 Answers
4
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Apply a function to every row of a matrix or a data frame
Suppose I have a n by 2 matrix and a function that takes a 2-vector as one of its arguments. I would like to apply the function to each row of the matrix and get a n-vector. How to do this in R?
...
Counting the occurrences / frequency of array elements
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LISTAGG in Oracle to return distinct values
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24 Answers
24
Active
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Array to Hash Ruby
...
a = ["item 1", "item 2", "item 3", "item 4"]
h = Hash[*a] # => { "item 1" => "item 2", "item 3" => "item 4" }
That's it. The * is called the splat operator.
One caveat per @Mike Lewis (in the comments): "Be very careful with this. Rub...
For each row return the column name of the largest value
...d() to make examples using sample reproducible):
DF <- data.frame(V1=c(2,8,1),V2=c(7,3,5),V3=c(9,6,4))
colnames(DF)[apply(DF,1,which.max)]
[1] "V3" "V1" "V2"
A faster solution than using apply might be max.col:
colnames(DF)[max.col(DF,ties.method="first")]
#[1] "V3" "V1" "V2"
...where ties...
How to drop columns by name in a data frame
... or the subset function. For example :
R> df <- data.frame(x=1:5, y=2:6, z=3:7, u=4:8)
R> df
x y z u
1 1 2 3 4
2 2 3 4 5
3 3 4 5 6
4 4 5 6 7
5 5 6 7 8
Then you can use the which function and the - operator in column indexation :
R> df[ , -which(names(df) %in% c("z","u"))]
x y
1 1...
Replacing NAs with latest non-NA value
...ample from the help page:
library(zoo)
az <- zoo(1:6)
bz <- zoo(c(2,NA,1,4,5,2))
na.locf(bz)
1 2 3 4 5 6
2 2 1 4 5 2
na.locf(bz, fromLast = TRUE)
1 2 3 4 5 6
2 1 1 4 5 2
cz <- zoo(c(NA,9,3,2,3,2))
na.locf(cz)
2 3 4 5 6
9 3 2 3 2
...