I have the following looking dataframe:
using DataFrames
df = DataFrame(
condition = [false, false, true, false, false, false, true, false, false, false],
time = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
Output:
10×2 DataFrame
Row │ condition time
│ Bool Int64
─────┼──────────────────
1 │ false 1
2 │ false 2
3 │ true 3
4 │ false 4
5 │ false 5
6 │ false 6
7 │ true 7
8 │ false 8
9 │ false 9
10 │ false 10
I would like to calculate the difference in rows with respect to a conditioned value (true/false). This means that for row 1 the nearest true is 2 rows a way. The conditioned rows with true should have a value of 0. Here is the desired output:
10×3 DataFrame
Row │ condition time diff
│ Bool Int64 Int64
─────┼─────────────────────────
1 │ false 1 2
2 │ false 2 1
3 │ true 3 0
4 │ false 4 1
5 │ false 5 2
6 │ false 6 1
7 │ true 7 0
8 │ false 8 1
9 │ false 9 2
10 │ false 10 3
So I was wondering if anyone knows how to calculate the difference in rows with the closest conditioned value in dataframe Julia?
transform(df, :condition =>
(w->((f,u)->min.(f(u),reverse(f(reverse(u)))))(
v->accumulate(
(x,y)->ifelse(y,0,x+1),
v;init=length(v)
),
w
)) => :diff)
(u,v,w are vectors. x,y are bool/int. f is a function)
Does the job with output:
10×3 DataFrame
Row │ condition time diff
│ Bool Int64 Int64
─────┼─────────────────────────
1 │ false 1 2
2 │ false 2 1
3 │ true 3 0
4 │ false 4 1
5 │ false 5 2
6 │ false 6 1
7 │ true 7 0
8 │ false 8 1
9 │ false 9 2
10 │ false 10 3
On my REPL it was 1-line as follows, but tried to make it more readable above:
transform(df, :condition => (v->((f, v)->min.(f(v),reverse(f(reverse(v)))))(v->accumulate((x, y)->ifelse(y, 0, x+1), v; init=length(v)), v)) => :diff)
It is not the clearest way, and also not the most efficient, but it is a short piece of code. To get clearer and more efficient result, a separate function should be defined.
Last thing, the column has to have one true value, otherwise the results are not meaningful (this can be checked easily with a bit more code, but not sure what OP wants in this case).
Related
I have the following dataframe called df:
using DataFrames
df = DataFrame(group = ["A", "A", "A", "A", "B", "B", "B", "B"],
value = [2,1,4,3,3,5,2,1])
8×2 DataFrame
Row │ group value
│ String Int64
─────┼───────────────
1 │ A 2
2 │ A 1
3 │ A 4
4 │ A 3
5 │ B 3
6 │ B 5
7 │ B 2
8 │ B 1
I would like to calculate the difference with previous values of consecutive rows in column value per group. The offset should have NaN, 0, or missing. Here is the desired output:
8×3 DataFrame
Row │ group value diff
│ String Int64 Float64
─────┼────────────────────────
1 │ A 2 NaN
2 │ A 1 -1.0
3 │ A 4 3.0
4 │ A 3 -1.0
5 │ B 3 NaN
6 │ B 5 2.0
7 │ B 2 -3.0
8 │ B 1 -2.0
So I was wondering if anyone knows how to calculate the difference between consecutive rows per group in Julia?
Using DataFrames.jl (you can replace missing by any value you like):
julia> select(groupby(df, :group),
:value => (x -> [missing; diff(x)]) => :diff)
8×2 DataFrame
Row │ group diff
│ String Int64?
─────┼─────────────────
1 │ A missing
2 │ A -1
3 │ A 3
4 │ A -1
5 │ B missing
6 │ B 2
7 │ B -3
8 │ B -1
Using DataFramesMeta.jl:
julia> #chain df begin
groupby(:group)
#select :diff = [missing; diff(:value)]
end
8×2 DataFrame
Row │ group diff
│ String Int64?
─────┼─────────────────
1 │ A missing
2 │ A -1
3 │ A 3
4 │ A -1
5 │ B missing
6 │ B 2
7 │ B -3
8 │ B -1
Normally diff in Julia like in e.g. R would produce one less row (and the syntax would be simpler:
julia> combine(groupby(df, :group), :value => diff => :diff)
6×2 DataFrame
Row │ group diff
│ String Int64
─────┼───────────────
1 │ A -1
2 │ A 3
3 │ A -1
4 │ B 2
5 │ B -3
6 │ B -1
julia> #chain df begin
groupby(:group)
#combine :diff = diff(:value)
end
6×2 DataFrame
Row │ group diff
│ String Int64
─────┼───────────────
1 │ A -1
2 │ A 3
3 │ A -1
4 │ B 2
5 │ B -3
6 │ B -1
Yet another way would be to use lag from ShiftedArrays.jl:
julia> using ShiftedArrays: lag
julia> #chain df begin
groupby(:group)
#combine :diff = :value - lag(:value)
end
8×2 DataFrame
Row │ group diff
│ String Int64?
─────┼─────────────────
1 │ A missing
2 │ A -1
3 │ A 3
4 │ A -1
5 │ B missing
6 │ B 2
7 │ B -3
8 │ B -1
I need to compare elements by rows of c1 and c2 columns in the DataFrame and return higher value in new column.
Column "Result" should return [6,5,4,4,5]
df = DataFrame(c1=[1,2,3,4,5], c2=[6,5,4,3,2])
println(df)
if broadcast(.>, df.c1, df.c2)
df[:, "Result"] .= df.c1
else
df[:, "Result"] .= df.c2
end
println(df5)
ERROR: TypeError: non-boolean (BitVector) used in boolean context
An alternative is:
julia> df.Result = max.(df.c1, df.c2)
5-element Vector{Int64}:
6
5
4
4
5
(as some users prefer such code than higher order functions presented excellently by #Shayan)
Using eachrow
julia> maximum.(eachrow(df))
5-element Vector{Int64}:
6
5
4
4
5
or as DataFrame
julia> DataFrame(new = maximum.(eachrow(df)))
5×1 DataFrame
Row │ new
│ Int64
─────┼───────
1 │ 6
2 │ 5
3 │ 4
4 │ 4
5 │ 5
or as a new column the DataFrame
julia> df.Result = maximum.(eachrow(df))
julia> df
5×3 DataFrame
Row │ c1 c2 Result
│ Int64 Int64 Int64
─────┼──────────────────────
1 │ 1 6 6
2 │ 2 5 5
3 │ 3 4 4
4 │ 4 3 4
5 │ 5 2 5
You can use select:
julia> select(df, All() => ByRow(max) => :Result)
5×1 DataFrame
Row │ Result
│ Int64
─────┼────────
1 │ 6
2 │ 5
3 │ 4
4 │ 4
5 │ 5
Another alternative is using DataFramesMeta.jl:
julia> #select(df, :Result = max.(:c1, :c2))
5×1 DataFrame
Row │ Result
│ Int64
─────┼────────
1 │ 6
2 │ 5
3 │ 4
4 │ 4
5 │ 5
# Alternatively, you can use the following line to avoid mentioning the column names manually:
#select(df, :Result = $(max.(propertynames(df)...)))
# Gives the same result.
If you want to make the change in place, then use select! or #select!.
If you prefer the returned dataframe to contain c1 and c2 columns as well, then you can go for transform (its alternative in-place operator is transform!):
julia> transform(df, All() => ByRow(max) => :Result)
5×3 DataFrame
Row │ c1 c2 Result
│ Int64 Int64 Int64
─────┼──────────────────────
1 │ 1 6 6
2 │ 2 5 5
3 │ 3 4 4
4 │ 4 3 4
5 │ 5 2 5
# And the same thing using DataFramesMeta.jl:
julia> #transform(df, :Result = $(max.(propertynames(df)...)))
5×3 DataFrame
Row │ c1 c2 Result
│ Int64 Int64 Int64
─────┼──────────────────────
1 │ 1 6 6
2 │ 2 5 5
3 │ 3 4 4
4 │ 4 3 3
5 │ 5 2 2
I have a DataFrame in Julia and I want to create a new column that represents the difference between consecutive rows in a specific column. In python pandas, I would simply use df.series.diff(). Is there a Julia equivelant?
For example:
data
1
2
4
6
7
# in pandas
df['diff_data'] = df.data.diff()
data diff_data
1 NaN
2 1
4 2
6 2
7 1
You can use ShiftedArrays.jl like this.
Declarative style:
julia> using DataFrames, ShiftedArrays
julia> df = DataFrame(data=[1, 2, 4, 6, 7])
5×1 DataFrame
Row │ data
│ Int64
─────┼───────
1 │ 1
2 │ 2
3 │ 4
4 │ 6
5 │ 7
julia> transform(df, :data => (x -> x - lag(x)) => :data_diff)
5×2 DataFrame
Row │ data data_diff
│ Int64 Int64?
─────┼──────────────────
1 │ 1 missing
2 │ 2 1
3 │ 4 2
4 │ 6 2
5 │ 7 1
Imperative style (in place):
julia> df = DataFrame(data=[1, 2, 4, 6, 7])
5×1 DataFrame
Row │ data
│ Int64
─────┼───────
1 │ 1
2 │ 2
3 │ 4
4 │ 6
5 │ 7
julia> df.data_diff = df.data - lag(df.data)
5-element Vector{Union{Missing, Int64}}:
missing
1
2
2
1
julia> df
5×2 DataFrame
Row │ data data_diff
│ Int64 Int64?
─────┼──────────────────
1 │ 1 missing
2 │ 2 1
3 │ 4 2
4 │ 6 2
5 │ 7 1
with diff you do not need extra packages and can do similarly the following:
julia> df.data_diff = [missing; diff(df.data)]
5-element Vector{Union{Missing, Int64}}:
missing
1
2
2
1
(the issue is that diff is a general purpose function that does change the length of vector from n to n-1 so you have to add missing manually in front)
Pandas df.diff() does it to the whole data frame at once and allows you to specify row-wise or column-wise. There might be a better way but this is what I used before (I like chaining or piping like in dplyr):
# using chain.jl
#chain df begin
eachcol()
diff.()
DataFrame(:auto)
rename!(names(df))
end
# OR base pipe
df |>
x -> eachcol(x) |>
x -> diff.(x) |>
x -> DataFrame(x, :auto) |>
x -> rename!(x, names(df)[2:end])
# OR without piping
rename!(DataFrame(diff.(eachcol(df)), :auto), names(df))
You might need to insert the starting row, which will now have missing values.
Suppose I have the following dataframe
using DataFrames
df = DataFrame(A = 1:10, B = ["a","a","b","b","b","c","c","c","c","d"])
grouped_df = groupby(df, "B")
I would have four groups. How can I drop the groups that have fewer than, say, 2 rows? For example, how can I keep only groups a,b, and c? I can easily do it with a for loop, but I don't think the optimal way.
If you want the result to be still grouped then filter is simplest:
julia> filter(x -> nrow(x) > 1, grouped_df)
GroupedDataFrame with 3 groups based on key: B
First Group (2 rows): B = "a"
Row │ A B
│ Int64 String
─────┼───────────────
1 │ 1 a
2 │ 2 a
⋮
Last Group (4 rows): B = "c"
Row │ A B
│ Int64 String
─────┼───────────────
1 │ 6 c
2 │ 7 c
3 │ 8 c
4 │ 9 c
If you want to get a data frame as a result of one operation then do e.g.:
julia> combine(grouped_df, x -> nrow(x) < 2 ? DataFrame() : x)
9×2 DataFrame
Row │ B A
│ String Int64
─────┼───────────────
1 │ a 1
2 │ a 2
3 │ b 3
4 │ b 4
5 │ b 5
6 │ c 6
7 │ c 7
8 │ c 8
9 │ c 9
For reasons, I have to use the do form (block form?) of DataFrames.transform
df = DataFrame(:a => [i for i in 1:10])
transform(df) do d
[d[i, :a] * 10 for i in 1:nrow(d)]
end
Row │ a x1
│ Int64 Int64
─────┼──────────────
1 │ 1 10
2 │ 2 20
3 │ 3 30
. | . ..
and that's all good, but I'd like that new column to be called something else other than x1. I can't see an obvious way in the docs aside from renaming it post-hoc. Is there a way of coming out of the block with a named column (and possibly even more than one?) instead of it being named xn?
You can return a DataFrame or a NamedTuple:
df = DataFrame(:a => [i for i in 1:10])
transform(df) do d
(result=[d[i, :a] * 10 for i in 1:nrow(d)],)
end
transform(df) do d
DataFrame(result=[d[i, :a] * 10 for i in 1:nrow(d)])
end
You can do:
df.something = map(eachrow(df)) do x
x.a * 10
end
Result:
julia> df
10×2 DataFrame
Row │ a something
│ Int64 Int64
─────┼──────────────────
1 │ 1 10
2 │ 2 20
3 │ 3 30
4 │ 4 40
5 │ 5 50
6 │ 6 60
7 │ 7 70
8 │ 8 80
9 │ 9 90
10 │ 10 100