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)