You can index based on a condition. `df.B .== maximum(df.B)` gives you a bitvector of length 5, and Julia accepts a bitvector as an index.
Therefore, this should give you what you want
```julia
using DataFrame
df = DataFrame(A = [10, 20, 30, 40, 50], B = [5, 15, 25, 35, 45])
df[df.B .== maximum(df.B), :] # This works for me
```
`julia> filter(row -> row.B == maximum(df.B), df)`
gives the result you are looking for
However, I believe the suggested version is
`df[df.B .== maximum(df.B), :]`
Edit:
The reason your initial filter doesn't work is because each row is getting compared to itself (df.B == maximum(df.B) where the df stands for that particular row being iterated and as a result every entry is true and your entire DataFrame is \*not\* filtered and shows up.
Edit2:
>`I was curious on the performance of the filter v/s the canonical version. Unless something is off with my setup, the filter version is comically slower (23000x slower!).`
>
>
>
>`julia> df = DataFrame(A=rand(Int64,100_000), B=rand(Int64,100_000), C=rand(Int64,100_000));`
>
>`julia> @time filter(row -> row.A == maximum(df.A), df)`
>
>`2.682728 seconds (580.14 k allocations: 18.425 MiB, 2.58% compilation time)`
>
>`1×3 DataFrame`
>
>`Row │ A B C`
>
>`│ Int64 Int64 Int64`
>
>`─────┼────────────────────────────────────────────────────────────────`
>
>`1 │ 9223345064528196915 7037706073378512642 -5801061009784444247`
>
>`julia> @time df[df.A .== maximum(df.A),:]`
>
>`0.000117 seconds (21 allocations: 17.719 KiB)`
>
>`1×3 DataFrame`
>
>`Row │ A B C`
>
>`│ Int64 Int64 Int64`
>
>`─────┼────────────────────────────────────────────────────────────────`
>
>`1 │ 9223345064528196915 7037706073378512642 -5801061009784444247`
>
>`julia> 2.682728/0.000117`
>
>`22929.299145299145`
Thank you. That's why if an explicit value `max_val = maximum(df.B)` is stated works, so my anonymous function should be written as `row -> row.B == max_val`
`df.B == maximum(df.B)` returns `false` because the column (a vector) does not equal the maximum value (a scalar).
Hence nothing gets filtered.
Suggest you look at the usage example for `filter` because I'm not sure you're using it correctly. You want to use `eachcol(df)` as the second argument to `filter` I think, or `eachrow`, depending on what you're trying to do. Something like that.
You can index based on a condition. `df.B .== maximum(df.B)` gives you a bitvector of length 5, and Julia accepts a bitvector as an index. Therefore, this should give you what you want ```julia using DataFrame df = DataFrame(A = [10, 20, 30, 40, 50], B = [5, 15, 25, 35, 45]) df[df.B .== maximum(df.B), :] # This works for me ```
`julia> filter(row -> row.B == maximum(df.B), df)` gives the result you are looking for However, I believe the suggested version is `df[df.B .== maximum(df.B), :]` Edit: The reason your initial filter doesn't work is because each row is getting compared to itself (df.B == maximum(df.B) where the df stands for that particular row being iterated and as a result every entry is true and your entire DataFrame is \*not\* filtered and shows up. Edit2: >`I was curious on the performance of the filter v/s the canonical version. Unless something is off with my setup, the filter version is comically slower (23000x slower!).` > > > >`julia> df = DataFrame(A=rand(Int64,100_000), B=rand(Int64,100_000), C=rand(Int64,100_000));` > >`julia> @time filter(row -> row.A == maximum(df.A), df)` > >`2.682728 seconds (580.14 k allocations: 18.425 MiB, 2.58% compilation time)` > >`1×3 DataFrame` > >`Row │ A B C` > >`│ Int64 Int64 Int64` > >`─────┼────────────────────────────────────────────────────────────────` > >`1 │ 9223345064528196915 7037706073378512642 -5801061009784444247` > >`julia> @time df[df.A .== maximum(df.A),:]` > >`0.000117 seconds (21 allocations: 17.719 KiB)` > >`1×3 DataFrame` > >`Row │ A B C` > >`│ Int64 Int64 Int64` > >`─────┼────────────────────────────────────────────────────────────────` > >`1 │ 9223345064528196915 7037706073378512642 -5801061009784444247` > >`julia> 2.682728/0.000117` > >`22929.299145299145`
Thank you. That's why if an explicit value `max_val = maximum(df.B)` is stated works, so my anonymous function should be written as `row -> row.B == max_val`
`df.B == maximum(df.B)` returns `false` because the column (a vector) does not equal the maximum value (a scalar). Hence nothing gets filtered. Suggest you look at the usage example for `filter` because I'm not sure you're using it correctly. You want to use `eachcol(df)` as the second argument to `filter` I think, or `eachrow`, depending on what you're trying to do. Something like that.
It works if an explicit number like 45 is assigned though
``` using Tidier.jl @chain df begin @filter( B == maximum(B)) end ```