Thanks for the answer. I tried "eachrow" but I have 2 problems :
1- I still have to do an array conversion, I think it is slow
julia> for r in eachrow(df)
println(mean(convert(Array,r)))
end
6.0
7.0
8.0
9.0
10.0
2- I do not manage to use a subset of the row, for example the 2 first
values :
julia> for r in eachrow(df)
println(mean(convert(Array,r)))
end
6.0
7.0
8.0
9.0
10.0
julia> for r in eachrow(df)
println(mean(convert(Array,r[1:2])))
end
WARNING: [a] concatenation is deprecated; use collect(a) instead
in depwarn at deprecated.jl:73
in oldstyle_vcat_warning at ./abstractarray.jl:29
[inlined code] from none:2
in anonymous at no file:0
while loading no file, in expression starting on line 0
4.0
Le samedi 21 novembre 2015 14:04:11 UTC+1, tshort a écrit :
>
> You can try `eachrow`. It probably won't be fast, though. Here's an
> example:
>
>
> https://github.com/JuliaStats/DataFrames.jl/blob/master/test/iteration.jl#L34
> On Nov 21, 2015 7:19 AM, "Fred" <fred.so...@gmail.com <javascript:>>
> wrote:
>
>> Hi,
>>
>> In DataFrames, it is easy to apply a function by columns using the
>> colwise() function. But I find very difficult and inefficient to apply a
>> function by rows.
>>
>> For example :
>>
>>
>>
>> julia> df = DataFrame(a=1:5, b=7:11, c=10:14)
>> 5x3 DataFrames.DataFrame
>> | Row | a | b | c |
>> |-----|---|----|----|
>> | 1 | 1 | 7 | 10 |
>> | 2 | 2 | 8 | 11 |
>> | 3 | 3 | 9 | 12 |
>> | 4 | 4 | 10 | 13 |
>> | 5 | 5 | 11 | 14 |
>>
>>
>>
>> julia> colwise(mean,df)
>> 3-element Array{Any,1}:
>> [3.0]
>> [9.0]
>> [12.0]
>>
>>
>> julia> colwise(mean,df[1,1:2])
>> 2-element Array{Any,1}:
>> [1.0]
>> [7.0]
>>
>>
>>
>> To calculate the mean of a row (or a subset), the only way I found is
>> this :
>>
>> julia> mean(convert(Array,df[1,1:3]))
>> 6.0
>>
>>
>>
>> I think this is inefficient and probably very slow. I there a better way
>> to apply a function by rows ?
>>
>> Thanks !
>>
>
