Update: I found a 1-line command to convert everything in a DataFrame into
a Float that seems to work generally:
data = data*1.0
So, for example,
julia> A=DataFrame([1 2 ; 3 4])
2x2 DataFrame:
x1 x2
[1,] 1 2
[2,] 3 4
julia> A=A*1.0
2x2 DataFrame:
x1 x2
[1,] 1.0 2.0
[2,] 3.0 4.0
Best,
Bradley
On Monday, August 18, 2014 10:02:02 AM UTC-5, Bradley Setzler wrote:
>
> Hi John,
>
> Thanks for your reply, I'm getting the following:
>
> julia> A=DataArray([1 2; 3 4])
> 2x2 DataArray{Int64,2}:
> 1 2
> 3 4
> julia> A*.5
> 2x2 DataArray{Float64,2}:
> 0.5 1.0
> 1.5 2.0
>
> julia> A/2.
> 2x2 DataArray{Float64,2}:
> 0.5 1.0
> 1.5 2.0
> julia> A/2
> InexactError()
>
>
> So it converts to Float if divided by Float, but does not convert if
> divided by Integer.
>
> Best,
> Bradley
>
>
>
> On Monday, August 18, 2014 9:31:43 AM UTC-5, John Myles White wrote:
>>
>> Hi Bradley,
>>
>> Would you consider using DataArrays for this? DataFrames no longer
>> support these operations, so any upgrade in your setup would turn all of
>> this code into errors.
>>
>> All of these operations work on DataArrays already.
>>
>> — John
>>
>> On Aug 18, 2014, at 7:28 AM, Bradley Setzler <[email protected]>
>> wrote:
>>
>> >
>> >
>> > Good morning,
>> >
>> > I am looking for a simple way to convert an Integer DataFrame to a
>> Float DataFrame. Here is an example of the problem:
>> >
>> > julia> using DataFrames
>> > julia> A=DataFrame([1 2; 3 4])
>> > 2x2 DataFrame:
>> > x1 x2
>> > [1,] 1 2
>> > [2,] 3 4
>> >
>> > With multiplication, there is no problem automatically converting to
>> Float:
>> >
>> > julia> A*.5
>> > 2x2 DataFrame:
>> > x1 x2
>> > [1,] 0.5 1.0
>> > [2,] 1.5 2.0
>> >
>> > But with division, for example, the conversion fails:
>> >
>> > julia> A/2
>> > InexactError()
>> >
>> > Ideally, there would be a one-line command so that we don't have to
>> worry about this issue, say DataFloat() of the form:
>> >
>> > julia> A=DataFloat(A)
>> > 2x2 DataFrame:
>> > x1 x2
>> > [1,] 1.0 2.0
>> > [2,] 3.0 4.0
>> >
>> > Does something like this exist?
>> >
>> > Thanks,
>> > Bradley
>> >
>>
>>
