DataFrames: replace function?

[Alexander Flyax]

In Julia's DataFrames, is there an equivalent of Pandas' replace()? If not, what's the most efficient way to accomplish its equivalent? Thanks...

E.g. (from Python/Pandas):

print(dating_df.head(3))

   miles     games  ice_cream  opinion
0  40920  8.326976   0.953952        3
1  14488  7.153469   1.673904        2
2  26052  1.441871   0.805124        1

dating_df.opinion.replace({1:'disliked', 2:'OK', 3:'liked'}, inplace=True)
print(dating_df.head(3))
   miles     games  ice_cream   opinion
0  40920  8.326976   0.953952     liked
1  14488  7.153469   1.673904        OK
2  26052  1.441871   0.805124  disliked

[Andrew Ellis]

in the absence of a built-in replace function, this might work:

using Compat, Distributions, DataFrames, DataFramesMeta

mapping = @Compat.Dict(1 => "a", 2 => "b", 3=> "c")

df[:opinion] = @with df begin

    :opinion = [mapping[i] for i in :opinion]

end

the @with isn't really necessary, you could just write:

df[:opinion] = [mapping[i] for i in df[:opinion]]

[Tom Short]

Another option is to convert the column to a PooledDataArray and then create a new PooledDataArray using the refs from the first PooledDataArray and a new pool that does the mapping you want.

Also, you can write Andrew's @with construct more concisely:

df[:opinion] = @with df [mapping[i] for i in :opinion]

On Tue, Jan 6, 2015 at 6:09 PM, Andrew Ellis <a.w....@gmail.com> wrote:

in the absence of a built-in replace function, this might work:

using Compat, Distributions, DataFrames, DataFramesMeta

mapping = @Compat.Dict(1 => "a", 2 => "b", 3=> "c")

df[:opinion] = @with df begin

    :opinion = [mapping[i] for i in :opinion]

end

the @with isn't really necessary, you could just write:

df[:opinion] =  df[:opinion] = [mapping[i] for i in df[:opinion]]

On Tuesday, January 6, 2015 12:47:23 AM UTC+1, Alexander Flyax wrote:

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[Alexander Flyax]

Thanks, Andrew and tshort. I thought of a similar solution using

dating_df[:opinion] = [ {1 => "bad", 2 => "OK", 3 => "good"}[i] for i in dating_df[:opinion] ]

The problem is when I feed it only, e.g., {1 => "bad", 2 => "OK", 3 => "good"}. Then, as expected: key not found: 3. In Pandas you can df.replace({0:Nan}).

I can write a for loop that will check for each value in dating_df[:opinion] or a function that will use that for loop, but I am just wondering if there is a more elegant solution. Something like:

mapping = {1 => "bad", 2 => "OK", 3 => "good"}
dating_df[:opinion] = [ mapping[i] for i in dating_df[:opinion] if i in mapping]

[Alexander Flyax]

Sorry, I meant if I feed it only {1 => "bad", 2 => "OK"} ...

[Alexander Flyax]

Actually, apparently, I can't write a for loop, because: 

`convert` has no method matching convert(::Type{Int64}, ::ASCIIString)

(I am trying to say df[:col][i] = "text".)