How to create a DataFrame with specific number of columns and rows and fill it with zeros?

General Usage
1

Hi!
I’m trying to create a dataframe where the first column have certain number of rows named in the following way:

df = DataFrame("Names" => names(df_rowname)[2:end])

Now I want to add n number of columns to df with labels names names(df_colname), and fill all the cells with zeros (or another initialization value).
What is the best way to do it? I can use the first line I wrote here to create the dataframe with the first column, but I’m not sure how to add the rest of them.

Thank you!

2

DataFrames are best used to represent data which has several attributes with heterogenous types. It seems that you are going to create some sort of a Matrix with a common type to the data in the columns.

If this is the case, and the names of the rows and columns are important, consider using NamedArrays.jl instead of a DataFrame.

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3

Try MWE:

using DataFrames

df_rowname = DataFrame([(a=1, b=2, c=3, d=4)])

df = DataFrame("Names" => names(df_rowname)[2:end])
foreach(x -> df[!, x] = zeros(nrow(df)), names(df_rowname))
df
4

Thanks both of you for the helpful insights!

5

Another option:

row_names = ["a","b","c","d"]
col_names = ["x","y","z"]
zeros_matrix = zeros(length(row_names), length(col_names))
base_df = DataFrame(zeros_matrix, col_names)
row_names_df = DataFrame(rows = row_names)
hcat(row_names_df, base_df)
1 Like
6

One more:

julia> DataFrame(:Names => [:a, :b, :c], ([:x, :y, :z] .=> 0.0)...)
3×4 DataFrame
 Row │ Names   x        y        z       
     │ Symbol  Float64  Float64  Float64 
─────┼───────────────────────────────────
   1 │ a           0.0      0.0      0.0
   2 │ b           0.0      0.0      0.0
   3 │ c           0.0      0.0      0.0

Or equivalently

DataFrame([:Names => [:a, :b, :c]; [:x, :y, :z] .=> 0.0])
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7

If you want to add columns to an existing data frame you can also use insertcols! or transform!.

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8 
insertcols!(describe(df_rowname)[2:end,1:1],(names(df_rowname).=>0)...)
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