# Iterating over row in a DataFrame

**URL:** <https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662>\
**Category:** New to Julia\
**Created:** [December 11, 2020, 12:30pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662 "2020-12-11T12:30:05Z")\
**Posts on this page:** 9\
**Page:** 1

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**Author:** ![tlorans](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/tlorans/32/19849_2.png) [@tlorans](https://discourse.julialang.org/u/tlorans)\
**Post date:** [December 11, 2020, 12:30pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/1 "2020-12-11T12:30:05Z")

</div>

Hello,

I wish I could implement one of my Python models on Julia, but have been stuck for hours on the basic iteration problem in the context of the Julia language.

Basically, I just want to iterate over each row of my DataFrame

```julia
#Step 1: declaration of endogenous variables
columnnames = ["A","B"]
T = 100
columns = [Symbol(col) => zeros(T) for col in columnnames]
y = DataFrame(columns...)
#I am launching my iteration
for t in 1:T
          if t == 0
#Step 2: Initial values are assigned
                y[1] = 1
          else
#Step 3: equations
                y[t] = y[t-1] + 1

```

No matter how hard I search through the different tutorials, I can’t find the solution to do this simple approach on Julia.  
I tried the following solution in particular:

Even the first step to replace the first line with a value doesn’t work…

```julia
y[:1,:] = 1.0

```

`ERROR: MethodError: no method matching setindex!(::DataFrame, ::Float64, ::Int64, ::UnitRange{Int64})`

Would you have a suggestion in my research please?

Best regards,

Thomas

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**Author:** ![oheil](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/oheil/32/220745_2.png) [@oheil](https://discourse.julialang.org/u/oheil)\
**Post date:** [December 11, 2020, 12:52pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/2 "2020-12-11T12:52:42Z")

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Seems to be very easy but your problem description seems to be overly complicated.

Even if I think the the answer you are looking for is very simple (I would give it if I would be sure which one it is), I think it is best if you first go through this

> [@Please read: make it easier to help you](https://discourse.julialang.org/t/psa-make-it-easier-to-help-you/14757):
>
> Welcome to the Julia Discourse! We are enthusiastic about helping Julia programmers, both beginner and experienced. This public service announcement (PSA) outlines best practices when asking for help. Following these points makes it easier for us to help you and more likely you’ll get a prompt, useful answer. Keywords are highlighted to make it easier to refer to specific points. Choose a descriptive title that captures the key part of your question, eg “plots with multiple axes” instead of …

and ask again.

Or, just skip the python and Country Code stuff and just ask what you want to do as a first step in Julia. We can go step by step until you are on the road…

---

<div class="post-metadata">

**Author:** ![tlorans](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/tlorans/32/19849_2.png) [@tlorans](https://discourse.julialang.org/u/tlorans)\
**Post date:** [December 11, 2020, 1:07pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/3 "2020-12-11T13:07:18Z")

</div>

Hello,

Of course, sorry for this and thank you for the recommandations. I’ve tried to reedit my question.

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<div class="post-metadata">

**Author:** ![oheil](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/oheil/32/220745_2.png) [@oheil](https://discourse.julialang.org/u/oheil)\
**Post date:** [December 11, 2020, 1:17pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/4 "2020-12-11T13:17:31Z")

</div>

Well done.  
Here is a slightly more Julia style version of the iteration (only changing column “A”). This part is still unclear, which column you want to change:

```julia
using DataFrames
t = 100
y = DataFrame("A"=>ones(t),"B"=>zeros(t))
for t in 2:t
    y[t,1] = y[t-1,1] + 1
end

```

I am not going into high efficiency, just more tutorial style and easy to comprehend.  
Note: Uppercase is style for types, variables should be lowercase starting.

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<div class="post-metadata">

**Author:** ![tlorans](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/tlorans/32/19849_2.png) [@tlorans](https://discourse.julialang.org/u/tlorans)\
**Post date:** [December 11, 2020, 1:27pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/5 "2020-12-11T13:27:35Z")

</div>

Thank you for your answer. Sorry for my unclear question.

Indeed, I would like to apply operations or assignments to all columns, such as:

```julia
using DataFrames
n = 100
y = DataFrame("A"=>zeros(n),"B"=>zeros(n))
for t in 1:n
    if t == 1
        y[t,1:end] = 1
    else
        y[t,1:end] = y[t-1,1:end] + 1
    end
end

```

However, trying this I’ve got the following error:

```julia
ERROR: MethodError: no method matching setindex!(::DataFrame, ::Float64, ::Int64, ::UnitRange{Int64})

```

Thank you for your help !

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<div class="post-metadata">

**Author:** ![oheil](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/oheil/32/220745_2.png) [@oheil](https://discourse.julialang.org/u/oheil)\
**Post date:** [December 11, 2020, 2:54pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/6 "2020-12-11T14:54:11Z")

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The Julia style solution would be broadcasting, but unfortunately this is currently not implemented over `DataFrameRow`. I found this discussion about this: [julia - Is there a way to subtract multiple dataframe columns at once? - Stack Overflow](https://stackoverflow.com/questions/63752879/is-there-a-way-to-subtract-multiple-dataframe-columns-at-once)

It would look like:

```julia
using DataFrames
n = 100
y = DataFrame("A"=>zeros(n),"B"=>zeros(n))
for t in 1:n
    if t == 1
        y[t,1:end] .= 1
    else
        y[t,1:end] .= ( y[t-1,1:end] .+ 1 )
    end
end

```

Which gives the error:

```julia
ERROR: ArgumentError: broadcasting over `DataFrameRow`s is reserved

```

For broadcast in general see: [Multi-dimensional Arrays · The Julia Language](https://docs.julialang.org/en/v1/manual/arrays/#Broadcasting)

The workaround (from above discussion) is:

```julia
using DataFrames
n = 100
y = DataFrame("A"=>zeros(n),"B"=>zeros(n))
for t in 1:n
    if t == 1
        y[t,1:end] .= 1
    else
        y[t,1:end] .= ( Vector(y[t-1,1:end]) .+ 1 )
    end
end

```

But I am not happy with this code. Depending on your real goal it is probably better just to do the processing for each column separately, as the columns seem to be independent from each other (but as I said, real peformance implementation needs the complete problem to know).

This is better because Julia arrays are column-major, see  
[https://docs.julialang.org/en/v1/manual/performance-tips/#man-performance-column-major](https://docs.julialang.org/en/v1/manual/performance-tips/#man-performance-column-major)

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<div class="post-metadata">

**Author:** ![pdeffebach](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/pdeffebach/32/10320_2.png) [@pdeffebach](https://discourse.julialang.org/u/pdeffebach)\
**Post date:** [December 11, 2020, 5:50pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/7 "2020-12-11T17:50:30Z")

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This is indeed a sub-optimal scenario, but your code looks good.

The reason you have to convert to vectors is because a `DataFrameRow` tries to have a very similar API as a `NamedTuple`. `NamedTuple`s currently do not support this kind of broadcasting, and we want to match that behavior for whatever they do eventually decide to do with broadcasting.

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**Author:** ![nilshg](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/nilshg/32/2283_2.png) [@nilshg](https://discourse.julialang.org/u/nilshg)\
**Post date:** [December 11, 2020, 7:30pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/8 "2020-12-11T19:30:37Z")

</div>

I would encourage you to post a more complete description of what you’re actually trying to achieve in order to avoid the danger of causing an [XY problem](https://en.wikipedia.org/wiki/XY_problem).

In particular, it feels to me like a DataFrame isn’t necessarily the right data structure for your use case - just because something was done in `pandas` doesn’t mean it has to be a DataFrame in Julia! You might be better off with a simple `Array{Float64, 2}`, or maybe a [`NamedArray`](https://github.com/davidavdav/NamedArrays.jl), or one of the many other low- or zero cost abstractions the Julia language offers to organise your data & algorithm.

---

<div class="post-metadata">

**Author:** ![tlorans](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/tlorans/32/19849_2.png) [@tlorans](https://discourse.julialang.org/u/tlorans)\
**Post date:** [December 12, 2020, 3:19pm UTC](https://discourse.julialang.org/t/iterating-over-row-in-a-dataframe/51662/9 "2020-12-12T15:19:51Z")

</div>

Thank you all for your answers.

Indeed, [NamedArrays.jl](https://github.com/davidavdav/NamedArrays.jl) does the job I need:

```julia
using NamedArrays

columnsnames = ["A","B"]
c = length(columnsnames)
n = 100
years = zeros(n)
start_date = 2020
years[1] = start_date

for t in 2:n
    years[t] = years[t-1] + 1
end

y = NamedArray((zeros(n,c)), (years, columnsnames)) 

for t in 1:n
    if t == 1
        y[t,1:end] .= 1
    else
        y[t,1:end] .= y[t-1,1:end] .+ 1
    end
end

println(y)

```