Q:

Transform in laravel collection

I'm trying to do some data transformation through Laravel collection in my application for graphs. I want to create labels and datasets
I'm having a data something like this:
[
    {
        "id": 1,
        "company_id": 1,
        "year": 25,
        "turnover": "3449",
        "profit": "3201",
        "turnover_range": 25,
        "financial_year": {
            "id": 25,
            "year": "2024-25"
        }
    },
    {
        "id": 2,
        "company_id": 1,
        "year": 33,
        "turnover": "5616",
        "profit": "5905",
        "turnover_range": 25,
        "financial_year": {
            "id": 33,
            "year": "2032-33"
        }
    },
    {
        "id": 3,
        "company_id": 1,
        "year": 1,
        "turnover": "4309",
        "profit": "8563",
        "turnover_range": 175,
        "financial_year": {
            "id": 1,
            "year": "2000-01"
        }
    },
    {
        "id": 4,
        "company_id": 1,
        "year": 14,
        "turnover": "5936",
        "profit": "8605",
        "turnover_range": 25,
        "financial_year": {
            "id": 14,
            "year": "2013-14"
        }
    },
    {
        "id": 5,
        "company_id": 1,
        "year": 29,
        "turnover": "7156",
        "profit": "3844",
        "turnover_range": 75,
        "financial_year": {
            "id": 29,
            "year": "2028-29"
        }
    },
    {
        "id": 6,
        "company_id": 1,
        "year": 6,
        "turnover": "5868",
        "profit": "633",
        "turnover_range": 75,
        "financial_year": {
            "id": 6,
            "year": "2005-06"
        }
    },
    {
        "id": 7,
        "company_id": 1,
        "year": 25,
        "turnover": "5809",
        "profit": "6831",
        "turnover_range": 575,
        "financial_year": {
            "id": 25,
            "year": "2024-25"
        }
    },
    {
        "id": 8,
        "company_id": 1,
        "year": 12,
        "turnover": "1",
        "profit": "1976",
        "turnover_range": 25,
        "financial_year": {
            "id": 12,
            "year": "2011-12"
        }
    },
    {
        "id": 9,
        "company_id": 1,
        "year": 30,
        "turnover": "680",
        "profit": "1222",
        "turnover_range": 25,
        "financial_year": {
            "id": 30,
            "year": "2029-30"
        }
    },
    {
        "id": 10,
        "company_id": 1,
        "year": 26,
        "turnover": "8197",
        "profit": "3687",
        "turnover_range": 25,
        "financial_year": {
            "id": 26,
            "year": "2025-26"
        }
    }
]

Which is coming from eloquent collection of my model:
$fRA =  FinancialAndRisk::whereHas('company', function($q) use($request){
    $q->where('slug', $request->slug);
})
    ->with('financialYear')
    ->get();

I want to pluck out all the year inside financial_year as labels so i tried:
$labels = $fRA->pluck('financial_year.year');

it is showing null.
Again I tried
$labels= $fRA->map(function ($item){
    $item->fin_year = $item->financial_year['year'];
    return $item;
})->pluck('fin_year');

Even if I do transform I am getting the same null result, 
Any ideas appreciated. Thanks
Edit:
Data format to be something like this:
labels = ['2000-01','2032-33','2024-25','2005-06'];

A:

You should just be able to do something like:
$labels = $fRA->map(function ($item) {
    return $item->financial_year->year;
})->unique();

Obviously, remove unique() if you want to include the duplicates (if there are any).
Alternatively, if you have the hasMany relationship set up in your FinancialYear model and you don't need $fRA after this you could just do:
$labels = FinancialYear::whereHas('FinancialAndRisk.company', function ($query) use($request) {
    $query->where('slug', $request->slug);
})->pluck('year');

