CODEWITHSUNDEEP
pythonaws-lambda
Alejandro Marín
Alejandro Marín

Reputation: 73

TypeError: unsupported operand type(s) for /: 'str' and 'str' (AWS lambda)

I am writing a function in AWS lambda to extract monthly unemployment data from some Spanish cities. The code when I try it in local works perfectly, but when I do it in lambda, I get that error. The only difference is that in local I use Python 3.7 and in lambda I have Python 3.6 (I had to do it this way to make it compatible with some modules that I included in the development package. After some research, it seems that the error comes from mixing operators and strings, but I can't think of another way to express it as I need to, and I emphasize that in local it works perfectly.

Specifically, the line of code that gives this error is:

data5.to_csv("s3://xxx/xxx/ '+ str(i).split("/")[-1] + str(fecha) + '.csv'")

I have to express it this way because I need you to save every csv with the name of the city and the date of the last updated data. The complete code to understand why I have expressed it this way is this one.

Row_list=[]
    MUNICIPIOS = ['https://datosmacro.expansion.com/paro/espana/municipios/andalucia/cadiz/cadiz', 'https://datosmacro.expansion.com/paro/espana/municipios/la-rioja/la-rioja/agoncillo', 'https://datosmacro.expansion.com/paro/espana/municipios/castilla-leon/leon/bembibre']
    for i in MUNICIPIOS:
        data5 = pd.read_html(i, header = 0, decimal=',', thousands='.')
        data5 = data5[0]
        data5 = data5[["Fecha", "Tasa de Paro Registrado", "Nº de parados registrados", "Población"]]
        for index, rows in data5.iterrows():
            my_list =[rows.Fecha] 
            Row_list.append(my_list)
        fecha=[Row_list[0]]
        data5.to_csv("s3://xxx/xxx/ '+ str(i).split("/")[-1] + str(fecha) + '.csv'")

Can you think of a way to fix it? I haven't been able to fix it in a while. Thank you very much in advance!

Upvotes: 0

Views: 872

Answers (1)

3ddavies
3ddavies

Reputation: 556

Your problem is mismatched quotes. Replace the line of code that gives you an error with this:

data5.to_csv("s3://xxx/xxx/ "+ str(i).split("/")[-1] + str(fecha) + ".csv")

It is important that you close strings with the same type of quote you opened them with.

Upvotes: 2

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