通过聚合特定的数据列生成新的数据列

Question

我的数据集有几个有趣的列，我想聚合这些列，并因此创建一个度量，我可以使用它进行更多的分析

我写的算法大约需要3秒钟才能完成，所以我想知道是否有更有效的方法来完成这项工作

def financial_score_calculation(df, dictionary_of_parameters):
    for parameter in dictionary_of_parameters:
        for i in dictionary_of_parameters[parameter]['target']:
            index = df.loc[df[parameter] == i].index
            for i in index:
                old_score = df.at[i, 'financialliteracyscore']
                new_score = old_score + dictionary_of_parameters[parameter]['score']
                df.at[i, 'financialliteracyscore'] = new_score
    for i in df.index:
        old_score = df.at[i, 'financialliteracyscore']
        new_score = (old_score/27.0)*100 #converting score to percent value
        df.at[i, 'financialliteracyscore'] = new_score

    return df

以下是\u参数字典\u的截断版本：

dictionary_of_parameters = {
    # money management parameters
    "SatisfactionLevelCurrentFinances": {'target': [8, 9, 10], 'score': 1},
    "WillingnessFinancialRisk": {'target': [8, 9, 10], 'score': 1},
    "ConfidenceLevelToEarn2000WithinMonth": {'target': [1], 'score': 1},
    "DegreeOfWorryAboutRetirement": {'target': [1], 'score': 1},
    "GoodWithYourMoney?": {'target': [7], 'score': 1}
    }

编辑：为df生成玩具数据

df = pd.DataFrame(columns = dictionary_of_parameters.keys())
df['financialliteracyscore'] = 0
for i in range(10):
    df.loc[i] = dict(zip(df.columns,2*i*np.ones(6)))

Answer 1

请注意，在Pandas中，您可以使用at以非元素方式进行索引。在下面的四行中，index是一个列表，然后可以用它来索引loc

for parameter in dictionary_of_parameters:
    index = df[df[parameter].isin(dictionary_of_parameters[parameter]['target'])].index
    df.loc[index,'financialliteracyscore'] += dictionary_of_parameters[parameter]['score']
df['financialliteracyscore'] = df['financialliteracyscore'] /27.0*100

这是一个参考，虽然我个人从来没有发现它在我的编程早期有用https://pandas.pydata.org/pandas-docs/stable/indexing.html