在Python中，如何用变量替换函数调用的一部分？

Question

我有一本字典，里面有一些函数表达式作为值。除了中间部分，每个值非常相似。在下面的示例中，长公式中只有earn_yld、free_cash_flow_yield和eps_growth不同

factor_bql = {
    "ltm_earnings_yield": bq.func.dropna(bq.data.earn_yld(as_of_date=bq.func.RANGE(params['start'],params['end']))),
    "ltm_fcf_yield": bq.func.dropna(bq.data.free_cash_flow_yield(as_of_date=bq.func.RANGE(params['start'],params['end']))),
    'ltm_eps_growth': bq.func.dropna(bq.data.eps_growth(as_of_date=bq.func.RANGE(params['start'],params['end'])))
}

有没有办法写一个函数或变量来简化字典中的值

def simple_formula(xyz):
    ... ...

factor_bql = {
    "ltm_earnings_yield": simple_formula('earn_yld'),
    "ltm_fcf_yield": simple_formula('free_cash_flow_yield'),
    'ltm_eps_growth': simple_formula('eps_growth')
}

Answer 1

假设bq.data是某个对象：

def simple_formula(xyz):
    method = getattr(bq.data, xyx) # get a method by its name
    return bq.func.dropna(method(as_of_date=bq.func.RANGE(params['start'],params['end'])))