Question

例如：

def update_condition(self, type, params):
    for condition in self.conditions:
        condition_loaded = json.loads(condition)
        if condition_loaded['type'] == type:
            condition_loaded['params'] = params
            condition = json.dumps(condition_loaded)

上述代码没有任何作用，因为条件不是通过引用。这样做的正确方法是什么？

Answer 1

您可以使用enumerate：

def update_condition(self, type, params):
    for i,condition in enumerate(self.conditions):
        condition_loaded = json.loads(condition)
        if condition_loaded['type'] == type:
            condition_loaded['params'] = params
            self.conditions[i] = json.dumps(condition_loaded)

但是，总的来说，这些东西对辅助函数和列表推导有点清晰：

def helper(condition,type,params)
    loaded = json.loads(condition)
    if loaded['type'] == type:
       loaded['params'] = params
       return json.dumps(loaded)
    return condition

...

def update_condition(self, type, params):
    self.conditions = [helper(c,type,params) for c in self.conditions]

应该注意的是，第二个解决方案不会更新列表 - 换句话说，如果您对此列表有其他引用，它们将不会受到影响。如果需要，可以使用切片分配轻松地进行替换：

def update_condition(self, type, params):
    self.conditions[:] = [helper(c,type,params) for c in self.conditions]

Answer 2

您可以使用列表理解：

def update_single_condition(self, condition, type, params):
    condition_loaded = json.loads(condition)
    if condition_loaded['type'] == type:
        condition_loaded['params'] = params
    return json.dumps(condition_loaded)

def update_conditions(self, type, params): 
    self.conditions = [self.update_single_condition(c, type, params) for c in self.conditions]

在Python中迭代列表的同时修改列表的值？

2 个答案: