Question

我需要以最有效的方式从字典中删除一些项目，现在是RIght，我正在使用stmt，如下所示。思考相同的事情应该在几行中完成。

for eachitem in dicta:
                del eachitem['NAME']
                del eachitem['STATE']
                del eachitem['COUNTRY']
                del eachitem['REGION']
                del eachitem['LNAME']

dicta = [{'name','Bob','STATE':'VA','COUNTRY':'US','REGION':'MIDWEST','LNAME':'Brian',Salary:6000}]

我只想删除字典中的工资项。任何意见都表示赞赏。

Answer 1

如果你的示例数据是你正在处理的，而不是删除元素，只需用那个单独的键重新创建你的dict

dicta = [{'Salary':e['Salary']} for e in dicta]

或者对我而言，更有意义的是，只创建一个列表而不是dicts列表

dicta = [e['Salary'] for e in dicta]

但取决于您计划实现的目标

Answer 2

我想你可以使用：

for eachitem in dicta:
    for k in ['NAME','STATE','COUNTRY','REGION','LNAME']:
        del eachitem[k]

或者，如果您只想要1个键：

for eachitem in dicta:
    salary = eachitem['SALARY']
    eachitem.clear()
    eachitem['SALARY'] = salary

这样可以确保您想要的所有内容 - 否则，您只需通过以下方式即可完成：

eachitem = {'SALARY':eachitem['SALARY']}

Answer 3

编辑：我说过dict理解最快：我错了。我不小心跑了%timeit test3而不是%timeit test3()。见下面的结果。

def test1():
    dicta = [{'NAME':'Bob','STATE':'VA','COUNTRY':'US','REGION':'MIDWEST','LNAME':'Brian','Salary':6000}]
    remove = ['NAME','STATE','COUNTRY','REGION','LNAME']
    for d in dicta:
        for r in remove:
            d.pop(r)

def test2():
    dicta = [{'NAME':'Bob','STATE':'VA','COUNTRY':'US','REGION':'MIDWEST','LNAME':'Brian','Salary':6000}]
    remove = ['NAME','STATE','COUNTRY','REGION','LNAME']
    for d in dicta:
        for r in remove:
            del d[r]

def test3():
    dicta = [{'NAME':'Bob','STATE':'VA','COUNTRY':'US','REGION':'MIDWEST','LNAME':'Brian','Salary':6000}]
    remove = ['NAME','STATE','COUNTRY','REGION','LNAME']    
    dicta = [{k:v for k,v in d.iteritems() if k not in remove } for d in dicta]

# this is really what OP was looking for, the other 3 tests are more generalized.    
def test4():    
    dicta = [{'NAME':'Bob','STATE':'VA','COUNTRY':'US','REGION':'MIDWEST','LNAME':'Brian','Salary':6000}]
    remove = ['NAME','STATE','COUNTRY','REGION','LNAME']    
    dicta = [e['Salary'] for e in dicta]

%timeit test1()
# 100000 loops, best of 3: 2.32 us per loop    
%timeit test2()
# 1000000 loops, best of 3: 1.68 us per loop    
%timeit test3()
# 100000 loops, best of 3: 3.23 us per loop
%timeit test4()
# 1000000 loops, best of 3: 1.46 us per loop

从python词典中删除元素

3 个答案: