Python 从映射列表中取出所有值

我想在下面换衣服

[{'course_id': '8'}, {'course_id': '16'}]

到

---

使用列表理解：

[x['course_id'] for x in data]

或使用：

---

根据Phantom的回答，这里是两种变化的速度比较：

import timeit
import operator
setup = "data = [{'course_id': '8'}, {'course_id': '16'}]; import operator"

print timeit.timeit( "[x['course_id'] for x in data]", setup, number=1000000 )
# [x['course_id'] for x in data]
# Result: 0.219176683582

print timeit.timeit( "map(operator.itemgetter('course_id'), data)", setup, number=1000000 )
# map(operator.itemgetter('course_id')
# Result: 0.476915375646

使用较长的词典，差异会变小：

# [x['course_id'] for x in data]
# Result: 1.65939845357

# map(operator.itemgetter('course_id')
# Result: 1.85912657123

这可能是函数调用的开销

import timeit
import operator
setup = "data = [{'course_id': '8'}, {'course_id': '16'}]; import operator"

print timeit.timeit( "[x['course_id'] for x in data]", setup, number=1000000 )
# [x['course_id'] for x in data]
# Result: 0.219176683582

print timeit.timeit( "map(operator.itemgetter('course_id'), data)", setup, number=1000000 )
# map(operator.itemgetter('course_id')
# Result: 0.476915375646

# [x['course_id'] for x in data]
# Result: 1.65939845357

# map(operator.itemgetter('course_id')
# Result: 1.85912657123