判断一个 list 是否为空
传统的方式:
if len(mylist):
# Do something with my list
else:
# The list is empty
由于一个空 list 本身等同于 False,所以可以直接:
if mylist:
# Do something with my list
else:
# The list is empty
遍历 list 的同时获取索引
传统的方式:
i = 0
for element in mylist:
# Do something with i and element
i += 1
这样更简洁些:
for i, element in enumerate(mylist):
# Do something with i and element
pass
list 排序
在包含某元素的列表中依据某个属性排序是一个很常见的操作。例如这里我们先创建一个包含 person 的 list:
class Person(object):
def __init__(self, age):
self.age = age
persons = [Person(age) for age in (14, 78, 42)]
传统的方式是:
def get_sort_key(element):
return element.age
for element in sorted(persons, key=get_sort_key):
print "Age:", element.age
更加简洁、可读性更好的方法是使用 Python 标准库中的 operator 模块:
from operator import attrgetter
for element in sorted(persons, key=attrgetter('age')):
print "Age:", element.age
attrgetter 方法优先返回读取的属性值作为参数传递给 sorted 方法。operator 模块还包括 itemgetter 和 methodcaller 方法,作用如其字面含义。
list解析
python有一个非常有意思的功能,就是list解析,就是这样的:
>>> squares = [x**2 for x in range(1,10)]
>>> squares
[1, 4, 9, 16, 25, 36, 49, 64, 81]
看到这个结果,看官还不惊叹吗?这就是python,追求简洁优雅的python!
其官方文档中有这样一段描述,道出了list解析的真谛:
List comprehensions provide a concise way to create lists. Common applications are to make new lists where each element is the result of some operations applied to each member of another sequence or iterable, or to create a subsequence of those elements that satisfy a certain condition.
还记得前面一讲中的那个问题吗?
找出100以内的能够被3整除的正整数。
我们用的方法是:
aliquot = []
for n in range(1,100):
if n%3 == 0:
aliquot.append(n)
print aliquot
好了。现在用list解析重写,会是这样的:
>>> aliquot = [n for n in range(1,100) if n%3==0]
>>> aliquot
[3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99]
震撼了。绝对牛X!
其实,不仅仅对数字组成的list,所有的都可以如此操作。请在平复了激动的心之后,默默地看下面的代码,感悟一下list解析的魅力。
>>> mybag = [' glass',' apple','green leaf '] #有的前面有空格,有的后面有空格
>>> [one.strip() for one in mybag] #去掉元素前后的空格
['glass', 'apple', 'green leaf']
enumerate
这是一个有意思的内置函数,本来我们可以通过for i in range(len(list))的方式得到一个list的每个元素编号,然后在用list[i]的方式得到该元素。如果要同时得到元素编号和元素怎么办?就是这样了:
>>> for i in range(len(week)):
... print week[i]+' is '+str(i) #注意,i是int类型,如果和前面的用+连接,必须是str类型
...
monday is 0
sunday is 1
friday is 2
python中提供了一个内置函数enumerate,能够实现类似的功能
>>> for (i,day) in enumerate(week):
... print day+' is '+str(i)
...
monday is 0
sunday is 1
friday is 2
算是一个有意思的内置函数了,主要是提供一个简单快捷的方法。
官方文档是这么说的:
Return an enumerate object. sequence must be a sequence, an iterator, or some other object which supports iteration. The next() method of the iterator returned by enumerate() returns a tuple containing a count (from start which defaults to 0) and the values obtained from iterating over sequence:
顺便抄录几个例子,供看官欣赏,最好实验一下。
>>> seasons = ['Spring', 'Summer', 'Fall', 'Winter']
>>> list(enumerate(seasons))
[(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]
>>> list(enumerate(seasons, start=1))
[(1, 'Spring'), (2, 'Summer'), (3, 'Fall'), (4, 'Winter')]
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