利用零散的时间，把python内置的函数一个一个过一遍。
Return the absolute value of a number.
The argument may be an integer or a floating point number.
If the argument is a complex number, its magnitude is returned.
返回一个数字的绝对值。
参数可以是整型，也可以是浮点数。
如果参数是复数，返回｀该复数与其共轭复数的正平方根｀ 额…
a = complex(3,4) # a = (3+4j)
abs(a) # 5.0
# Return True if all elements of the iterable are true
# (or if the iterable is empty).
# Equivalent to:
def all(iterable):
for element in iterable:
if not element:
return False
return True
如果可迭代的每一个元素都是True，返回True
a = [1,2,3]
all(a) # True
a = [1,2,0]
all(a) # False
# Return True if any element of the iterable is true.
# If the iterable is empty, return False. Equivalent to:
def any(iterable):
for element in iterable:
if element:
return True
return False
可迭代的元素有一个不为空，返回True, 所有为空False 例子
a = (1,2,0,"test")
print any(a) # True
b = (0,0,"")
print any(b) # False
filter函数,filter(f, l),用函数f将列表l的元素筛选出来
filter(lambda x : type(x) == int, [3,2,0,"",True]) #[3,2,0]
获取对象属性
# Return the value of the named attribute of object.
# name must be a string.
# If the string is the name of one of the object’s attributes,
# the result is the value of that attribute.
# For example, getattr(x, 'foobar') is equivalent to x.foobar.
# If the named attribute does not exist,
# default is returned if provided,
# otherwise AttributeError is raised.
class Play:
def play(self):
return "let's play"
p = Play()
action = getattr(p, "play)
action() # let's play
对象是否有指定的属性
# The arguments are an object and a string.
# The result is True
# if the string is the name of one of the object’s attributes,
# False if not.
# (This is implemented by calling getattr(object, name)
# and seeing whether it raises an exception or not.)
class Play:
def play(self):
return "let's play"
hasattr(Play, "play") # True
# The method join() returns a string
# in which the string elements of sequence
# have been joined by str separator.
a1 = ["a","b","c"]
a2 = "abcde"
b1 = ",".join(a1)
b2 = ",".join(a2)
print b1 # "a,b,c"
print b2 # "a,b,c,d,e"
可迭代的字符元素用指定的字符连接起来
map函数,对列表中的每个元素进行操作
f = lambda x : x * 10
l = [2,3,1]
new_l = map(f,l) # f是函数,l是列表; new_l = [20,30,10)
归纳列表
from operator import add, mul
l = [2,3,1,4]
reduce(add, l) #10, 全部相加
reduce(add, l, 10) #20, 加上初始值
reduce(mul, l) #24, 全部相乘
从每个可迭代的参数中取出元素组成元组的列表
a = [1,2]
b = [3,4]
zip(a,b) #[(1,3), (2,4)]