python可变参数求和

在Python函数中，还可以定义可变参数。顾名思义，可变参数就是传入的参数个数是可变的，可以是1个、2个到任意个，还可以是0个。

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我们以数学题为例子，给定一组数字a，b，c……，请计算a^2 + b^2 + c^2 + ……。

def calc(*numbers):

sum = 0

for n in numbers:

sum = sum + n * n

return sum

定义可变参数和定义一个list或tuple参数相比，仅仅在参数前面加了一个*号。在函数内部，参数numbers接收到的是一个tuple，因此，函数代码完全不变。但是，调用该函数时，可以传入任意个参数，包括0个参数：

calc(1, 2)

5

calc()

Python的函数参数总结

import math

a = abs

print(a(-1))

n1 = 255

print(str(hex(n1)))

def my_abs(x):

# 增加了参数的检查

if not isinstance(x, (int, float)):

raise TypeError('bad operand type')

if x = 0:

return x

else:

return -x

print(my_abs(-3))

def nop():

pass

if n1 = 255:

pass

def move(x, y, step, angle=0):

nx = x + step * math.cos(angle)

ny = y - step * math.sin(angle)

return nx, ny

x, y = move(100, 100, 60, math.pi / 6)

print(x, y)

tup = move(100, 100, 60, math.pi / 6)

print(tup)

print(isinstance(tup, tuple))

def quadratic(a, b, c):

k = b * b - 4 * a * c

# print(k)

# print(math.sqrt(k))

if k 0:

print('This is no result!')

return None

elif k == 0:

x1 = -(b / 2 * a)

x2 = x1

return x1, x2

else:

x1 = (-b + math.sqrt(k)) / (2 * a)

x2 = (-b - math.sqrt(k)) / (2 * a)

return x1, x2

print(quadratic(2, 3, 1))

def power(x, n=2):

s = 1

while n 0:

n = n - 1

s = s * x

return s

print(power(2))

print(power(2, 3))

def enroll(name, gender, age=8, city='BeiJing'):

print('name:', name)

print('gender:', gender)

print('age:', age)

print('city:', city)

enroll('elder', 'F')

enroll('android', 'B', 9)

enroll('pythone', '6', city='AnShan')

def add_end(L=[]):

L.append('end')

return L

print(add_end())

print(add_end())

print(add_end())

def add_end_none(L=None):

if L is None:

L = []

L.append('END')

return L

print(add_end_none())

print(add_end_none())

print(add_end_none())

def calc(*nums):

sum = 0

for n in nums:

sum = sum + n * n

return sum

print(calc(1, 2, 3))

print(calc())

l = [1, 2, 3, 4]

print(calc(*l))

def foo(x, y):

print('x is %s' % x)

print('y is %s' % y)

foo(1, 2)

foo(y=1, x=2)

def person(name, age, **kv):

print('name:', name, 'age:', age, 'other:', kv)

person('Elder', '8')

person('Android', '9', city='BeiJing', Edu='人民大学')

extra = {'city': 'Beijing', 'job': 'Engineer'}

person('Jack', 24, **extra)

def person2(name, age, *, city, job):

print(name, age, city, job)

person2('Pthon', 8, city='BeiJing', job='Android Engineer')

def person3(name, age, *other, city='BeiJing', job='Android Engineer'):

print(name, age, other, city, job)

person3('Php', 18, 'test', 1, 2, 3)

person3('Php2', 28, 'test', 1, 2, 3, city='ShangHai', job='Pyhton Engineer')

def test2(a, b, c=0, *args, key=None, **kw):

print('a =', a, 'b =', b, 'c =', c, 'args =', args, 'key=', key, 'kw =', kw)

test2(1, 2, 3, 'a', 'b', 'c', key='key', other='extra')

args = (1, 2, 3, 4)

kw = {'d': 99, 'x': '#'}

test2(*args, **kw)

python代码求解释

def calc(numbers):

sum = 0

for n in numbers:

sum = sum + n * n

print (n,numbers)

return sum

结果：

1 [1, 2, 3]

2 [1, 2, 3]

3 [1, 2, 3]

for n(循环次数)，numbers（需要的处理的数组）

list2 = [1, 2, 3, 4, 5, 6, 7 ];

print (calc(list2))

list直接放numbers就可以了

Python菜鸟求助 calc函数要怎么穿进去才不报错啊？

def calc(average):

sum=0

for i in average:

sum+=i

return (sum/3)

print(calc([1,2,3]))

文章标题：calc函数python calc函数matlab
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