问题:如何将列表中的所有项目与Python相乘?
我需要编写一个函数,该函数接受一个数字列表并将它们相乘。例子:
[1,2,3,4,5,6]
会给我的1*2*3*4*5*6
。我真的可以用你的帮助。
I need to write a function that takes
a list of numbers and multiplies them together. Example:
[1,2,3,4,5,6]
will give me 1*2*3*4*5*6
. I could really use your help.
回答 0
Python 3:使用functools.reduce
:
>>> from functools import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
Python 2:使用reduce
:
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
为了与2和3兼容,请使用pip install six
:
>>> from six.moves import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
Python 3: use functools.reduce
:
>>> from functools import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
Python 2: use reduce
:
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
For compatible with 2 and 3 use pip install six
, then:
>>> from six.moves import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
回答 1
您可以使用:
import operator
import functools
functools.reduce(operator.mul, [1,2,3,4,5,6], 1)
有关说明,请参见reduce
和operator.mul
文档。
您需要import functools
Python 3+中的代码行。
You can use:
import operator
import functools
functools.reduce(operator.mul, [1,2,3,4,5,6], 1)
See reduce
and operator.mul
documentations for an explanation.
You need the import functools
line in Python 3+.
回答 2
我会使用numpy.prod
来执行任务。见下文。
import numpy as np
mylist = [1, 2, 3, 4, 5, 6]
result = np.prod(np.array(mylist))
I would use the numpy.prod
to perform the task. See below.
import numpy as np
mylist = [1, 2, 3, 4, 5, 6]
result = np.prod(np.array(mylist))
回答 3
如果要避免导入任何内容并避免使用更复杂的Python区域,则可以使用简单的for循环
product = 1 # Don't use 0 here, otherwise, you'll get zero
# because anything times zero will be zero.
list = [1, 2, 3]
for x in list:
product *= x
If you want to avoid importing anything and avoid more complex areas of Python, you can use a simple for loop
product = 1 # Don't use 0 here, otherwise, you'll get zero
# because anything times zero will be zero.
list = [1, 2, 3]
for x in list:
product *= x
回答 4
从开始Python 3.8
,.prod
函数已经包含math
在标准库的模块中:
math.prod(iterable, *, start=1)
该方法返回一个start
值(默认值:1)乘以数字可迭代的乘积:
import math
math.prod([1, 2, 3, 4, 5, 6])
>>> 720
如果iterable为空,则将产生1
(或start
提供值,如果提供)。
Starting Python 3.8
, a .prod
function has been included to the math
module in the standard library:
math.prod(iterable, *, start=1)
The method returns the product of a start
value (default: 1) times an iterable of numbers:
import math
math.prod([1, 2, 3, 4, 5, 6])
>>> 720
If the iterable is empty, this will produce 1
(or the start
value, if provided).
回答 5
这是我机器上的一些性能指标。与在长时间运行的循环中对较小的输入执行此操作有关:
import functools, operator, timeit
import numpy as np
def multiply_numpy(iterable):
return np.prod(np.array(iterable))
def multiply_functools(iterable):
return functools.reduce(operator.mul, iterable)
def multiply_manual(iterable):
prod = 1
for x in iterable:
prod *= x
return prod
sizesToTest = [5, 10, 100, 1000, 10000, 100000]
for size in sizesToTest:
data = [1] * size
timerNumpy = timeit.Timer(lambda: multiply_numpy(data))
timerFunctools = timeit.Timer(lambda: multiply_functools(data))
timerManual = timeit.Timer(lambda: multiply_manual(data))
repeats = int(5e6 / size)
resultNumpy = timerNumpy.timeit(repeats)
resultFunctools = timerFunctools.timeit(repeats)
resultManual = timerManual.timeit(repeats)
print(f'Input size: {size:>7d} Repeats: {repeats:>8d} Numpy: {resultNumpy:.3f}, Functools: {resultFunctools:.3f}, Manual: {resultManual:.3f}')
结果:
Input size: 5 Repeats: 1000000 Numpy: 4.670, Functools: 0.586, Manual: 0.459
Input size: 10 Repeats: 500000 Numpy: 2.443, Functools: 0.401, Manual: 0.321
Input size: 100 Repeats: 50000 Numpy: 0.505, Functools: 0.220, Manual: 0.197
Input size: 1000 Repeats: 5000 Numpy: 0.303, Functools: 0.207, Manual: 0.185
Input size: 10000 Repeats: 500 Numpy: 0.265, Functools: 0.194, Manual: 0.187
Input size: 100000 Repeats: 50 Numpy: 0.266, Functools: 0.198, Manual: 0.185
您会看到Numpy在较小的输入上要慢得多,因为它在执行乘法之前会分配一个数组。另外,请注意Numpy中的溢出。
Here’s some performance measurements from my machine. Relevant in case this is performed for small inputs in a long-running loop:
import functools, operator, timeit
import numpy as np
def multiply_numpy(iterable):
return np.prod(np.array(iterable))
def multiply_functools(iterable):
return functools.reduce(operator.mul, iterable)
def multiply_manual(iterable):
prod = 1
for x in iterable:
prod *= x
return prod
sizesToTest = [5, 10, 100, 1000, 10000, 100000]
for size in sizesToTest:
data = [1] * size
timerNumpy = timeit.Timer(lambda: multiply_numpy(data))
timerFunctools = timeit.Timer(lambda: multiply_functools(data))
timerManual = timeit.Timer(lambda: multiply_manual(data))
repeats = int(5e6 / size)
resultNumpy = timerNumpy.timeit(repeats)
resultFunctools = timerFunctools.timeit(repeats)
resultManual = timerManual.timeit(repeats)
print(f'Input size: {size:>7d} Repeats: {repeats:>8d} Numpy: {resultNumpy:.3f}, Functools: {resultFunctools:.3f}, Manual: {resultManual:.3f}')
Results:
Input size: 5 Repeats: 1000000 Numpy: 4.670, Functools: 0.586, Manual: 0.459
Input size: 10 Repeats: 500000 Numpy: 2.443, Functools: 0.401, Manual: 0.321
Input size: 100 Repeats: 50000 Numpy: 0.505, Functools: 0.220, Manual: 0.197
Input size: 1000 Repeats: 5000 Numpy: 0.303, Functools: 0.207, Manual: 0.185
Input size: 10000 Repeats: 500 Numpy: 0.265, Functools: 0.194, Manual: 0.187
Input size: 100000 Repeats: 50 Numpy: 0.266, Functools: 0.198, Manual: 0.185
You can see that Numpy is quite a bit slower on smaller inputs, since it allocates an array before multiplication is performed. Also, watch out for the overflow in Numpy.
回答 6
我个人对将通用列表的所有元素相乘的函数喜欢这样:
def multiply(n):
total = 1
for i in range(0, len(n)):
total *= n[i]
print total
它紧凑,使用简单的东西(变量和for循环),并且对我来说很直观(看起来就像我对问题的看法,只用一个,乘以它,然后乘以下一个,依此类推! )
I personally like this for a function that multiplies all elements of a generic list together:
def multiply(n):
total = 1
for i in range(0, len(n)):
total *= n[i]
print total
It’s compact, uses simple things (a variable and a for loop), and feels intuitive to me (it looks like how I’d think of the problem, just take one, multiply it, then multiply by the next, and so on!)
回答 7
简单的方法是:
import numpy as np
np.exp(np.log(your_array).sum())
The simple way is:
import numpy as np
np.exp(np.log(your_array).sum())
回答 8
Numpy
具有prod()
返回列表乘积的函数,或者在这种情况下,因为它是numpy,所以它是给定轴上数组的乘积:
import numpy
a = [1,2,3,4,5,6]
b = numpy.prod(a)
…否则您可以只导入numpy.prod()
:
from numpy import prod
a = [1,2,3,4,5,6]
b = prod(a)
Numpy
has the prod()
function that returns the product of a list, or in this case since it’s numpy, it’s the product of an array over a given axis:
import numpy
a = [1,2,3,4,5,6]
b = numpy.prod(a)
…or else you can just import numpy.prod()
:
from numpy import prod
a = [1,2,3,4,5,6]
b = prod(a)
回答 9
今天发现了这个问题,但我注意到列表中没有的情况None
。因此,完整的解决方案将是:
from functools import reduce
a = [None, 1, 2, 3, None, 4]
print(reduce(lambda x, y: (x if x else 1) * (y if y else 1), a))
对于加法,我们有:
print(reduce(lambda x, y: (x if x else 0) + (y if y else 0), a))
Found this question today but I noticed that it does not have the case where there are None
‘s in the list. So, the complete solution would be:
from functools import reduce
a = [None, 1, 2, 3, None, 4]
print(reduce(lambda x, y: (x if x else 1) * (y if y else 1), a))
In the case of addition, we have:
print(reduce(lambda x, y: (x if x else 0) + (y if y else 0), a))
回答 10
nums = str(tuple([1,2,3]))
mul_nums = nums.replace(',','*')
print(eval(mul_nums))
nums = str(tuple([1,2,3]))
mul_nums = nums.replace(',','*')
print(eval(mul_nums))
回答 11
我想通过以下方式:
def product_list(p):
total =1 #critical step works for all list
for i in p:
total=total*i # this will ensure that each elements are multiplied by itself
return total
print product_list([2,3,4,2]) #should print 48
I would like this in following way:
def product_list(p):
total =1 #critical step works for all list
for i in p:
total=total*i # this will ensure that each elements are multiplied by itself
return total
print product_list([2,3,4,2]) #should print 48
回答 12
这是我的代码:
def product_list(list_of_numbers):
xxx = 1
for x in list_of_numbers:
xxx = xxx*x
return xxx
print(product_list([1,2,3,4]))
结果:(’1 * 1 * 2 * 3 * 4’,24)
This is my code:
def product_list(list_of_numbers):
xxx = 1
for x in list_of_numbers:
xxx = xxx*x
return xxx
print(product_list([1,2,3,4]))
result : (‘1*1*2*3*4’, 24)
回答 13
如何使用递归?
def multiply(lst):
if len(lst) > 1:
return multiply(lst[:-1])* lst[-1]
else:
return lst[0]
How about using recursion?
def multiply(lst):
if len(lst) > 1:
return multiply(lst[:-1])* lst[-1]
else:
return lst[0]
回答 14
我的解决方案:
def multiply(numbers):
a = 1
for num in numbers:
a *= num
return a
pass
My solution:
def multiply(numbers):
a = 1
for num in numbers:
a *= num
return a
pass
回答 15
”’了解循环逻辑使用的唯一简单方法”’
对膝中的i而言,lap = [2,5,7,7,9] x = 1:x = i * x print(x)
”’the only simple method to understand the logic use for loop”’
Lap=[2,5,7,7,9] x=1 for i in Lap: x=i*x print(x)
回答 16
很简单,不导入任何内容。这是我的代码。这将定义一个将列表中所有项目相乘并返回其乘积的函数。
def myfunc(lst):
multi=1
for product in lst:
multi*=product
return product
It is very simple do not import anything. This is my code. This will define a function that multiplies all the items in a list and returns their product.
def myfunc(lst):
multi=1
for product in lst:
multi*=product
return product
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