You can just use a list comprehension:
my_list = [1, 2, 3, 4, 5]
my_new_list = [i * 5 for i in my_list]
>>> print[my_new_list]
[5, 10, 15, 20, 25]
Note that a list comprehension is generally a more efficient way to do a for
loop:
my_new_list = []
for i in my_list:
my_new_list.append[i * 5]
>>> print[my_new_list]
[5, 10, 15, 20, 25]
As an alternative, here is a solution using the popular Pandas package:
import pandas as pd
s = pd.Series[my_list]
>>> s * 5
0 5
1 10
2 15
3 20
4 25
dtype: int64
Or, if you just want the list:
>>> [s * 5].tolist[]
[5, 10, 15, 20, 25]
Finally, one could use map
, although this is generally
frowned upon.
my_new_list = map[lambda x: x * 5, my_list]
Using map
, however, is generally less efficient. Per a comment from ShadowRanger on a deleted answer to this question:
The reason "no one" uses it is that, in general, it's a performance pessimization. The only time it's worth considering
map
in CPython is if you're using a built-in function implemented in C as the mapping function; otherwise,map
is going to run equal to or slower than the more Pythonic listcomp or genexpr [which are also more explicit about whether they're lazy generators or eagerlist
creators; on Py3, your code wouldn't work without wrapping themap
call inlist
]. If you're usingmap
with alambda
function, stop, you're doing it wrong.
And another one of his comments posted to this reply:
Please don't teach people to use
map
withlambda
; the instant you need alambda
, you'd have been better off with a list comprehension or generator expression. If you're clever, you can makemap
work withoutlambda
s a lot, e.g. in this case,map[[5].__mul__, my_list]
, although in this particular case, thanks to some optimizations in the byte code interpreter for simpleint
math,[x * 5 for x in my_list]
is faster, as well as being more Pythonic and simpler.
Given a list, print the value obtained after multiplying all numbers in a list.
Examples:
Input : list1 = [1, 2, 3] Output : 6 Explanation: 1*2*3=6
Input : list1 = [3, 2, 4] Output : 24
Method 1: Traversal
Initialize the value of the product to 1[not 0 as 0 multiplied with anything returns zero]. Traverse till the end of the list, multiply every number with the product. The value stored in the product at the end will give you your final answer.
Below is the Python implementation of the above approach:
Python
def
multiplyList[myList]:
result
=
1
for
x
in
myList:
result
=
result
*
x
return
result
list1
=
[
1
,
2
,
3
]
list2
=
[
3
,
2
,
4
]
print
[multiplyList[list1]]
print
[multiplyList[list2]]
Method 2: Using numpy.prod[]
We can use numpy.prod[] from import numpy to get the multiplication of all the numbers in the list. It returns an integer or a float value depending on the multiplication result.
Below is the Python3 implementation of the above approach:
Python3
import
numpy
list1
=
[
1
,
2
,
3
]
list2
=
[
3
,
2
,
4
]
result1
=
numpy.prod[list1]
result2
=
numpy.prod[list2]
print
[result1]
print
[result2]
Output:
6 24
Method 3 Using lambda function: Using numpy.array
Lambda’s definition does not include a “return” statement, it always contains an expression that is returned. We can also put a lambda definition anywhere a function is expected, and we don’t have to assign it to a variable at all. This is the simplicity of lambda functions. The reduce[] function in Python takes in a function and a list as an argument. The function is called with a lambda function and a list and a new reduced result is returned. This performs a repetitive operation over the pairs of the list.
Below is the Python3 implementation of the above approach:
Python3
from
functools
import
reduce
list1
=
[
1
,
2
,
3
]
list2
=
[
3
,
2
,
4
]
result1
=
reduce
[[
lambda
x, y: x
*
y], list1]
result2
=
reduce
[[
lambda
x, y: x
*
y], list2]
print
[result1]
print
[result2]
Method 4 Using prod function of math library: Using math.prod
Starting Python 3.8, a prod function has been included in the math module in the standard library, thus no need to install external libraries.
Below is the Python3 implementation of the above approach:
Python3
import
math
list1
=
[
1
,
2
,
3
]
list2
=
[
3
,
2
,
4
]
result1
=
math.prod[list1]
result2
=
math.prod[list2]
print
[result1]
print
[result2]
Output:
6 24
Method 5: Using mul[] function of operator module.
First we have to import the operator module then using the mul[] function of operator module multiplying the all values in the list.
Python3
from
operator
import
*
list1
=
[
1
,
2
,
3
]
m
=
1
for
i
in
list1:
m
=
mul[i, m]
print
[m]
Method 6: Using traversal by index
Python3
def
multiplyList[myList] :
result
=
1
for
i
in
range
[
0
,
len
[myList]]:
result
=
result
*
myList[i]
return
result
list1
=
[
1
,
2
,
3
]
list2
=
[
3
,
2
,
4
]
print
[multiplyList[list1]]
print
[multiplyList[list2]]
Method 7: Using itertools.accumulate
Python
from
itertools
import
accumulate
list1
=
[
1
,
2
,
3
]
list2
=
[
3
,
2
,
4
]
result1
=
list
[accumulate[list1, [
lambda
x, y: x
*
y]]]
result2
=
list
[accumulate[list2, [
lambda
x, y: x
*
y]]]
print
[result1[
-
1
]]
print
[result2[
-
1
]]
Output:
6 24