如何有效地修改列表列表的所有元素并将其添加到现有列表列表中

Question

Suppose I have a list of list that looks like this:

myList = [[1.,1.,6.],[2.,4.,4.],[3.,3.,3.]]

Now I want to apply a certain function to each list in myList which - for the sake of simplicity - looks like this (it just divides each element of the list by the sum of this list).

def changeVal(l):
    return map(lambda x: x/sum(l),l)

Applying it to myList gives me:

modList = map(lambda x: changeVal(x), myList)
[[0.125, 0.125, 0.75],
[0.2, 0.4, 0.4],
[0.3333333333333333, 0.3333333333333333, 0.3333333333333333]]

But what I actually want is to add all elements of modList to myList . I could do this in a for-loop:

for sl in modList:
    myList.append(sl)

which gives me the desired output:

[[1.0, 1.0, 6.0],
 [2.0, 4.0, 4.0],
 [3.0, 3.0, 3.0],
 [0.125, 0.125, 0.75],
 [0.2, 0.4, 0.4],
 [0.3333333333333333, 0.3333333333333333, 0.3333333333333333]]

However, I want to do this without using a for-loop since append is slow. I tried:

myList.append(*modList)
myList.extend(*modList)

which both give me an TypeError :

TypeError: extend() takes exactly one argument (3 given)

Here is the code again and the desired output:

myList = [[1.,1.,6.],[2.,4.,4.],[3.,3.,3.]]
def changeVal(l):
    return map(lambda x: x/sum(l),l)
modList = map(lambda x: changeVal(x), myList)

How to add the elements of modList to myList in order to get the following output?

[[1.0, 1.0, 6.0],
 [2.0, 4.0, 4.0],
 [3.0, 3.0, 3.0],
 [0.125, 0.125, 0.75],
 [0.2, 0.4, 0.4],
 [0.3333333333333333, 0.3333333333333333, 0.3333333333333333]]

Answer 1

The correct function to use here would be extend() , but extend accepts the list you want to extend with, not the elements of that list. Try something like -

myList.extend(modList)

Example -

>>> l = [[1,2],[3,4]]
>>> l1 = [[5,6],[7,8]]
>>> l.extend(l1)
>>>
>>> l
[[1, 2], [3, 4], [5, 6], [7, 8]]

Though if modList is not used anywhere else, you can avoid creation of the intermediate list altogether, By doing -

myList.extend(map(lambda x: changeVal(x), myList))

Timeit results for both method -

In [42]: def func1():
   ....:     l = [[1,2],[3,4]]
   ....:     l1 = [[5,6],[7,8]]
   ....:     for i in l1:
   ....:         l.append(i)
   ....:     return l
   ....:

In [43]: def func2():
   ....:     l = [[1,2],[3,4]]
   ....:     l1 = [[5,6],[7,8]]
   ....:     l.extend(l1)
   ....:     return l
   ....:

In [44]:

In [44]: %timeit func1()
The slowest run took 8.35 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1 µs per loop

In [45]: %timeit func2()
The slowest run took 9.11 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 794 ns per loop

In [47]: %timeit func1()
The slowest run took 7.74 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 983 ns per loop

In [46]: %timeit func2()
1000000 loops, best of 3: 799 ns per loop

So using .extend() is a little bit faster than for loop and .append() .