python list comprehension to produce two values in

问题:

I want to generate a list in python as follows -

[1, 1, 2, 4, 3, 9, 4, 16, 5, 25 .....]

You would have figured out, it is nothing but n, n*n

I tried writing such a list comprehension in python as follows -

lst_gen = [i, i*i for i in range(1, 10)]

But doing this, gives a syntax error.

What would be a good way to generate the above list via list comprehension?

回答1:

Use itertools.chain.from_iterable:

>>> from itertools import chain
>>> list(chain.from_iterable((i, i**2) for i in xrange(1, 6)))
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25]

Or you can also use a generator function:

>>> def solve(n):
...     for i in xrange(1,n+1):
...         yield i
...         yield i**2

>>> list(solve(5))
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25]

回答2:

The question is old, but just for the curious reader, i propose another possibility: As stated on first post, you can easily make a couple (i, i**2) from a list of numbers. Then you want to flatten this couple. So just add the flatten operation in your comprehension.

[x for i in range(1, 10) for x in (i,i**2)]

回答3:

A little-known trick: list comprehensions can have multiple for clauses.

For example:

>>> [10*x+y for x in range(4) for y in range(3)]
[0, 1, 2, 10, 11, 12, 20, 21, 22, 30, 31, 32]

In your particular case, you could do:

>>> [x*x if y else x for x in range(5) for y in range(2)]
[0, 0, 1, 1, 2, 4, 3, 9, 4, 16]

回答4:

lst_gen = sum([(i, i*i) for i in range(1, 10)],())

oh I should mention the sum probably breaks the one iteration rule :(

回答5:

List comprehensions generate one element at a time. Your options are, instead, to change your loop to only generate one value at a time:

[(i//2)**2 if i % 2 else i//2 for i in range(2, 20)]

or to produce tuples then flatten the list using itertools.chain.from_iterable():

from itertools import chain

list(chain.from_iterable((i, i*i) for i in range(1, 10)))

Output:

>>> [(i//2)**2 if i % 2 else i//2 for i in range(2, 20)]
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]
>>> list(chain.from_iterable((i, i*i) for i in range(1, 10)))
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]

回答6:

You can create a list of lists then use reduce to join them.

print [[n,n*n] for n in range (10)]

[[0, 0], [1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]

print reduce(lambda x1,x2:x1+x2,[[n,n*n] for n in range (10)])

[0, 0, 1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]

 print reduce(lambda x1,x2:x1+x2,[[n**e for e in range(1,4)]\
 for n in range (1,10)])

[1, 1, 1, 2, 4, 8, 3, 9, 27, 4, 16, 64, 5, 25, 125, 6, 36, 216, 7, 49, 343, 8, 64, 512, 9, 81, 729]

Reduce takes a callable expression that takes two arguments and processes a sequence by starting with the first two items. The result of the last expression is then used as the first item in subsequent calls. In this case each list is added one after another to the first list in the list of lists and then that list is returned as a result.

List comprehensions implicitly call map with a lambda expression using the variable and sequence defined by the "for var in sequence" expression. The following is the same sort of thing.

map(lambda n:[n,n*n],range(1,10))

[[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]

I am unaware of a more natural python expression for reduce.

回答7:

Another option, might seem perverse to some

>>> from itertools import izip, tee
>>> g = xrange(1, 11)
>>> x, y = tee(g)
>>> y = (i**2 for i in y)
>>> z = izip(x, y)
>>> output = []
>>> for k in z:
...     output.extend(k)
... 
>>> print output
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81, 10, 100]

回答8:

Another option:

reduce(lambda x,y: x + [y, y*y], range(1,10), [])

回答9:

>>> lst_gen = [[i, i*i] for i in range(1, 10)]
>>> 
>>> lst_gen
[[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]
>>> 
>>> [num for elem in lst_gen for num in elem]
[1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81]

Here is my reference http://docs.python.org/2/tutorial/datastructures.html

回答10:

Try this two liner

lst = [[i, i*i] for i in range(10)]
[lst.extend(i) for i in lst]

Change math as necessary.

EVEN BETTER

#Change my_range to be the number you want range() function of
start = 1
my_range = 10
lst = [i/2 if i % 2 == 0 else ((i-1)/2)**2 for i in range(start *2, my_range*2 - 1)]

回答11:

As mentioned, itertools is the way to go. Here's how I would do it, I find it more clear:

[i if turn else i*i for i,turn in itertools.product(range(1,10), [True, False])]

回答12:

Lots of tricks in this thread. Here is another using a one liner generator without imports

x = (lamdba : [[(yield i), (yield i**2)] for i in range(10)])()

EDIT: This will raise DeprecatedWarning in Python 3.7 and SyntaxError in Python 3.8: https://docs.python.org/dev/whatsnew/3.7.html#deprecated-python-behavior