i would like to know how to convert a generator with only one element into list.For example if f is a generator with one element,list(f) will raise "not iterable".how to solve this problem ?thanks in advance!
nx.adamic_adar_index(g, (0, 1))
list(nx.adamic_adar_index(g, (0, 1)))
this will raise TypeError: 'int' object is not iterable,the following is OK
nx.adamic_adar_index(g, [(0, 1),(2,3)])
list(nx.adamic_adar_index(g, [(0, 1),(2,3)]))
the result is [(0, 1, 2.3878841007006875), (2, 3, 0.9282626109897467)]
If I understand correctly, this: [(0, 1),(2,3)] is a list of two elements (each being a tuple of two elements) but (0, 1) is a single element, could it be that you need to be a list of a single element, like [(0, 1)]?
I'd guess that:
list(nx.adamic_adar_index(g, [(0, 1)]))
will do what you want.
Could this be it?
The generator must be evaluated to be listed. You can't list the generator itself.
>>> def g():
... yield 1
...
>>> list(g())
[1]
>>> list(g)
Traceback (most recent call last):
File "<interactive input>", line 1, in <module>
TypeError: 'function' object is not iterable
I have two class instances which have list attributes, e.g.:
foo.range = [1,2]
bar.range = [3,4]
I also have a function which takes multiple arguments:
def permutations(*args):
return list(itertools.product(arg.range for arg in args))
I want permutations(foo, bar) to return all permutations of those two (or more) lists (i.e. [(1,3), (1,4) …]) but I actually get [([1, 2],), ([3, 4],)]
Could someone please help me to understand where am I going wrong and how to achieve the result I'm hoping for?
itertools.product expects the iterables as individual arguments(itertools.product(*iterables[, repeat])), so you need to unpack the items to it using the the splat(*) operator:
>>> def permutations(*args):
return list(itertools.product(*(arg.range for arg in args)))
>>> permutations(foo, bar)
[(1, 3), (1, 4), (2, 3), (2, 4)]
Consider this simple function, it will collect all positional arguments passed to it in args:
>>> def func(*args):
print args
print list(args[0])
...
When we pass it a generator expression it is collected as one individual item in the args tuple:
>>> func(x for x in range(5))
(<generator object <genexpr> at 0x7fda2ab4e780>,)
[0, 1, 2, 3, 4]
To fix this we need to unpack our generator expression, also as somethine like *x for x in range(5) is not a valid syntax in Python, we need to add extra parenthesis around the generator expression:
>>> func(*(x for x in range(5)))
(0, 1, 2, 3, 4)
# Failed as expected for args[0]
Traceback (most recent call last):
File "<ipython-input-215-13af4340dad1>", line 1, in <module>
func(*(x for x in range(5)))
File "<ipython-input-213-08941e2a3c06>", line 3, in func
print list(args[0])
TypeError: 'int' object is not iterable
I have a list of lists with 4 elements in each of them.
LoL=[[1,1,1,1],[4,2,3,[1,3]],[4,5,3,[0,4]]]
The 4th elements can be a list of two parts like [0,4] in [4,5,3,[0,4]].
I need to use its elements as keys for a dictionary,
Pseudo code:
dic = { [1,1,1,1]:'a',[4,2,3,[1,3]]:'b',[4,5,3,[0,4]]:'c' }
so tried to change them to tuples.
It works for simple lists (like [1,1,1,1]), but for the ones containing another list (like [4,5,3,[0,4]]) it raises an error:
dic[tuple([1,1,1,1])]=['bla','blah']
print dic
{(1, 1, 1, 1): ['bla', 'blah']}
dic[tuple([4, 2, 3, [1, 3]])]=['blablah']
TypeError: unhashable type: 'list'
I need to reuse the keys as lists later. So trying to change elements of LoL to strings (e.g. using repr()) is not an option!
Edit:
I know why lists cannot be used as dictionary keys. Here they are not changed while in the dic. I just need some way to pass them to another module to extract them.
Just convert your nested lists to nested tuples. Here's a quick demo. It's not perfect, but it works.
#! /usr/bin/env python
LoL = [[1,1,1,1],[4,2,3,[1,3]],[4,5,3,[0,4]]]
def nested_list_to_tuple(nlist):
return tuple([nested_list_to_tuple(i) if isinstance(i, list) else i for i in nlist])
ToT = nested_list_to_tuple(LoL)
print ToT
output
((1, 1, 1, 1), (4, 2, 3, (1, 3)), (4, 5, 3, (0, 4)))
Just use tuples:
a = {}
a[(4, 2, 3, (1, 3))] = ['blablah']
print(a)
Output:
{(4, 2, 3, (1, 3)): ['blablah']}
type(3,) returns the int type, while
t = 3,
type(t)
returns the tuple type.
Why?
Inside the parentheses that form the function call operator, the comma is not for building tuples, but for separating arguments. Thus, type(3, ) is equivalent to type(3). An additional comma at the end of the argument list is allowed by the grammar. You need an extra pair of parens to build a tuple:
>>> def f(x):
... print x
...
>>> f(3)
3
>>> f(3,)
3
>>> f((3,))
(3,)
The builtin type() is a function, so the comma is being parsed as an argument separator rather than a tuple constructor.
>>> type(3,)
<type 'int'>
>>> type((3,))
<type 'tuple'>
I suspect Python ignores a trailing comma in function arguments:
def f (a):
print a
print type(a)
>>> f(3,)
3
<type 'int'>
Using comma-separated values without parentheses to create a tuple is a trick that doesn't work everywhere. List comprehensions is a good example:
>>> [a,a*2 for a in range(4)]
File "<stdin>", line 1
[a,a*2 for a in range(4)]
^
You have to do this:
>>> [(a,a*2) for a in range(4)]
[(0, 0), (1, 2), (2, 4), (3, 6)]
List unpacking works fine, though, so it's a bit variable where unbounded, comma-separated values are allowed:
>>> [(a,b) for a, b in zip(range(4),range(4))]
[(0, 0), (1, 1), (2, 2), (3, 3)]
In func(bunch_of_args) you are allowed to follow the last arg with a comma, just like in
alist = [1, 2, 3, ]
s = [1,2,3,4,5,6,7,8,9]
n = 3
list(zip(*[iter(s)]*n)) # returns [(1,2,3),(4,5,6),(7,8,9)]
How does zip(*[iter(s)]*n) work? What would it look like if it was written with more verbose code?
iter() is an iterator over a sequence. [x] * n produces a list containing n quantity of x, i.e. a list of length n, where each element is x. *arg unpacks a sequence into arguments for a function call. Therefore you're passing the same iterator 3 times to zip(), and it pulls an item from the iterator each time.
x = iter([1,2,3,4,5,6,7,8,9])
print(list(zip(x, x, x)))
The other great answers and comments explain well the roles of argument unpacking and zip().
As Ignacio and ujukatzel say, you pass to zip() three references to the same iterator and zip() makes 3-tuples of the integers—in order—from each reference to the iterator:
1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9
^ ^ ^
^ ^ ^
^ ^ ^
And since you ask for a more verbose code sample:
chunk_size = 3
L = [1,2,3,4,5,6,7,8,9]
# iterate over L in steps of 3
for start in range(0,len(L),chunk_size): # xrange() in 2.x; range() in 3.x
end = start + chunk_size
print L[start:end] # three-item chunks
Following the values of start and end:
[0:3) #[1,2,3]
[3:6) #[4,5,6]
[6:9) #[7,8,9]
FWIW, you can get the same result with map() with an initial argument of None:
>>> map(None,*[iter(s)]*3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
For more on zip() and map(): http://muffinresearch.co.uk/archives/2007/10/16/python-transposing-lists-with-map-and-zip/
I think one thing that's missed in all the answers (probably obvious to those familiar with iterators) but not so obvious to others is -
Since we have the same iterator, it gets consumed and the remaining elements are used by the zip. So if we simply used the list and not the iter
eg.
l = range(9)
zip(*([l]*3)) # note: not an iter here, the lists are not emptied as we iterate
# output
[(0, 0, 0), (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6), (7, 7, 7), (8, 8, 8)]
Using iterator, pops the values and only keeps remaining available, so for zip once 0 is consumed 1 is available and then 2 and so on. A very subtle thing, but quite clever!!!
iter(s) returns an iterator for s.
[iter(s)]*n makes a list of n times the same iterator for s.
So, when doing zip(*[iter(s)]*n), it extracts an item from all the three iterators from the list in order. Since all the iterators are the same object, it just groups the list in chunks of n.
One word of advice for using zip this way. It will truncate your list if it's length is not evenly divisible. To work around this you could either use itertools.izip_longest if you can accept fill values. Or you could use something like this:
def n_split(iterable, n):
num_extra = len(iterable) % n
zipped = zip(*[iter(iterable)] * n)
return zipped if not num_extra else zipped + [iterable[-num_extra:], ]
Usage:
for ints in n_split(range(1,12), 3):
print ', '.join([str(i) for i in ints])
Prints:
1, 2, 3
4, 5, 6
7, 8, 9
10, 11
Unwinding layers of "cleverness", you may find this equivalent spelling easier to follow:
x = iter(s)
for a, b, c in zip(*([x] * n)):
print(a, b, c)
which is, in turn, equivalent to the even less-clever:
x = iter(accounts_iter)
for a, b, c in zip(x, x, x):
print(a, b, c)
Now it should start to become clear. There is only a single iterator object, x. On each iteration, zip(), under the covers, calls next(x) 3 times, once for each iterator object passed to it. But it's the same iterator object here each time. So it delivers the first 3 next(x) results, and leaves the shared iterator object waiting to deliver its 4th result next. Lather, rinse, repeat.
BTW, I suspect you're parsing *([iter(x)]*n) incorrectly in your head. The trailing *n happens first, and then the prefix * is applied to the n-element list *n created. f(*iterable) is a shortcut for calling f() with a variable number of arguments, one for each object iterable delivers.
I needed to break down each partial step to really internalize how it is working. My notes from the REPL:
>>> # refresher on using list multiples to repeat item
>>> lst = list(range(15))
>>> lst
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
>>> # lst id value
>>> id(lst)
139755081359872
>>> [id(x) for x in [lst]*3]
[139755081359872, 139755081359872, 139755081359872]
# replacing lst with an iterator of lst
# It's the same iterator three times
>>> [id(x) for x in [iter(lst)]*3 ]
[139755085005296, 139755085005296, 139755085005296]
# without starred expression zip would only see single n-item list.
>>> print([iter(lst)]*3)
[<list_iterator object at 0x7f1b440837c0>, <list_iterator object at 0x7f1b440837c0>, <list_iterator object at 0x7f1b440837c0>]
# Must use starred expression to expand n arguments
>>> print(*[iter(lst)]*3)
<list_iterator object at 0x7f1b4418b1f0> <list_iterator object at 0x7f1b4418b1f0> <list_iterator object at 0x7f1b4418b1f0>
# by repeating the same iterator, n-times,
# each pass of zip will call the same iterator.__next__() n times
# this is equivalent to manually calling __next__() until complete
>>> iter_lst = iter(lst)
>>> ((iter_lst.__next__(), iter_lst.__next__(), iter_lst.__next__()))
(0, 1, 2)
>>> ((iter_lst.__next__(), iter_lst.__next__(), iter_lst.__next__()))
(3, 4, 5)
>>> ((iter_lst.__next__(), iter_lst.__next__(), iter_lst.__next__()))
(6, 7, 8)
>>> ((iter_lst.__next__(), iter_lst.__next__(), iter_lst.__next__()))
(9, 10, 11)
>>> ((iter_lst.__next__(), iter_lst.__next__(), iter_lst.__next__()))
(12, 13, 14)
>>> ((iter_lst.__next__(), iter_lst.__next__(), iter_lst.__next__()))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
# all together now!
# continuing with same iterator multiple times in list
>>> print(*[iter(lst)]*3)
<list_iterator object at 0x7f1b4418b1f0> <list_iterator object at 0x7f1b4418b1f0> <list_iterator object at 0x7f1b4418b1f0>
>>> zip(*[iter(lst)]*3)
<zip object at 0x7f1b43f14e00>
>>> list(zip(*[iter(lst)]*3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 14)]
# NOTE: must use list multiples. Explicit listing creates 3 unique iterators
>>> [iter(lst)]*3 == [iter(lst), iter(lst), iter(lst)]
False
>>> list(zip(*[[iter(lst), iter(lst), iter(lst)]))
[(0, 0, 0), (1, 1, 1), (2, 2, 2), (3, 3, 3), ....
It is probably easier to see what is happening in python interpreter or ipython with n = 2:
In [35]: [iter("ABCDEFGH")]*2
Out[35]: [<iterator at 0x6be4128>, <iterator at 0x6be4128>]
So, we have a list of two iterators which are pointing to the same iterator object. Remember that iter on a object returns an iterator object and in this scenario, it is the same iterator twice due to the *2 python syntactic sugar. Iterators also run only once.
Further, zip takes any number of iterables (sequences are iterables) and creates tuple from i'th element of each of the input sequences. Since both iterators are identical in our case, zip moves the same iterator twice for each 2-element tuple of output.
In [41]: help(zip)
Help on built-in function zip in module __builtin__:
zip(...)
zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]
Return a list of tuples, where each tuple contains the i-th element
from each of the argument sequences. The returned list is truncated
in length to the length of the shortest argument sequence.
The unpacking (*) operator ensures that the iterators run to exhaustion which in this case is until there is not enough input to create a 2-element tuple.
This can be extended to any value of n and zip(*[iter(s)]*n) works as described.
x = [1,2,3,4,5,6,7,8,9]
zip(*[iter(x)] * 3)
is the same as:
x = [1,2,3,4,5,6,7,8,9]
iter_var = iter(x)
zip(iter_var,iter_var,iter_var)
Each time zip() gets the next value in iter_var it moves to the next value of x.
Try running next(iter_var) to see how this works.