Can anyone help me step through the logic of the program shown below? I tried using the Python debugger. This did not help me that much though.
I do not understand the following:
preorder_traversal()
For instance at the yield (parent, root) line of code; does the function return these values as a generator at this point to the caller or does it return the generator and then keep going inside the preorder_traversal() function?
Also, mind completely melts when trying to wrap my head around the recursive call to preorder_traversal(). Does anyone know of a way to understand this? Like a truth table or something like that that I can use to manually step through the program with a pen and paper or notepad or whatever. I think the most complicated part of this is the nesting and the recursion.
I do not understand the Node inside a Node inside a Node, etc. Or the whole adding and edge part which adds a Node to a list.
Code
class Node(object):
"""A simple digraph where each node knows about the other nodes
it leads to.
"""
def __init__(self, name):
self.name = name
self.connections = []
return
def add_edge(self, node):
"Create an edge between this node and the other."
self.connections.append(node)
return
def __iter__(self):
return iter(self.connections)
def preorder_traversal(root, seen=None, parent=None):
"""Generator function to yield the edges via a preorder traversal."""
if seen is None:
seen = set()
yield (parent, root)
if root in seen:
return
seen.add(root)
for node in root:
for (parent, subnode) in preorder_traversal(node, seen, root):
yield (parent, subnode)
return
def show_edges(root):
"Print all of the edges in the graph."
for parent, child in preorder_traversal(root):
if not parent:
continue
print '%5s -> %2s (%s)' % (parent.name, child.name, id(child))
# Set up the nodes.
root = Node('root')
a = Node('a')
b = Node('b')
c = Node('c')
# Add edges between them.
root.add_edge(a)
root.add_edge(b)
a.add_edge(b)
b.add_edge(a)
b.add_edge(c)
a.add_edge(a)
print 'ORIGINAL GRAPH:'
show_edges(root)
Thank-you for reading this.
As for the yield operator, yield allows the function to be a generator, thus lazy. For this particular example, the need for a generator is not needed and its only benefit is much better readability (i.e. for _ in _). Abstractly, yield (parent, root) is returned by using the next() operation on the generator. Then, when next() is called again, the generator continues to dynamically execute the remaining code in the function.
As for the recursive call, this is fairly common when doing any type of graph traversal. Furthermore, a graph is a recursive data structure.
Here is a good resource for understanding graph traversals.
Below is a slightly modified version of the preorder_traversal() (easier to read) which has some comments:
def preorder_traversal(root, seen=set(), parent=None):
"""Generator function to yield the edges via a preorder traversal."""
yield (parent, root)
# avoid cycle
if root not in seen:
seen.add(root)
# for each neighbor of the root
for node in root:
# for each (parent, neighbor) pair in the subgraph
# not containing the nodes already seen
for (parent, subnode) in preorder_traversal(node, seen, root):
yield (parent, subnode)
To demonstrate the lazy nature of a Python generator, consider the custom irange() generator where irange(n) == xrange(n+1):
def irange(limit):
current_number = 0
while current_number <= limit:
yield current_number
current_number += 1
If you do a = irange(9999999999999999999999), no code in irange() is performed until next() is called on it.
To understand recursion with generators, consider the custom rrange() generator where rrange(n) == reversed(irange(n)):
def rrange(limit):
if limit >= 0:
yield limit
for num in rrange(limit - 1):
yield num
I'm attempting to work with Vagrant to perform some automation in spinning up Docker containers. Vagrantfiles are essentially Ruby and thus I should be able to apply Ruby logic to assist with this issue.
I am reading through a conf.d directory filled with YAML files containing configuration data and then pushing a hash of configuration items into an array. Once done I am itering through the array with .each and applying the configuration to each entry in the array based on the values of some of the keys inside the hash. One of these keys is "link". The value of link will correlate to the value of another key "name".
I essentially need to ensure that the hash with link => 'name' is in the array prior to the hash with name => 'value'.
Example of input and expected output:
Input
containers = [{"name"=>"foo", "ports"=>["80:80", "443:443"], "links"=>["bar", "baz"]}, {"name"=>"bar", "ports"=>["8888:8888"]}, {"name"=>"baz","ports"=>"80:80"}]
Expected Output
containers = [{"name"=>"bar", "ports"=>["8888:8888"]}, {"name"=>"baz", "ports"=>"80:80"}, {"name"=>"foo", "ports"=>["80:80", "443:443"], "links"=>["bar", "baz"]}]
The end result is that any entry with "link" appears after an entry in the array where the hash's name key matches it. (Basically dependency ordering based on the link key.)
Note it may occur that a linked container links to another linked container.
It's been puzzling me a bit as I have the ideas of what I need to do but lack the technical chops to actually figure out "How?" :)
Thanks in advance for any assistance.
This should work for you:
def order_containers(containers)
unordered = containers.dup
ordered = []
names_from_ordered = {}
name_is_ordered = names_from_ordered.method(:[])
until unordered.empty?
container = unordered.find do |c|
c.fetch('links', []).all? &name_is_ordered
end
raise 'container ordering impossible' if !container
ordered << container
unordered.delete(container)
names_from_ordered[container.fetch('name')] = true
end
ordered
end
containers = [
{ 'name'=>'foo', 'links'=>['bar'] },
{ 'name'=>'a', 'links'=>['goo'] },
{ 'name'=>'bar' },
{ 'name'=>'goo', 'links'=>['foo'] },
]
containers = order_containers(containers)
require 'pp'
pp containers
# => [{"name"=>"bar"},
# {"name"=>"foo", "links"=>["bar"]},
# {"name"=>"goo", "links"=>["foo"]},
# {"name"=>"a", "links"=>["goo"]}]
The basic idea is that we use a loop, and each iteration of the loop will find one container from the input list that is suitable for adding to the output list. A container is suitable for adding to the output list if all the containers it depends on have already been added to the output list. The container is then removed from the input list and added to the output list.
This loop can terminate in two main ways:
when the input list is empty, which would indicate success, or
when we cannot find a container that we are able to start, which would be an error caused by a circular dependency.
Seems to me the simplest thing would be something like:
linkless_configs = []
linked_configs = []
if config_hash.has_key?("links")
linked_configs.push(config_hash)
else
linkless_configs.push(config_hash)
end
then you can iterator over linkless_configs + linked_configs and be guaranteed that each linked config comes after the corresponding link-less config.
Alternatively, if you must sort, you could
containers.sort_by { |config| config.has_key?("links") ? 1 : 0 }
[Edit: #DavidGrayson has pointed out a flaw with my answer. I'll see if I can find a fix, but if I cannot, and I fear that may be the case, I'll delete the answer. [Edit#2: Oh, my! Someone upvoted my answer after my initial edit. I'm not sure I can delete it now, but to be truthful, I'd already decided not to do so, mainly because my explanation has implications for any proposed solution to the OP's problem. With 10 points in the balance, leaving it up is now even more compelling. 2#tidE]
I believe I understand the problem. sort requires a total order, which is a partial order in which a <= b or a <= b for every pair of elements. ref The latter is not a problem, but the partial order requirement is. A partial order must satisfy axioms of:
reflexivity (x ≤ x),
antisymmetry (if x ≤ y and y ≤ x then x = y) and
transitivity (if x ≤ y and y ≤ z, then x ≤ z).
My ordering only satisfies the reflexivity axiom. David gives the counter-example:
containers = [h0, h1, h2]
where
h0 = {'name'=>'foo', 'links'=>['bar']},
h1 = {'name'=>'a'},
h2 = {'name'=>'bar'},
containers.sort
#=> [{"name"=>"foo", "links"=>["bar"]},
# {"name"=>"a"}, {"name"=>"bar"}]
My method Hash#<=> establishes:
h0 = h1
h0 > h2
h1 = h2
If sort were to find that h0 = h1 = h2, it would conclude, by transitivity, that h0 = h2 (and not check h0 <=> h2), which may result in an incorrect result.
David also points out that o.follows?(self) should raise an exception because I have defined it as private. As I have not yet encountered an exception, I conclude that statement has not been executed, but I have not traced the reason for that, but that's a minor point (though no doubt a useful clue).
I'm grateful to David for identifying the problem. Incorrect answers need to be exposed, of course, but I feel I've learned something useful as well.
tidE]
If I understand the question correctly, and the data provides a valid ordering, I think you could do it as follows.
class Hash
def <=>(o)
case
when follows?(o) then 1
when o.follows?(self) then -1
else 0
end
end
private
def follows?(o)
key?("links") && self["links"].include?(o["name"])
end
end
containers = [{"name"=>"foo", "ports"=>["80:80", "443:443"],
"links"=>["bar", "baz"]},
{"name"=>"bar", "ports"=>["8888:8888"]},
{"name"=>"baz","ports"=>"80:80"}]
containers.sort
#=> [{"name"=>"baz", "ports"=>"80:80"},
# {"name"=>"bar", "ports"=>["8888:8888"]},
# {"name"=>"foo", "ports"=>["80:80", "443:443"],
# "links"=>["bar", "baz"]}]
Addendum
Although I prefaced with the assumption that the data provides a valid ordering, #Ajedi32 asks what happens when there is a circular reference. Let's find out:
containers = [{"name"=>"foo", "links"=>["bar"]},
{"name"=>"bar", "links"=>["baz"]},
{"name"=>"baz", "links"=>["foo"]}]
containers.sort
#=> [{ "name"=>"baz", "links"=>["foo"] },
# { "name"=>"bar", "links"=>["baz"] },
# { "name"=>"foo", "links"=>["bar"] }]
containers = [{"name"=>"foo", "links"=>["bar"]},
{"name"=>"bar", "links"=>["foo"]}]
containers.sort
#=> [{ "name"=>"bar", "links"=>["foo"] },
# { "name"=>"foo", "links"=>["bar"] }]
This shows that if one were not certain that there were no circular references, one should check for that before sorting.
How can I treat the last element of the input specially, when iterating with a for loop? In particular, if there is code that should only occur "between" elements (and not "after" the last one), how can I structure the code?
Currently, I write code like so:
for i, data in enumerate(data_list):
code_that_is_done_for_every_element
if i != len(data_list) - 1:
code_that_is_done_between_elements
How can I simplify or improve this?
Most of the times it is easier (and cheaper) to make the first iteration the special case instead of the last one:
first = True
for data in data_list:
if first:
first = False
else:
between_items()
item()
This will work for any iterable, even for those that have no len():
file = open('/path/to/file')
for line in file:
process_line(line)
# No way of telling if this is the last line!
Apart from that, I don't think there is a generally superior solution as it depends on what you are trying to do. For example, if you are building a string from a list, it's naturally better to use str.join() than using a for loop “with special case”.
Using the same principle but more compact:
for i, line in enumerate(data_list):
if i > 0:
between_items()
item()
Looks familiar, doesn't it? :)
For #ofko, and others who really need to find out if the current value of an iterable without len() is the last one, you will need to look ahead:
def lookahead(iterable):
"""Pass through all values from the given iterable, augmented by the
information if there are more values to come after the current one
(True), or if it is the last value (False).
"""
# Get an iterator and pull the first value.
it = iter(iterable)
last = next(it)
# Run the iterator to exhaustion (starting from the second value).
for val in it:
# Report the *previous* value (more to come).
yield last, True
last = val
# Report the last value.
yield last, False
Then you can use it like this:
>>> for i, has_more in lookahead(range(3)):
... print(i, has_more)
0 True
1 True
2 False
Although that question is pretty old, I came here via google and I found a quite simple way: List slicing. Let's say you want to put an '&' between all list entries.
s = ""
l = [1, 2, 3]
for i in l[:-1]:
s = s + str(i) + ' & '
s = s + str(l[-1])
This returns '1 & 2 & 3'.
if the items are unique:
for x in list:
#code
if x == list[-1]:
#code
other options:
pos = -1
for x in list:
pos += 1
#code
if pos == len(list) - 1:
#code
for x in list:
#code
#code - e.g. print x
if len(list) > 0:
for x in list[:-1]:
#process everything except the last element
for x in list[-1:]:
#process only last element
The 'code between' is an example of the Head-Tail pattern.
You have an item, which is followed by a sequence of ( between, item ) pairs. You can also view this as a sequence of (item, between) pairs followed by an item. It's generally simpler to take the first element as special and all the others as the "standard" case.
Further, to avoid repeating code, you have to provide a function or other object to contain the code you don't want to repeat. Embedding an if statement in a loop which is always false except one time is kind of silly.
def item_processing( item ):
# *the common processing*
head_tail_iter = iter( someSequence )
head = next(head_tail_iter)
item_processing( head )
for item in head_tail_iter:
# *the between processing*
item_processing( item )
This is more reliable because it's slightly easier to prove, It doesn't create an extra data structure (i.e., a copy of a list) and doesn't require a lot of wasted execution of an if condition which is always false except once.
If you're simply looking to modify the last element in data_list then you can simply use the notation:
L[-1]
However, it looks like you're doing more than that. There is nothing really wrong with your way. I even took a quick glance at some Django code for their template tags and they do basically what you're doing.
you can determine the last element with this code :
for i,element in enumerate(list):
if (i==len(list)-1):
print("last element is" + element)
This is similar to Ants Aasma's approach but without using the itertools module. It's also a lagging iterator which looks-ahead a single element in the iterator stream:
def last_iter(it):
# Ensure it's an iterator and get the first field
it = iter(it)
prev = next(it)
for item in it:
# Lag by one item so I know I'm not at the end
yield 0, prev
prev = item
# Last item
yield 1, prev
def test(data):
result = list(last_iter(data))
if not result:
return
if len(result) > 1:
assert set(x[0] for x in result[:-1]) == set([0]), result
assert result[-1][0] == 1
test([])
test([1])
test([1, 2])
test(range(5))
test(xrange(4))
for is_last, item in last_iter("Hi!"):
print is_last, item
We can achieve that using for-else
cities = [
'Jakarta',
'Surabaya',
'Semarang'
]
for city in cities[:-1]:
print(city)
else:
print(' '.join(cities[-1].upper()))
output:
Jakarta
Surabaya
S E M A R A N G
The idea is we only using for-else loops until n-1 index, then after the for is exhausted, we access directly the last index using [-1].
You can use a sliding window over the input data to get a peek at the next value and use a sentinel to detect the last value. This works on any iterable, so you don't need to know the length beforehand. The pairwise implementation is from itertools recipes.
from itertools import tee, izip, chain
def pairwise(seq):
a,b = tee(seq)
next(b, None)
return izip(a,b)
def annotated_last(seq):
"""Returns an iterable of pairs of input item and a boolean that show if
the current item is the last item in the sequence."""
MISSING = object()
for current_item, next_item in pairwise(chain(seq, [MISSING])):
yield current_item, next_item is MISSING:
for item, is_last_item in annotated_last(data_list):
if is_last_item:
# current item is the last item
Is there no possibility to iterate over all-but the last element, and treat the last one outside of the loop? After all, a loop is created to do something similar to all elements you loop over; if one element needs something special, it shouldn't be in the loop.
(see also this question: does-the-last-element-in-a-loop-deserve-a-separate-treatment)
EDIT: since the question is more about the "in between", either the first element is the special one in that it has no predecessor, or the last element is special in that it has no successor.
I like the approach of #ethan-t, but while True is dangerous from my point of view.
data_list = [1, 2, 3, 2, 1] # sample data
L = list(data_list) # destroy L instead of data_list
while L:
e = L.pop(0)
if L:
print(f'process element {e}')
else:
print(f'process last element {e}')
del L
Here, data_list is so that last element is equal by value to the first one of the list. L can be exchanged with data_list but in this case it results empty after the loop. while True is also possible to use if you check that list is not empty before the processing or the check is not needed (ouch!).
data_list = [1, 2, 3, 2, 1]
if data_list:
while True:
e = data_list.pop(0)
if data_list:
print(f'process element {e}')
else:
print(f'process last element {e}')
break
else:
print('list is empty')
The good part is that it is fast. The bad - it is destructible (data_list becomes empty).
Most intuitive solution:
data_list = [1, 2, 3, 2, 1] # sample data
for i, e in enumerate(data_list):
if i != len(data_list) - 1:
print(f'process element {e}')
else:
print(f'process last element {e}')
Oh yes, you have already proposed it!
There is nothing wrong with your way, unless you will have 100 000 loops and wants save 100 000 "if" statements. In that case, you can go that way :
iterable = [1,2,3] # Your date
iterator = iter(iterable) # get the data iterator
try : # wrap all in a try / except
while 1 :
item = iterator.next()
print item # put the "for loop" code here
except StopIteration, e : # make the process on the last element here
print item
Outputs :
1
2
3
3
But really, in your case I feel like it's overkill.
In any case, you will probably be luckier with slicing :
for item in iterable[:-1] :
print item
print "last :", iterable[-1]
#outputs
1
2
last : 3
or just :
for item in iterable :
print item
print iterable[-1]
#outputs
1
2
3
last : 3
Eventually, a KISS way to do you stuff, and that would work with any iterable, including the ones without __len__ :
item = ''
for item in iterable :
print item
print item
Ouputs:
1
2
3
3
If feel like I would do it that way, seems simple to me.
Use slicing and is to check for the last element:
for data in data_list:
<code_that_is_done_for_every_element>
if not data is data_list[-1]:
<code_that_is_done_between_elements>
Caveat emptor: This only works if all elements in the list are actually different (have different locations in memory). Under the hood, Python may detect equal elements and reuse the same objects for them. For instance, for strings of the same value and common integers.
Google brought me to this old question and I think I could add a different approach to this problem.
Most of the answers here would deal with a proper treatment of a for loop control as it was asked, but if the data_list is destructible, I would suggest that you pop the items from the list until you end up with an empty list:
while True:
element = element_list.pop(0)
do_this_for_all_elements()
if not element:
do_this_only_for_last_element()
break
do_this_for_all_elements_but_last()
you could even use while len(element_list) if you don't need to do anything with the last element. I find this solution more elegant then dealing with next().
For me the most simple and pythonic way to handle a special case at the end of a list is:
for data in data_list[:-1]:
handle_element(data)
handle_special_element(data_list[-1])
Of course this can also be used to treat the first element in a special way .
Better late than never. Your original code used enumerate(), but you only used the i index to check if it's the last item in a list. Here's an simpler alternative (if you don't need enumerate()) using negative indexing:
for data in data_list:
code_that_is_done_for_every_element
if data != data_list[-1]:
code_that_is_done_between_elements
if data != data_list[-1] checks if the current item in the iteration is NOT the last item in the list.
Hope this helps, even nearly 11 years later.
if you are going through the list, for me this worked too:
for j in range(0, len(Array)):
if len(Array) - j > 1:
notLast()
Instead of counting up, you can also count down:
nrToProcess = len(list)
for s in list:
s.doStuff()
nrToProcess -= 1
if nrToProcess==0: # this is the last one
s.doSpecialStuff()
I will provide with a more elegant and robust way as follows, using unpacking:
def mark_last(iterable):
try:
*init, last = iterable
except ValueError: # if iterable is empty
return
for e in init:
yield e, True
yield last, False
Test:
for a, b in mark_last([1, 2, 3]):
print(a, b)
The result is:
1 True
2 True
3 False
If you are looping the List,
Using enumerate function is one of the best try.
for index, element in enumerate(ListObj):
# print(index, ListObj[index], len(ListObj) )
if (index != len(ListObj)-1 ):
# Do things to the element which is not the last one
else:
# Do things to the element which is the last one
Delay the special handling of the last item until after the loop.
>>> for i in (1, 2, 3):
... pass
...
>>> i
3
There can be multiple ways. slicing will be fastest. Adding one more which uses .index() method:
>>> l1 = [1,5,2,3,5,1,7,43]
>>> [i for i in l1 if l1.index(i)+1==len(l1)]
[43]
If you are happy to be destructive with the list, then there's the following.
We are going to reverse the list in order to speed up the process from O(n^2) to O(n), because pop(0) moves the list each iteration - cf. Nicholas Pipitone's comment below
data_list.reverse()
while data_list:
value = data_list.pop()
code_that_is_done_for_every_element(value)
if data_list:
code_that_is_done_between_elements(value)
else:
code_that_is_done_for_last_element(value)
This works well with empty lists, and lists of non-unique items.
Since it's often the case that lists are transitory, this works pretty well ... at the cost of destructing the list.
Assuming input as an iterator, here's a way using tee and izip from itertools:
from itertools import tee, izip
items, between = tee(input_iterator, 2) # Input must be an iterator.
first = items.next()
do_to_every_item(first) # All "do to every" operations done to first item go here.
for i, b in izip(items, between):
do_between_items(b) # All "between" operations go here.
do_to_every_item(i) # All "do to every" operations go here.
Demo:
>>> def do_every(x): print "E", x
...
>>> def do_between(x): print "B", x
...
>>> test_input = iter(range(5))
>>>
>>> from itertools import tee, izip
>>>
>>> items, between = tee(test_input, 2)
>>> first = items.next()
>>> do_every(first)
E 0
>>> for i,b in izip(items, between):
... do_between(b)
... do_every(i)
...
B 0
E 1
B 1
E 2
B 2
E 3
B 3
E 4
>>>
The most simple solution coming to my mind is:
for item in data_list:
try:
print(new)
except NameError: pass
new = item
print('The last item: ' + str(new))
So we always look ahead one item by delaying the the processing one iteration. To skip doing something during the first iteration I simply catch the error.
Of course you need to think a bit, in order for the NameError to be raised when you want it.
Also keep the `counstruct
try:
new
except NameError: pass
else:
# continue here if no error was raised
This relies that the name new wasn't previously defined. If you are paranoid you can ensure that new doesn't exist using:
try:
del new
except NameError:
pass
Alternatively you can of course also use an if statement (if notfirst: print(new) else: notfirst = True). But as far as I know the overhead is bigger.
Using `timeit` yields:
...: try: new = 'test'
...: except NameError: pass
...:
100000000 loops, best of 3: 16.2 ns per loop
so I expect the overhead to be unelectable.
Count the items once and keep up with the number of items remaining:
remaining = len(data_list)
for data in data_list:
code_that_is_done_for_every_element
remaining -= 1
if remaining:
code_that_is_done_between_elements
This way you only evaluate the length of the list once. Many of the solutions on this page seem to assume the length is unavailable in advance, but that is not part of your question. If you have the length, use it.
One simple solution that comes to mind would be:
for i in MyList:
# Check if 'i' is the last element in the list
if i == MyList[-1]:
# Do something different for the last
else:
# Do something for all other elements
A second equally simple solution could be achieved by using a counter:
# Count the no. of elements in the list
ListLength = len(MyList)
# Initialize a counter
count = 0
for i in MyList:
# increment counter
count += 1
# Check if 'i' is the last element in the list
# by using the counter
if count == ListLength:
# Do something different for the last
else:
# Do something for all other elements
Just check if data is not the same as the last data in data_list (data_list[-1]).
for data in data_list:
code_that_is_done_for_every_element
if data != data_list[- 1]:
code_that_is_done_between_elements
So, this is definitely not the "shorter" version - and one might digress if "shortest" and "Pythonic" are actually compatible.
But if one needs this pattern often, just put the logic in to a
10-liner generator - and get any meta-data related to an element's
position directly on the for call. Another advantage here is that it will
work wit an arbitrary iterable, not only Sequences.
_sentinel = object()
def iter_check_last(iterable):
iterable = iter(iterable)
current_element = next(iterable, _sentinel)
while current_element is not _sentinel:
next_element = next(iterable, _sentinel)
yield (next_element is _sentinel, current_element)
current_element = next_element
In [107]: for is_last, el in iter_check_last(range(3)):
...: print(is_last, el)
...:
...:
False 0
False 1
True 2
This is an old question, and there's already lots of great responses, but I felt like this was pretty Pythonic:
def rev_enumerate(lst):
"""
Similar to enumerate(), but counts DOWN to the last element being the
zeroth, rather than counting UP from the first element being the zeroth.
Since the length has to be determined up-front, this is not suitable for
open-ended iterators.
Parameters
----------
lst : Iterable
An iterable with a length (list, tuple, dict, set).
Yields
------
tuple
A tuple with the reverse cardinal number of the element, followed by
the element of the iterable.
"""
length = len(lst) - 1
for i, element in enumerate(lst):
yield length - i, element
Used like this:
for num_remaining, item in rev_enumerate(['a', 'b', 'c']):
if not num_remaining:
print(f'This is the last item in the list: {item}')
Or perhaps you'd like to do the opposite:
for num_remaining, item in rev_enumerate(['a', 'b', 'c']):
if num_remaining:
print(f'This is NOT the last item in the list: {item}')
Or, just to know how many remain as you go...
for num_remaining, item in rev_enumerate(['a', 'b', 'c']):
print(f'After {item}, there are {num_remaining} items.')
I think the versatility and familiarity with the existing enumerate makes it most Pythonic.
Caveat, unlike enumerate(), rev_enumerate() requires that the input implement __len__, but this includes lists, tuples, dicts and sets just fine.