I am trying to understand what is a semantically right way to use map. As map can behave the same way as each, you could modify the array any way you like. But I've been told by my colleague that after map is applied, array should have
the same order and the same size.
For example, that would mean using the map to return an updated array won't be the right way to use map:
array = [1,2,3,4]
array.map{|num| num unless num == 2 || num == 4}.compact
I've been using map and other Enumerator methods for ages and never thought about this too much. Would appreciate advice from experienced Ruby Developers.
In Computer Science, map according to Wikipedia:
In many programming languages, map is the name of a higher-order
function that applies a given function to each element of a list,
returning a list of results in the same order
This statement implies the returned value of map should be of the same length (because we're applying the function to each element). And the returned-elements are to be in the same order. So when you use map, this is what the reader expects.
How not to use map
arr.map {|i| arr.pop } #=> [3, 2]
This clearly betrays the intention of map since we have a different number of elements returned and they are not even in the original order of application. So don't use map like this. See "How to use ruby's value_at to get subhashes in a hash" and subsequent comments for further clarification and thanks to #meager for originally pointing this out to me.
Meditate on this:
array = [1,2,3,4]
array.map{|num| num unless num == 2 || num == 4} # => [1, nil, 3, nil]
.compact # => [1, 3]
The intermediate value is an array of the same size, however it contains undesirable values, forcing the use of compact. The fallout of this is CPU time is wasted generating the nil values, then deleting them. In addition, memory is being wasted generating another array that is the same size when it shouldn't be. Imagine the CPU and memory cost in a loop that is processing thousands of elements in an array.
Instead, using the right tool cleans up the code and avoids wasting CPU or memory:
array.reject { |num| num == 2 || num == 4 } # => [1, 3]
I've been using map and other Enumerator methods for ages and never thought about this too much.
I'd recommend thinking about it. It's the little things like this that can make or break code or a system, and everything we do when programming needs to be done deliberately, avoiding all negative side-effects we can foresee.
I can't seem to find a definitive answer on this and I want to make sure I understand this to the "n'th level" :-)
a = { "a" => "Hello", "b" => "World" }
a.count # 2
a.size # 2
a.length # 2
a = [ 10, 20 ]
a.count # 2
a.size # 2
a.length # 2
So which to use? If I want to know if a has more than one element then it doesn't seem to matter but I want to make sure I understand the real difference. This applies to arrays too. I get the same results.
Also, I realize that count/size/length have different meanings with ActiveRecord. I'm mostly interested in pure Ruby (1.92) right now but if anyone wants to chime in on the difference AR makes that would be appreciated as well.
Thanks!
For arrays and hashes size is an alias for length. They are synonyms and do exactly the same thing.
count is more versatile - it can take an element or predicate and count only those items that match.
> [1,2,3].count{|x| x > 2 }
=> 1
In the case where you don't provide a parameter to count it has basically the same effect as calling length. There can be a performance difference though.
We can see from the source code for Array that they do almost exactly the same thing. Here is the C code for the implementation of array.length:
static VALUE
rb_ary_length(VALUE ary)
{
long len = RARRAY_LEN(ary);
return LONG2NUM(len);
}
And here is the relevant part from the implementation of array.count:
static VALUE
rb_ary_count(int argc, VALUE *argv, VALUE ary)
{
long n = 0;
if (argc == 0) {
VALUE *p, *pend;
if (!rb_block_given_p())
return LONG2NUM(RARRAY_LEN(ary));
// etc..
}
}
The code for array.count does a few extra checks but in the end calls the exact same code: LONG2NUM(RARRAY_LEN(ary)).
Hashes (source code) on the other hand don't seem to implement their own optimized version of count so the implementation from Enumerable (source code) is used, which iterates over all the elements and counts them one-by-one.
In general I'd advise using length (or its alias size) rather than count if you want to know how many elements there are altogether.
Regarding ActiveRecord, on the other hand, there are important differences. check out this post:
Counting ActiveRecord associations: count, size or length?
There is a crucial difference for applications which make use of database connections.
When you are using many ORMs (ActiveRecord, DataMapper, etc.) the general understanding is that .size will generate a query that requests all of the items from the database ('select * from mytable') and then give you the number of items resulting, whereas .count will generate a single query ('select count(*) from mytable') which is considerably faster.
Because these ORMs are so prevalent I following the principle of least astonishment. In general if I have something in memory already, then I use .size, and if my code will generate a request to a database (or external service via an API) I use .count.
In most cases (e.g. Array or String) size is an alias for length.
count normally comes from Enumerable and can take an optional predicate block. Thus enumerable.count {cond} is [roughly] (enumerable.select {cond}).length -- it can of course bypass the intermediate structure as it just needs the count of matching predicates.
Note: I am not sure if count forces an evaluation of the enumeration if the block is not specified or if it short-circuits to the length if possible.
Edit (and thanks to Mark's answer!): count without a block (at least for Arrays) does not force an evaluation. I suppose without formal behavior it's "open" for other implementations, if forcing an evaluation without a predicate ever even really makes sense anyway.
I found a good answare at http://blog.hasmanythrough.com/2008/2/27/count-length-size
In ActiveRecord, there are several ways to find out how many records
are in an association, and there are some subtle differences in how
they work.
post.comments.count - Determine the number of elements with an SQL
COUNT query. You can also specify conditions to count only a subset of
the associated elements (e.g. :conditions => {:author_name =>
"josh"}). If you set up a counter cache on the association, #count
will return that cached value instead of executing a new query.
post.comments.length - This always loads the contents of the
association into memory, then returns the number of elements loaded.
Note that this won't force an update if the association had been
previously loaded and then new comments were created through another
way (e.g. Comment.create(...) instead of post.comments.create(...)).
post.comments.size - This works as a combination of the two previous
options. If the collection has already been loaded, it will return its
length just like calling #length. If it hasn't been loaded yet, it's
like calling #count.
Also I have a personal experience:
<%= h(params.size.to_s) %> # works_like_that !
<%= h(params.count.to_s) %> # does_not_work_like_that !
We have a several ways to find out how many elements in an array like .length, .count and .size. However, It's better to use array.size rather than array.count. Because .size is better in performance.
Adding more to Mark Byers answer. In Ruby the method array.size is an alias to Array#length method. There is no technical difference in using any of these two methods. Possibly you won't see any difference in performance as well. However, the array.count also does the same job but with some extra functionalities Array#count
It can be used to get total no of elements based on some condition. Count can be called in three ways:
Array#count # Returns number of elements in Array
Array#count n # Returns number of elements having value n in Array
Array#count{|i| i.even?} Returns count based on condition invoked on each element array
array = [1,2,3,4,5,6,7,4,3,2,4,5,6,7,1,2,4]
array.size # => 17
array.length # => 17
array.count # => 17
Here all three methods do the same job. However here is where the count gets interesting.
Let us say, I want to find how many array elements does the array contains with value 2
array.count 2 # => 3
The array has a total of three elements with value as 2.
Now, I want to find all the array elements greater than 4
array.count{|i| i > 4} # =>6
The array has total 6 elements which are > than 4.
I hope it gives some info about count method.
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.