I've been studying a few searching algorithms and my last problem comes down to binary searching. I watched a few youtube videos to understand the concept and then tried to solve the problem, but keep getting an endless loop error. I've looked through stack overflow, and reddit, and wherever Google would lead me, but I can't quite find a solution that fits my method of coding. Also, please excuse the term 'monkey patching', it's been brought to my attention that the technical term is called 'extending' so the fault lies on my instructors for teaching it to us as 'monkey patching'.
Here's my code:
class Array
def my_bsearch(target)
return nil if self.empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx + 1)
return middle_idx if self[middle_idx] == target
until self[middle_idx] == target || self.nil? == nil
if self[middle_idx] < target
right.my_bsearch(target)
elsif self[middle_idx] > target
left.my_bsearch(target)
end
end
end
end
I have a solution, but I don't want to just memorize it-- and I'm having trouble understanding it; as I'm trying to translate it, learn from it, and implement what I'm missing into my own code.
class Array
def my_bsearch(target)
return nil if size == 0
mid = size/2
case self[mid] <=> target
when 0
return mid
when 1
return self.take(mid).my_bsearch(target)
else
search_res = self.drop(mid+1).my_bsearch(target)
search_res.nil? ? nil : mid + 1 + search_res
end
end
end
I guess I understand case/when despite not use to using it. I've tried following it with debugger, but I think I'm hung up on what's going on in the ELSE section. The syntactic sugar, while making this obviously more concise than my logic, isn't straight-forward/clean to someone of my ruby literacy level. So, yeah, my ignorance is most of the problem I guess.
Is there someone who is a little more literate, and patient, able to help me break this down into something I can understand a bit better so I can learn from this?
First, take and drop have sufficiently similar interfaces that you don't actually want your + 1 for drop. It will disregard one element in the array if you do.
Next, self.nil? will always be false (and never nil) for instances of this class. In fact, .nil? is a method exactly to avoid having to ever compare against nil with ==.
You want self.empty?. Furthermore, with the exception of setters, in Ruby messages are sent to self by default. In other words, the only time self. is a useful prefix is when the message ends in = and operates as an lvalue, as in self.instance_var = 'a constant', since without the self., the tokens instance_var = would be interpreted as a local variable rather than an instance variable setting. That's not the case here, so empty? will suffice just as well as self.empty?
So I figured it out, and I decided to answer my own post in hopes to help someone else out if they run into this issue.
So, if I have an Array and the target is the middle_element, then it will report middle_element_idx. That's fine. What if the target is less than middle_element? It recursively searches the left-side of the original Array. When it finds it, it reports the left_side_idx. There's no problem with that because elements in an array are sequentially counted left to right. So, it starts at 0 and goes up.
But what if the target is on the right side of the middle element?
Well, searching for the right side is easy. Relatively the same logic as searching left. Done recursively. And it will return a target_idx if it's found on that right side --however that's the target's idx as it was found in the right-side array! So, you need to take that returned target_idx and add 1 to it and the original middle_element_idx. See below:
def my_bsearch(target)
return nil if self.empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx + 1)
if self[middle_idx] == target
return middle_idx
elsif self[middle_idx] > target
return left.my_bsearch(target)
else
searched_right_side = 1 + right.my_bsearch(target)
return nil if searched_right_side.nil? == true
return searched_right_side + middle_idx
end
end
end
Notice how many more lines this solution is? The spaceship operator used in conjunction with case/when and a ternary method will reduce the number of lines significantly.
Based on suggestions/feedback from Tim, I updated it to:
def my_bsearch(target)
return nil if empty?
middle_idx = self.length/2
left = self.take(middle_idx)
right = self.drop(middle_idx)
if self[middle_idx] == target
return middle_idx
elsif self[middle_idx] > target
return left.my_bsearch(target)
else
searched_right_side = right.my_bsearch(target)
return nil if searched_right_side.nil?
return searched_right_side + middle_idx
end
end
end
My CSV contains about 60 million rows. The 10th column contains some alphanumeric entries, some of which repeat, that I want to convert into integers with a one-to-one mapping. That is, I don't want the same entry in Original.csv to have multiple corresponding integer values in Processed.csv. So, initially, I wrote the following code:
require 'csv'
udids = []
CSV.open('Original.csv', "wb") do |csv|
CSV.foreach('Processed.csv', :headers=>true) do |row|
unless udids.include?(row[9])
udids << row[9]
end
udid = udids.index(row[9]) + 1
array = [udid]
csv<<array
end
end
But, the program was taking a lot of time, which I soon realized was because it had to check all the previous rows to make sure only the new values get assigned a new integer value, and the existing ones are not assigned any new value.
So, I thought of hashing them, because when exploring the web about this issue, I learnt that hashing is faster than sequential comparing, somehow (I have not read the details about the how, but anyway...) So, I wrote the following code to hash them:
arrayUDID=[]
arrayUser=[]
arrayHash=[]
array1=[]
f = File.open("Original.csv", "r")
f.each_line { |line|
row = line.split(",");
arrayUDID<<row[9]
arrayUser<<row[9]
}
arrayUser = arrayUser.uniq
arrayHash = []
for i in 0..arrayUser.size-1
arrayHash<<arrayUser[i]
arrayHash<<i
end
hash = Hash[arrayHash.each_slice(2).to_a]
array1=hash.values_at *arrayUDID
logfile = File.new("Processed.csv","w")
for i in 0..array1.size-1
logfile.print("#{array1[i]}\n")
end
logfile.close
But here again, I observed that the program was taking a lot of time, which I realized must be due to the hash array (or hash table) running out of memory.
So, can you kindly suggest any method that will work for my huge file in a reasonable amount of time? By reasonable amount, I mean within 10 hours, because I realize that it's going to take some hours at least as it took about 5 hours to extract that dataset from an even bigger dataset. So, with my aforementioned codes, it was not getting finished even after 2 days of running the programs. So, if you can suggest a method which can do the task by leaving the computer on overnight, that would be great. Thanks.
I think this should work:
udids = {}
unique_count = 1
output_csv = CSV.open("Processed.csv", "w")
CSV.foreach("Original.csv").with_index do |row, i|
output_csv << row and next if i == 0 # skip first row (header info)
val = row[9]
if udids[val.to_sym]
row[9] = udids[val.to_sym]
else
udids[val.to_sym] = unique_count
row[9] = unique_count
unique_count += 1
end
output_csv << row
end
output_csv.close
The performance depends heavily on how many duplicates there are (the more the better), but basically it keeps track of each value as a key in a hash, and checks to see if it has encountered that value yet or not. If so, it uses the corresponding value, and if not it increments a counter, stores that count as the new value for that key and continues.
I was able to process a 10 million line test CSV file in about 3 minutes.
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.