Itertools en Python

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itertools , . « », Python.

itertools Python.

: 

• ;

• ;

• .

. :

• «next»

• for

• list()

:

count(), cycle(), repeat()

1. count()

, , , start. start 0, step -1. . .

import itertools
c=itertools.count()
#next() function returns next item in the iterator.By default starts with number 0 and increments by 1.
print (next(c))#Output:0
print (next(c))#Output:1
print (next(c))#Output:2
print (next(c))#Output:3


#Returns an infinite iterator starting with number 5 and incremented by 10. The values in the iterator are accessed by next()
c1=itertools.count(5,10)
print(next(c1))#Output:5
print(next(c1))#Output:10
print(next(c1))#Output:15


#accessing values in the iterator using for loop.step argument can be float values also.
c2=itertools.count(2.5,2.5)
for i in c2:
    #including terminating condition, else loop will keep on going.(infinite loop)
    if i>25:
        break
    else:
        print (i,end=" ") #Output:2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0
        

#step can be negative numbers also.negative numbers count backwards.
c3=itertools.count(2,-2.5)
print (next(c3))#Output:2
print (next(c3))#Output:-0.5
print (next(c3))#Output:-3.0

map() zip().

zip() – Python, zip, . itertools.count() zip().

import itertools
#itertools.count() used in zip()
l1=[5,15,25]
l2=zip(itertools.count(),l1)
#It will return zip object which is an iterable instance of zip class
print (l2)#Output:<zip object at 0x032C92C8>


#we can convert zip object to list.
print (list(l2))#Output:[(0, 5), (1, 15), (2, 25)]


# We can access the zip object by using "for loop".This is more efficient way to access large sequences.It won't be memory intensive.
l3=zip(itertools.count(),l1)
for i in l3:
    print (i)
'''
Output:
(0, 5)
(1, 15)
(2, 25)
'''

map() map, . itertools.count() map().

import itertools
#lambda function and itertools.count() is given as argument in map()function.
l1=map(lambda x:x**2,itertools.count())
#It returns a map object which is an iterator.
print(l1)#Output:<map object at 0x00E2E610>

#iterate thorugh map object using for loop
for i in l1:
    if i>50:
        break
    else:
        print (i ,end=" ")#Output:0 1 4 9 16 25 36 49

2. cycle()

, . , . .

itertools.cycle(iterable)

import itertools
l1=[1,2,3]
#using list iterable as an argument in itertools.cycle()
l2=itertools.cycle(l1)
print (l2)#Output:<itertools.cycle object at 0x02F794E8>

count=0
for i in l2:
    #It will end in infinite loop. So have to define terminating condition.
    if count > 15:
        break
    else:
        print (i,end=" ")#Output:1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1
        count+=1

        
#string is given as an argument in itertools.cycle() function.
s1="hello"
l3=itertools.cycle(s1)
#accessing the iterator using next()
print (next(l3))#Output:h
print (next(l3))#Output:e
print (next(l3))#Output:l
print (next(l3))#Output:l
print (next(l3))#Output:o

3. repeat():

, . , times.

map() zip().

itertools.repeat(object,times)

import itertools
#times argument is not mentioned. It will result in infinite loop.
l1=itertools.repeat(2)
print (next(l1))#Output:2
print (next(l1))#Output:2


#string is used as an argument.times argument is mentioned as 10.It will repeat the string 10 times.
l2=itertools.repeat("hello",times=10)
for i in l2:
    print (i,end=" ")#Output:hello hello hello hello hello hello hello hello hello hello
print (" ")


#list is used as argument
l3=itertools.repeat([1,2,3],times=3)
for i in l3:
    print (i,end=" ")#Output:[1, 2, 3] [1, 2, 3] [1, 2, 3]
print (" ")


#tuple is used as an argument
l4=itertools.repeat(('red','blue'),times=3)
for i in l4:
    print (i,end=" ")#Output:('red', 'blue') ('red', 'blue') ('red', 'blue')

map().

import itertools
#It will return square of numbers from 0 to 9.
l1=map(pow,range(10),itertools.repeat(2))
print(l1)#Output:<map object at 0x011CEC10>

#iterate through map object using for loop
for i in l1:
    print (i,end=" ") #Output:0 1 4 9 16 25 36 49 64 81

zip().

import itertools
#itertools.repeat() used in zip()
l1=[5,15,25]
l2=zip(itertools.repeat(2),l1)
#It will return zip object which is an iterable instance of zip class
print (l2)#Output:<zip object at 0x032794E8>


#we can convert zip object to list.
print (list(l2))#Output:[(2, 5), (2, 15), (2, 25)]


# We can access zip object by using for loop.This is more efficient way to access large sequences.It won't be memory intensive.
l3=zip(itertools.repeat("hello"),l1)
for i in l3:
    print (i)
'''
Output:
('hello', 5)
('hello', 15)
('hello', 25)
'''

:

1.accumulate():

, , func. 

. initial, , , . 

functools.reduce() . 

initial Python 3.8.

itertools.accumulate(iterable,func,*,initial=None)

import itertools
#using add and mul operator,so importing operator module
import operator
#using reduce(),so importing reduce() from functools module
from functools import reduce

#If func parameter is not mentioned,by default it will perform addition operation)
l1=itertools.accumulate([1,2,3,4,5])
print (l1)#Output:<itertools.accumulate object at 0x02CD94C8>
#Converting iterator to list object.
print (list(l1))#Output:[1, 3, 6, 10, 15]
#using reduce() for same function
r1=reduce(operator.add,[1,2,3,4,5])
print (r1)#Output:15


#If initial parameter is mentioned, it will start accumulating from the initial value.
#It will contain more than one element in the ouptut iterable.
l2=itertools.accumulate([1,2,3,4,5],operator.add,initial=10)
print (list(l2))#Output:[10, 11, 13, 16, 20, 25]


#it takes operator mul as function
# It will perform multiplication on first two elements, then it will perform multiplication of next two element in the iterable.
l3=itertools.accumulate([1,2,3,5,5],operator.mul)
print (list(l3))#Output:[1, 2, 6, 30, 150]
#using reduce() for same function mul.
r2=reduce(operator.mul,[1,2,3,4,5])
print (r2)#Output:120


l4=itertools.accumulate([2,4,6,3,1],max)
print (list(l4))#Output:[2, 4, 6, 6, 6]
#using reduce for same function max
r3=reduce(max,[2,4,6,3,1])
print (r3)#Output:6


#It takes min function as parameter.
# It will compare two elements and find the minimum element,then compare the other elements and find the mininum element.
l5=itertools.accumulate([2,4,6,3,1],min)
print (list(l5))#Output:[2, 2, 2, 2, 1]
#using reduce() for same function min
r4=reduce(min,[2,4,6,3,1])
print (r4)#Output:1

2. chain():

, , , . , , . 

itertools.chain(*iterables)

import itertools
l1=itertools.chain(["red","blue"],[1,2,3],"hello")
#Returns an iterator object
print (l1)#Output:<itertools.chain object at 0x029FE4D8>
#converting iterator object to list object
print(list(l1))#Output:['red', 'blue', 1, 2, 3, 'h', 'e', 'l', 'l', 'o']


l2=itertools.chain("ABC","DEF","GHI")
print (list(l2))#Output:['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I']

3. chain.from_iterable()

«» , . , , , TypeError.

chain.from_iterable(iterable)

import itertools
l1=itertools.chain.from_iterable(["ABC","DEF","GHI"])
print (list(l1))#Output:['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I']

l2=itertools.chain(["ABC","DEF","GHI"])
print (list(l2))#Output:['ABC', 'DEF', 'GHI']

#all elements in the input iterable should be iterable.Otherwise it will raise TypeError.
l3=itertools.chain.from_iterable([1,2,3])
print (list(l3))#Output:TypeError: 'int' object is not iterable

4. compress()

, data, , (selectors), True. , , . 

itertools.compress(data,selectors)

import itertools
selectors=[True,False,True,False]
l1=itertools.compress([1,2,3,4],selectors)
#Only returns element whose corresponding selector is True.
print (list(l1))#Output:[1,3]


#filter - instead of passing an iterable of True and False. function is used to determine the value "True or False"
l2=filter(lambda x:x%2!=0,[1,2,3,4])
print (list(l2))#Output:[1,3]

5. dropwhile()

, , (predicate) True, . , False.

itertools.dropwhile(predicate,iterable)

6. takewhile()

, , True.

itertools.takewhile(predicate,iterable)

import itertools
##iterator will start displaying after condition is False.
l1=itertools.dropwhile(lambda x:x>4,[5,6,7,8,9,1,2,3])
print (l1)#Output:<itertools.dropwhile object at 0x02E592C8>
print (list(l1))#Output:[1, 2, 3]
#iterator used to print values till condition is False.
l4=itertools.takewhile(lambda x:x>4,[5,6,7,8,9,1,2,3])
print (list(l4))#Output:[5, 6, 7, 8, 9]


#iterator will start displaying after condition is False.
l2=itertools.dropwhile(lambda x:x%2==0,[2,4,6,8,10])
print (list(l2))#Output:[]
#iterator used to print values till condition is False.
l3=itertools.takewhile(lambda x:x%2==0,[2,4,6,8,10])
print (list(l3))#Output:[2, 4, 6, 8, 10]

7.filterfalse():

, , , False. None, , False. 

filter() , , True.

itertools.filterfalse(predicate,iterables)

import itertools
#iterator will filter the elements from the iterable which returns False for the given function
l1=itertools.filterfalse(lambda x:x>4,[1,2,3,4,5,6,7,8,9])
print (l1)#Output:<itertools.filterfalse object at 0x0083E658>
print (list(l1))#Output:[1, 2, 3,4]
#filter() function will  filter the elements from the iterable which returns True for the given function
l4=filter(lambda x:x>4,[1,2,3,4,5,6,7,8,9])
print (list(l4))#Output:[5, 6, 7, 8, 9]


#iterator will filter the elements from the iterable which returns False for the given function
l2=itertools.filterfalse(lambda x:x%2==0,[2,4,6,8,10])
print (list(l2))#Output:[]
#filter() function will filter the elements from the iterable which returns True for the given function
l3=filter(lambda x:x%2==0,[2,4,6,8,10])
print (list(l3))#Output:[2, 4, 6, 8, 10]


#If predicate is None, returns the items that are False.
l5=itertools.filterfalse(None,[0,1,2,3,4,5])
print (list(l5))#Output:[0]
#If predicate is None,filter() function returns the items that are True.
l6=filter(None,[0,1,2,3,4])
print (list(l6))#Output:[1, 2, 3, 4]

8. zip_longest()

, . , fillvalue. , .

zip() , .

itertools.zip_longest(*iterables,fillvalue=None)

import itertools
#fillvalue is not given,by default it will be None.
#iteration continues until longest iterable is exhausted.
z1=itertools.zip_longest([1,2,3,4,5],['a','b','c'])
print (z1)#Output:<itertools.zip_longest object at 0x00AA3BE0>
#we can iterate through zip object using for loop or we can convert to list object.
print (list(z1))#Output:[(1, 'a'), (2, 'b'), (3, 'c'), (4, None), (5, None)]


#fillvalue is mentioned
z2=itertools.zip_longest([1,2,3,4,5],['a','b','c'],fillvalue="z")
print (list(z2))#Output:[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'z'), (5, 'z')]


#using zip(),iteration continues until shorest iterable is exhausted.
z3=zip([1,2,3,4,5],['a','b','c'])
print (list(z3))#Output:[(1, 'a'), (2, 'b'), (3, 'c')]

9. starmap()

, , . map(), ( ).

itertools.starmap(function,iterable)

import itertools
l1=itertools.starmap(pow,[(0,2),(1,2),(2,2)])
print (l1)#Output:<itertools.starmap object at 0x00C8E4D8>
#We can iterate through starmap object,using for loop or using next() function. We can also convert to list object.
print (list(l1))#Output:[0, 1, 4]


#using map()
l2=map(pow,[0,1,2],[2,2,2])
print (list(l2))#Output:[0, 1, 4]


a1=map(lambda x:x**2,[1,2,3])
print (list(a1))#Output:[1, 4, 9]

#If elements inside the iterable are not iterable, it will raise TypeError.
a2=itertools.starmap(lambda x:x**2,[1,2,3])
print (list(a2))#Output:TypeError: 'int' object is not iterable

a3=itertools.starmap(lambda x,y:x+y,[(0,1),(1,2),(2,3)])
print (list(a3))#Output:[1, 3, 5]

10. islice()

, . start None, . step None, 1. stop None, , , . . islice() start, stop step.

itertools.islice(iterable,stop)

itertools.islice(iterable, start, stop[, step])

import itertools
#if only one argument is mentioned other than iterable,it is stop value.
l1=itertools.islice([1,2,3,4,5,6,7,8,9,10],5)
print (list(l1))#Output:[1, 2, 3, 4, 5]


#start=2 and stop=5 mentioned. It will start from second index and ends at fifth index
l2=itertools.islice([1,2,3,4,5,6,7,8,9,10],2,5)
print (list(l2))#Output:[3,4,5]


#start=2,step=3.It will start from second index and increment by 3.
l3=itertools.islice([1,2,3,4,5,6,7,8,9,10],2,None,3)
print (list(l3))#Output:[3,6,9]


#step is given as negative value,it raises ValueError
l4=itertools.islice([1,2,3,4,5,6,7,8,9,10],2,None,-2)
print (list(l4))#Output:ValueError: Step for islice() must be a positive integer or None.

11. tee()

n .

itertools.tee(iterable,n=2)

import itertools
l1=[1,2,3,4,5]
l2=itertools.tee(l1,3)
print (l2)#Output:(<itertools._tee object at 0x028794C8>, <itertools._tee object at 0x028792C8>, <itertools._tee object at 0x02879528>)
for i in l2:
    print (list(i))
'''
Output:    
[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5]
'''

12. groupby():

, .

key – , . None, , . 

itertools.groupby(iterable,key=None)

import itertools
l1=[('color','red'),('color','blue'),('color','green'),('Numbers',1),('Numbers',5)]
l2=itertools.groupby(l1,key=lambda x:x[0])
print (l2)#Output:<itertools.groupby object at 0x0305F528>
for key,group in l2:
    result={key:list(group)}
    print (result)
'''
Output:
{'color': [('color', 'red'), ('color', 'blue'), ('color', 'green')]}
{'Numbers': [('Numbers', 1), ('Numbers', 5)]}
'''

#list doesn't contain sorted data based on the key function.It breaks the group every time,when value of key function changes.
l1=[('color','red'),('color','blue'),('color','green'),('Numbers',1),('Numbers',5),('color','purple')]
l2=itertools.groupby(l1,key=lambda x:x[0])
print (l2)#Output:<itertools.groupby object at 0x028AF5A0>

for key,group in l2:
    result={key:list(group)}
    print (result)
'''
Output:
{'color': [('color', 'red'), ('color', 'blue'), ('color', 'green')]}
{'Numbers': [('Numbers', 1), ('Numbers', 5)]}
{'color': [('color', 'purple')]}
'''


#key is not mentioned
l1=[('color','red'),('color','blue'),('color','green'),('Numbers',1),('Numbers',5),('color','purple')]
l2=itertools.groupby(l1)
print (l2)#Output:<itertools.groupby object at 0x028AF578>
for key,group in l2:
    result={key:list(group)}
    print (result)
'''
Output:
{('color', 'red'): [('color', 'red')]}
{('color', 'blue'): [('color', 'blue')]}
{('color', 'green'): [('color', 'green')]}
{('Numbers', 1): [('Numbers', 1)]}
{('Numbers', 5): [('Numbers', 5)]}
{('color', 'purple'): [('color', 'purple')]}

'''

:

1. product()

, .

: X Y – , (x, y), x X, y Y.

, repeat. , product(A, repeat=4) – , product(A, A, A, A).

itertools.product(*iterables,repeat)

import itertools
#Only one iterable is given
l1=itertools.product("ABCD")
print (list(l1))#Output:[('A',), ('B',), ('C',), ('D',)]


#two iterables are given
l2=itertools.product("ABC",[1,2])
print (list(l2))#Output:[('A', 1), ('A', 2), ('B', 1), ('B', 2), ('C', 1), ('C', 2)]


#one iterable and repeat is mentioned.
l3=itertools.product("xy",repeat=2)
print (list(l3))#Output:[('x', 'x'), ('x', 'y'), ('y', 'x'), ('y', 'y')]


l4=itertools.product("aa",repeat=2)
print (list(l4))#Output:[('a', 'a'), ('a', 'a'), ('a', 'a'), ('a', 'a')]


#More than two iterables is mentioned
l5=itertools.product([1,2],[3,4],[5,6])
print (list(l5))#Output:[(1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6)]

2. permutations()

r . r None, r . . , , .

, . , , . 

itertools.permutations(iterable,r=None)

import itertools
l1=itertools.permutations("ABC")
print (list(l1))#Output:[('A', 'B', 'C'), ('A', 'C', 'B'), ('B', 'A', 'C'), ('B', 'C', 'A'), ('C', 'A', 'B'), ('C', 'B', 'A')]


l2=itertools.permutations([3,2,1])
print (list(l2))#Output:[(3, 2, 1), (3, 1, 2), (2, 3, 1), (2, 1, 3), (1, 3, 2), (1, 2, 3)]


#elements are treated as unique based on their position and not by their value.
l3=itertools.permutations([1,1])
print (list(l3))#Output:[(1, 1), (1, 1)]


l4=itertools.permutations(["ABC"])
print (list(l4))#Output:[('ABC',)]


#r value is mentioned as 2. It will return all different permutations in 2 values.
l5=itertools.permutations([1,2,3,4],2)
print (list(l5))#Output:[(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)]

 : .

3. combinations()

r , . 

. , , .

– . 

, . , , .

itertools.combinations(iterable, r)

4. combinations_with_replacement():

r , , . 

itertools.combinations_with_replacement (iterable, r)

import itertools
l1=itertools.combinations("ABC",2)
print (list(l1))#Output:[('A', 'B'), ('A', 'C'), ('B', 'C')]
l1=itertools.combinations_with_replacement("ABC",2)
print (list(l1))#Output:[('A', 'A'), ('A', 'B'), ('A', 'C'), ('B', 'B'), ('B', 'C'), ('C', 'C')]


l2=itertools.combinations([3,2,1],3)
print (list(l2))#Output:[(3, 2, 1)]
l2=itertools.combinations_with_replacement([3,2,1],3)
print(list(l2))#Output:[(3, 3, 3), (3, 3, 2), (3, 3, 1), (3, 2, 2), (3, 2, 1), (3, 1, 1), (2, 2, 2), (2, 2, 1), (2, 1, 1), (1, 1, 1)]


#elements are treated as unique based on their position and not by their value.
l3=itertools.combinations([1,1],2)
print (list(l3))#Output:[(1, 1)]
l3=itertools.combinations_with_replacement([1,1],2)
print (list(l3))#Output:[(1, 1), (1, 1), (1, 1)]


#since list contains only one element, given r value is 2. So it returns empty list.
l4=itertools.combinations(["ABC"],2)
print (list(l4))#Output:[]
#In combinations_with_replacement,it allows repeated element.
l4=itertools.combinations_with_replacement(["ABC"],2)
print (list(l4))#Output:[('ABC', 'ABC')]


#r value is not mentioned. It will raise TypeError
#l5=itertools.combinations([1,2,3,4])
#print (list(l5))#Output:TypeError: combinations() missing required argument 'r' (pos 2)
l5=itertools.combinations_with_replacement([1,2,3,4])
print (list(l5))#Output:TypeError: combinations_with_replacement() missing required argument 'r' (pos 2)

:

1. map() zip():

• count() 

• repeat()

repeat() - map() zip().

count() - map() zip().

2. cycle() repeat():

cycle() – ;

repeat() – .

3. reduce() itertools.accumulate():

reduce():

• ;

• , – .

accumulate():

• . , .

• , – .

4. filter() itertools.compress():

• filter() , , True .

• compress() . True/False. 

5. filter() itertools.filterfalse():

• filter(): , True.

• filterfalse(): , False.

6. zip() itertools.zip_longest():

• zip(): , .

• zip_longest(): , .

7. itertools.islice():

• ;

• islice() – . , . 

8. itertools.permutations() itertools.combinations():

• itertools.permutations(): ;

• itertools.combinations(): . 

.

9. itertools.combinations() itertools.combination_swith_replacement:

• combinations(): , .

• combinations_with_replacement(): , .

10. itertools.takewhile() itertools.dropwhile():

• takewhile(): , , True.

• dropwhile(): , , True, .

:

Itertools