I reimplemented the set in python but i have some problem with multiple intersection.... I followed the book Learning Python but i have problem with my code
class Set:
def __init__(self,value=[]):
self.data = []
self.remDupli(value)
def remDupli(self,val):
for i in val:
if i not in self.data:
self.data.append(i)
def intersect(self,other):
val=[]
for i in self.data:
for k in other:
if i == k:
val.append(i)
return Set(val)
def union(self,other):
val=self.data
for i in other:
if i not in self.data:
val.append(i)
return Set(val)
def __or__(self,a): return self.union(a)
def __and__(self,a): return self.intersect(a)
def __len__(self): return len(self.data)
def __getitem__(self,key): return self.data[key]
def __repr__(self): return 'Set: ' +repr(self.data)
class Extend(Set):
def intersect(self, *others):
val = []
for i in self:
for x in others:
if i in x:
val.append(i)
return Set(val)
but when I run this:
x = Extend([1,2,3,4])
y = Extend([3,4,5])
z = Extend([0,1,2])
print(x & y & z)
print(x.intersect(y, z))
I have two different behavior
Set: []
Set: [1, 2, 3, 4]
I don't understand because the second is different, in my opinion they should have the same behavior, anyone can help me?
Extend.intersect does not calculate intersection between many sets. It calculates intersection between self and union of others.
The results are different because x & y & z calls Extend.intersect(Extend.intersect(x,y), z), while x.intersect(y,z) calls Extend.intersect(x, *[y,z]) and given what Extend.intersect actually does, those happen to be different operations.
Related
Let say I have a class:
class MATH(object):
def __init__(self):
self.results = [0, 1, 2]
def add(self, value):
# Add amount 'value' to every element in the results list
def minus(self, value):
# Subtract amount 'value' from every element in the results list
def compute(self):
# Perform computation
Is there a way to do something like:
m = MATH()
m.add(5).minus(2).add(7) # This would be a lazy and not actually compute
m.compute() # This would actually run the computations in order
How do I do something like this in python?
Personally, I would have .add(), et al, push the operator and the operand onto a list and then have .compute() walk through the list, computing the answer as it goes.
Operator chaining is easily accomplished by having each operator return self as its final instruction.
For example:
class MATH(object):
def __init__(self):
self.results = [0, 1, 2]
self.operations = []
def add(self, value):
# Add amount 'value' to every element in the results list
self.operations.append(('+', value))
return self
def minus(self, value):
# Subtract amount 'value' from every element in the results list
self.operations.append(('-', value))
return self
def compute(self):
results = []
for x in self.results:
for op, value in self.operations:
if op == '+':
x += value
elif op == '-':
x -= value
results.append(x)
return results
m = MATH()
m.add(5).minus(2).add(7) # This would be a lazy and not actually compute
print(m.compute()) # This would actually run the computations in order
Wow, you guys are fast!
Here is another go also with a stack, but manipulating the results-list:
class MATH(object):
def __init__(self):
self.results = [0, 1, 2]
self.stack = []
def add(self, value):
self.stack.append(value)
return self
def minus(self, value):
self.stack.append(-value)
return self
def compute(self):
for s in self.stack:
for index, _ in enumerate(self.results):
self.results[index] += s
m = MATH()
m.add(5).minus(2).add(7) # This would be a lazy and not actually compute
m.compute() # This would actually run the computations in order
print m.results
[10, 11, 12]
As #Rob pointed out, you will need some way to store the operators so that the final compute method can be utilized correctly. This solution uses __add__ and __sub__, with a decorator to store the operators. Note, however, that it would be much more efficient to keep a running total of the values that have been pushed to the stack:
import operator as op
from collections import deque
def operator(f):
def wrapper(cls, _):
cls.operators.append(f.__name__.replace('__', ''))
return f(cls, _)
return wrapper
class Math:
def __init__(self):
self.stack = []
self.operators = deque()
#operator
def __sub__(self, _val):
self.stack.append(_val)
return self
#operator
def __add__(self, _val):
self.stack.append(_val)
return self
def compute(self):
_result = 0
while self.stack:
a, *c = self.stack
_result = getattr(op, self.operators.popleft())(_result, a)
self.stack = c
return _result
m = Math()
m1 = m + 5 - 2 + 7
print([m1.stack, m1.operators])
print(m1.compute())
Output:
[[5, 2, 7], ['add', 'sub', 'add']]
10
Here's a string-based approach which requires little brainpower.
class Math:
def __init__(self):
self.stack = '0'
#staticmethod
def wrap(expr):
return '(' + expr + ')'
def _op(self, other, op):
self.stack = ' '.join([Math.wrap(self.stack), op, str(other)])
def add(self, other):
self._op(other, '+')
return self
def mul(self, other):
self._op(other, '*')
return self
def compute(self):
return eval(self.stack)
m = Math()
print(m.add(2).mul(3).compute())
from math import pi
class Circle(object):
'Circle(x,y,r)'
def __init__(self, x=0, y=0, r=1):
self._r = r
self._x = x
self._y = y
def __repr__(self):
return 'Circle({},{},{})'.\
format(self.getx(), self.gety(),\
self.getr())
#silly, but has a point: str can be different from repr
def __str__(self):
return 'hello world'
def __contains__(self, item):
'point in circle'
px, py = item
return (self.getx() - px)**2 + \
(self.gety() - py)**2 < self.getr()**2
def getr(self):
'radius'
return self._r
def getx(self):
'x'
self._lst.append(self._x)
return self._x
def gety(self):
'y'
self._lst.append(self._y)
return self._y
def setr(self,r):
'set r'
self._r = r
def setx(self,x):
'set x'
self._x = x
def sety(self,y):
'set y'
self._y = y
def move(self,x,y):
self._x += x
self._y += y
def concentric(self, d):
d = self._list
def area(self):
'area of circle'
return (self.getr())**2*pi
def circumference(self):
'circumference of circle'
return 2*self.getr()*pi
My question is worded kinda awkwardly but what I am trying to do is check if 2 different circles have the same center (x,y). I think the easiest way to solve this would be to input the 2 points into a list but I am not sure how to compare the 2 lists as every time i try my code it adds everything to the same list
Add the following method to your Circle class.
def equal_center(self, other):
'check if another circle has same center'
return (self._x == other._x) & (self._y == other._y)
Usage
C1 = Circle(3, 5, 8)
C2 = Circle(3, 5, 10)
C3 = Circle(3, 2, 1)
C1.equal_center(C2) # True
C1.equal_center(C3) # False
I would recommend creating a function which takes two circle objects and returns if the coordinates are the same or not by comparing the x and y values of each object:
def same_center(circle_1, circle_2):
if circle_1.getx() == circle_2.getx() and circle_1.gety() == circle_2.gety():
return True
else:
return False
This solution is much easier than using lists and should be easy to implement.
If you have two instances of the class...
a = Circle(0,0,1)
b = Circle(0,0,1)
You could add them to a list of circles...
circles = [a,b]
And loop through the list, checking their values...
for i in circles:
for j in filter(lambda x : x != i, circles):
if i._x == j._x and i._y == j._y:
return True #two circles have same center
This should work for n instances of the class, though if its only two you want to check
if a._x == b._x and a._y == a._y:
return True
I have a list which I want to sort by multiple keys, like:
L = [ ... ]
L.sort(key = lambda x: ( f(x), g(x) ))
This works fine. However, this results with unnecessary calls to g, which I would like to avoid (for being potentially slow). In other words, I want to partially and lazily evaluate the key.
For example, if f is unique over L (i.e. len(L) == len(set(map(f,L)))) no calls to g should be made.
What would be the most elegant/pythonic way to do this?
One way I can think of is to define a custom cmp function (L.sort(cmp=partial_cmp)), but IMO this is less elegant and more complicated than using the key parameter.
Another way would be to define a key-wrapper class which takes a generator expression to generate the different parts of the key, and override the comparison operators to compare one-by-one. However, I'm feeling there must be a simpler way...
EDIT: I'm interested in a solution for the general problem of sorting by multiple functions, not only two as in my example above.
You can try using itertools.groupby:
result = []
for groupKey, group in groupby(sorted(L, key=f), key=f):
sublist = [y for y in group]
if len(sublist) > 1:
result += sorted(sublist, key=g)
else:
result += sublist
Another possibility, even less elegant, but in place:
L.sort(key = f)
start = None
end = None
for i,x in enumerate(L):
if start == None:
start = i
elif f(x) == f(L[start]):
end = i
elif end == None:
start = i
else:
L[start:end+1] = sorted(L[start:end+1], key=g)
start = None
if start != None and end != None:
L[start:end+1] = sorted(L[start:end+1], key=g)
First version generalized to any number of functions:
def sortBy(l, keyChain):
if not keyChain:
return l
result = []
f = keyChain[0]
for groupKey, group in groupby(sorted(l, key=f), key=f):
sublist = [y for y in group]
if len(sublist) > 1:
result += sortBy(sublist, keyChain[1:])
else:
result += sublist
return result
The second version generalized to any number of functions (not fully in place though):
def subSort(l, start, end, keyChain):
part = l[start:end+1]
sortBy(part, keyChain[1:])
l[start:end+1] = part
def sortBy(l, keyChain):
if not keyChain:
return
f = keyChain[0]
l.sort(key = f)
start = None
end = None
for i,x in enumerate(l):
if start == None:
start = i
elif f(x) == f(l[start]):
end = i
elif end == None:
start = i
else:
subSort(l, start, end, keyChain)
start = i
end = None
if start != None and end != None:
subSort(l, start, end, keyChain)
Given a function, you could create a LazyComparer class like this:
def lazy_func(func):
class LazyComparer(object):
def __init__(self, x):
self.x = x
def __lt__(self, other):
return func(self.x) < func(other.x)
def __eq__(self, other):
return func(self.x) == func(other.x)
return lambda x: LazyComparer(x)
To make a lazy key function out of multiple functions, you could create a utility function:
def make_lazy(*funcs):
def wrapper(x):
return [lazy_func(f)(x) for f in funcs]
return wrapper
And together they could be used like this:
def countcalls(f):
"Decorator that makes the function count calls to it."
def _f(*args, **kwargs):
_f._count += 1
return f(*args, **kwargs)
_f._count = 0
return _f
#countcalls
def g(x): return x
#countcalls
def f1(x): return 0
#countcalls
def f2(x): return x
def report_calls(*funcs):
print(' | '.join(['{} calls to {}'.format(f._count, f.func_name)
for f in funcs]))
L = range(10)[::-1]
L.sort(key=make_lazy(f1, g))
report_calls(f1, g)
g._count = 0
L.sort(key=make_lazy(f2, g))
report_calls(f2, g)
which yields
18 calls to f1 | 36 calls to g
36 calls to f2 | 0 calls to g
The #countcalls decorator above is used to connfirm that when f1 returns a lot
of ties, g is called to break the ties, but when f2 returns distinct values,
g does not get called.
NPE's solution adds memoization within the Key class. With the solution above,
you could add memoization outside (independent of) the LazyComparer class:
def memo(f):
# Author: Peter Norvig
"""Decorator that caches the return value for each call to f(args).
Then when called again with same args, we can just look it up."""
cache = {}
def _f(*args):
try:
return cache[args]
except KeyError:
cache[args] = result = f(*args)
return result
except TypeError:
# some element of args can't be a dict key
return f(*args)
_f.cache = cache
return _f
L.sort(key=make_lazy(memo(f1), memo(g)))
report_calls(f1, g)
which results in fewer calls to g:
10 calls to f1 | 10 calls to g
You could use a key object that would lazily evaluate and cache g(x):
class Key(object):
def __init__(self, obj):
self.obj = obj
self.f = f(obj)
#property
def g(self):
if not hasattr(self, "_g"):
self._g = g(self.obj)
return self._g
def __cmp__(self, rhs):
return cmp(self.f, rhs.f) or cmp(self.g, rhs.g)
Here is an example of use:
def f(x):
f.count += 1
return x // 2
f.count = 0
def g(x):
g.count += 1
return x
g.count = 0
L = [1, 10, 2, 33, 45, 90, 3, 6, 1000, 1]
print sorted(L, key=Key)
print f.count, g.count
I am trying to get values from my dictionary VALUES. My program creates combination of possible positions and gets the last position. Then I want to get the value. Everything works well here except indicated .get_value method. When I execute this code I receive:
AttributeError: 'Combination' object has no attribute 'get_value'
Theoretically it should be easy but I am new to OOP and I don't see what is wrong here.
X = ['A','B','C']
Y = ['1','2','3']
VALUES = {'A':10, 'B': 50, 'C':-20}
class Combination:
def __init__(self,x,y):
if (x in X) and (y in Y):
self.x = x
self.y = y
else:
print "WRONG!!"
def __repr__ (self):
return self.x+self.y
def get_x(self):
return self.x
def get_y(self):
return self.y
class Position:
def __init__(self):
self.xy = []
for i in X:
for j in Y:
self.xy.append(Combination(i,j))
def choose_last(self):
return self.xy.pop()
def __str__(self):
return "List contains: " + str(self.xy)
class Operation1:
def __init__(self):
self.operation1 = []
def __str__(self):
s = str(self.operation1)
return s
def get_value(self):
V = VALUES.get(self)
return V
pos = Position()
print pos
last_item = pos.choose_last()
print "Last item:", last_item, pos
last_value = last_item.get_value() # <---- Here is a problem
How can I obtain value of my position? Value is determined by the X value - this is A,B or C. In the dictionary I have a numeral value for the letter.
You are appending objects of Combination into xy of Position. When you say choose_last, it will return the last Combination object inserted into xy. And you are trying to invoke get_value method on a Combination object, which doesnt have that method. Thats why you are getting that error.
Always use new style classes.
I need to operate on two separate infinite list of numbers, but could not find a way to generate, store and operate on it in python.
Can any one please suggest me a way to handle infinite Arithmetic Progession or any series and how to operate on them considering the fact the minimal use of memory and time.
Thanks every one for their suggestions in advance.
You are looking for a python generator instead:
def infinitenumbers():
count = 0
while True:
yield count
count += 1
The itertools package comes with a pre-built count generator.
>>> import itertools
>>> c = itertools.count()
>>> next(c)
0
>>> next(c)
1
>>> for i in itertools.islice(c, 5):
... print i
...
2
3
4
5
6
This is where the iterator comes in. You can't have an infinite list of numbers, but you can have an infinite iterator.
import itertools
arithmetic_progression = itertools.count(start,step) #from the python docs
The docs for Python2 can be found here
I have another python3 solution (read SICP chapter 3.5)
class Stream:
def __init__(self, head, tail):
self.head = head
self.tail = tail
self.memory = None
self.isDone = False
def car(self):
return self.head
def cdr(self):
if self.isDone:
return self.memory
self.memory = self.tail()
self.isDone = True
return self.memory
def __getitem__(self, pullFrom):
if pullFrom < 1 or self.memory == []:
return []
return [self.car()] + self.cdr()[pullFrom - 1]
def __repr__(self):
return "[" + repr(self.car()) + " x " + repr(self.tail) + "]"
def map(self, func):
if self.memory == []:
return []
return Stream(func(self.car()), lambda: Stream.map(self.cdr(), func))
def from_list(lst):
if lst == []:
return []
return Stream(lst[0], lambda:
Stream.from_list(lst[1:]))
def filter(self, pred):
if self.memory == []:
return []
elif pred(self.car()):
return Stream(self.car(), lambda: Stream.filter(self.cdr(), pred))
else:
return self.cdr().filter(pred)
def sieve(self):
return Stream(self.car(), lambda: self.cdr().filter(lambda n: n % self.car() > 0).sieve())
def foreach(self, action, pull = None):
if pull is None:
action(self.car())
self.cdr().foreach(action, pull)
elif pull <= 0:
return
else:
action(self.car())
self.cdr().foreach(action, pull-1)and run:
a = Stream(0, lambda: a.map((lambda x: x + 1)))
print(a[10])
which returns:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9] .
But streams are lazily evaluated, so:
>>> a = Stream(0, lambda: a.map((lambda x: x + 1)))
>>> print(a)
prints:
[0 x [...]]
To create an object that acts like a "mutable" infinite list, you can overload the __getitem__ and __setitem__ methods in a class:
class infinite_list():
def __init__(self, func):
self.func = func
self.assigned_items = {}
def __getitem__(self, key):
if key in self.assigned_items:
return self.assigned_items[key]
else:
return self.func(key)
def __setitem__(self, key , value):
self.assigned_items[key] = value
Then, you can initialize the "infinite list" with a lambda expression and modify an item in the list:
infinite_thing = infinite_list(lambda a: a*2)
print(infinite_thing[1]) #prints "2"
infinite_thing[1] = infinite_thing[2]
print(infinite_thing[1]) #prints "4"
Similarly, it is possible to create an "infinite dictionary" that provides a default value for each missing key.
Perhaps the natural way to generate an infinite series is using a generator:
def arith(a, d):
while True:
yield a
a += d
This can be used like so:
print list(itertools.islice(arith(10, 2), 100))
My solution is:
from hofs import *
def cons_stream(head,tail):
return [head,tail,False,False]
def stream_cdr(strm):
if strm[2]:
return strm[3]
strm[3] = strm[1]()
strm[2] = True
return strm[3]
def show_stream(stream, num = 10):
if empty(stream):
return []
if num == 0:
return []
return adjoin(stream[0], show_stream(stream_cdr(stream), num - 1))
def add_streams(a , b):
if empty(a):
return b
if empty(b):
return a
return cons_stream(a[0] + b[0] , lambda : add_streams( stream_cdr(a), stream_cdr(b)))
def stream_filter( pred , stream ):
if empty(stream):
return []
if pred(stream[0]):
return cons_stream(stream[0], lambda : stream_filter(pred, stream_cdr(stream)))
else:
return stream_filter( pred , stream_cdr( stream ))
def sieve(stream):
return cons_stream(stream[0] , lambda : sieve(stream_filter(lambda x : x % stream[0] > 0 , stream_cdr(stream))))
ones = cons_stream(1, lambda : ones)
integers = cons_stream(1, lambda : add_streams(ones, integers))
primes = sieve(stream_cdr(integers))
print(show_stream(primes))
Copy the Python code above.
When I tried it, i got [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] which is 10 of an infinite list of primes.
You need hofs.py to be
def empty(data):
return data == []
def adjoin(value,data):
result = [value]
result.extend(data)
return result
def map(func, data):
if empty(data):
return []
else:
return adjoin(func(data[0]), map(func, data[1:]))
def keep(pred, data):
if empty(data):
return []
elif pred(data[0]):
return adjoin( data[0] , keep(pred, data[1:]))
else:
return keep(pred, data[1:])
I assume you want a list of infinite numbers within a range. I have a similar problem, and here is my solution:
c = 0
step = 0.0001 # the difference between the numbers
limit = 100 # The upper limit
myInfList = []
while c <= limit:
myInfList.append(c)
c = c + step
print(myInfList)