集合介绍
Python中的一个set是独特且不可变（不可更改）对象的集合。
要了解一个集合的工作方式，请看下面的代码：
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# start with an empty set
my_set = set()
print(my_set)
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# add a few elements
my_set.add('a')
my_set.add('b')
my_set.add('c')
my_set.add(1)
my_set.add(2)
my_set.add(3)
print(my_set)
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# like a dictionary, a set is UNORDERED.
# We can still loop through a set though.
for element in my_set:
print(element)
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# let's see how many elements are in this set...
print("there are", len(my_set), "elements in my_set")
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# can we make the set bigger by adding more "copies"
# of existing elements?
my_set.add("a")
my_set.add("a")
my_set.add("a")
my_set.add("a")
my_set.add("a")
my_set.add("a")
my_set.add("a")
my_set.add("a")
print("there are", len(my_set), "elements in my_set")
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# there are still only 6 elements...
#
# that's because sets only care about UNIQUE elements.
# They do not allow for multiple "copies"
print(my_set)
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# and they haven't changed. What if we remove "a"
my_set.remove("a")
print("there are", len(my_set), "elements in my_set")
print(my_set)
集合的力量
该set受到一组称为集合论的数学分支的启发而产生。
集合如此有用的原因在于它们可以让我们利用“文氏图逻辑”。
例如，这里有两个集合，一个是包含小于10的质数的primes；一个是包含小于10的奇数的odds。寻找这两个集合之间关系的方法之一是使用 文氏图。
在此图中，蓝色区域包含奇数，红色包含素数，重叠的紫色区域包含同时属于奇数和素数的数字。
注意 - 在这里要慢一点哦！
接下来的几个单元格展示了一些关于集合的一些方法。仔细阅读后，尝试将每种方法与上述的文氏图进行关联。
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# Initializing two sets
odds = set([1,3,5,7,9])
primes = set([2,3,5,7])
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# Demonstration of the "intersection" between two sets
# The intersection corresponds to the overlapping region
# in the Venn Diagram above.
odd_AND_prime = odds.intersection(primes)
print(odd_AND_prime)
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# Demonstration of the "union" of two sets. The union
# of sets A and B includes ANY element that is in A OR B
# or both.
odd_OR_prime = odds.union(primes)
print(odd_OR_prime)
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# Demonstration of the "set difference" between two sets.
# What do you expect odds-primes to return?
odd_not_prime = odds - primes
print(odd_not_prime)
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# Another demo of "set difference"
prime_not_odd = primes - odds
print(prime_not_odd)
并集 vs 交集
两个集合A与B的并集包含A或*B中的元素或* 包含两者的元素。交集包含两者中的元素。
A-B vs B-A
A-B 包含在A中但不在B中的元素。 B-A 包含B中的元素，但不包含A中的元素。
TODO - 练习：A或B，但不能同时包含
编写一个函数，将两个集合（set_a和set_b）作为输入，并返回一个新的集合，其中包含set_a或set_b中的元素，但不包含两者兼有的元素。
在上面的文氏图中，除了重叠的中间区域之外，将包括图表中的所有内容。在这个示例中，将是数字9、1和2。
注 - 尝试在答案中使用以下所有集合操作：
- intersection
- union
- difference
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def a_or_b_but_not_both(set_a, set_b):
"""Returns a set which contains any element that is
a member of set_a OR a member of set_b but NOT a member
of both."""
return
# testing code
assert a_or_b_but_not_both(odds, primes) == set([9,1,2])
print("Nice job! Your function works correctly!")
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#
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# Spoiler Alert! Solution Below!
#
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# Solution 1
def a_or_b_but_not_both(set_a, set_b):
"""Returns a set which contains any element that is
a member of set_a OR a member of set_b but NOT a member
of both."""
a_and_b = set_a.intersection(set_b)
a_or_b = set_a.union(set_b)
return a_or_b - a_and_b
assert a_or_b_but_not_both(odds, primes) == set([9,1,2])
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# Solution 2
def a_or_b_but_not_both(set_a, set_b):
"""Returns a set which contains any element that is
a member of set_a OR a member of set_b but NOT a member
of both."""
a_without_b = set_a - set_b
b_without_a = set_b - set_a
return a_without_b.union(b_without_a)
assert a_or_b_but_not_both(odds, primes) == set([9,1,2])
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# Solution 3
#
# There is actually a REALLY succinct way of writing this
# function using a single character. I'm not going to tell
# you what that is now, but at the end of this notebook
# you'll find a link to Python documentation that will...
太好了，但标签与标记呢？
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def consolidate_labels(t1_labels, t2_labels):
"""
Combines labels from two tickets without duplication.
Given t1_labels and t2_labels (both lists), return
a consolidated list of labels without duplicates.
"""
# TODO - rewrite this function to use sets. You should
# be able to replace all the code below with 1 or 2
# lines if you use sets appropriately.
#
# NOTE - to convert a set back to a list, you can
# use the list() function (demonstrated in the
# cell below).
combined = []
for label in t1_labels:
if label not in combined:
combined.append(label)
for label in t2_labels:
if label not in combined:
combined.append(label)
return combined
# testing code
labels_1 = ["python", "bug", "localization", "bug"]
labels_2 = ["planning", "localization"]
combined_labels = consolidate_labels(labels_1, labels_2)
assert( set(combined_labels) == set(["python", "bug",
"localization", "planning"]))
print("Nice job! Your function works correctly!")
# demonstration of Python's list() and set() functions
odds = [1,3,5,7,9]
odds_as_set = set(odds)
odds_back_to_list = list(odds_as_set)
print("odds: ", odds)
print("as set: ", odds_as_set)
print("back to list:", odds_back_to_list)
# andy's solution
def consolidate_labels(t1_labels, t2_labels):
"""
Combines labels from two tickets without duplication.
Given t1_labels and t2_labels (both lists), return
a consolidated list of labels without duplicates.
"""
return set(t1_labels).union(set(t2_labels))
labels_1 = ["python", "bug", "localization", "bug"]
labels_2 = ["planning", "localization"]
combined_labels = consolidate_labels(labels_1, labels_2)
assert( set(combined_labels) == set(["python", "bug",
"localization", "planning"]))
