WWPD部分省略
Print If
返回列表中满足函数f的元素
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def print_if(s, f):
"""Print each element of s for which f returns a true value.
>>> print_if([3, 4, 5, 6], lambda x: x > 4)
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6
>>> result = print_if([3, 4, 5, 6], lambda x: x % 2 == 0)
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6
>>> print(result) # print_if should return None
None
"""
for x in s:
if f(x):
print(x)
return NoneClose
返回列表中元素大小与下标差的绝对值小于规定值的元素个数
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def close(s, k):
"""Return how many elements of s that are within k of their index.
>>> t = [6, 2, 4, 3, 5]
>>> close(t, 0) # Only 3 is equal to its index
1
>>> close(t, 1) # 2, 3, and 5 are within 1 of their index
3
>>> close(t, 2) # 2, 3, 4, and 5 are all within 2 of their index
4
>>> close(list(range(10)), 0)
10
"""
count = 0
for i in range(len(s)): # Use a range to loop over indices
if abs(s[i]-i)<=k :
count+=1
return countClose List
返回一个列表，元素为给定列表中元素大小与下标差的绝对值小于规定值的元素
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def close_list(s, k):
"""Return a list of the elements of s that are within k of their index.
>>> t = [6, 2, 4, 3, 5]
>>> close_list(t, 0) # Only 3 is equal to its index
[3]
>>> close_list(t, 1) # 2, 3, and 5 are within 1 of their index
[2, 3, 5]
>>> close_list(t, 2) # 2, 3, 4, and 5 are all within 2 of their index
[2, 4, 3, 5]
"""
return [s[i] for i in range(len(s)) if s[i]-i<=k]Squares Only
返回一个列表，元素为给定列表中为完全平方数的元素
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from math import sqrt
def squares(s):
"""Returns a new list containing square roots of the elements of the
original list that are perfect squares.
>>> seq = [8, 49, 8, 9, 2, 1, 100, 102]
>>> squares(seq)
[7, 3, 1, 10]
>>> seq = [500, 30]
>>> squares(seq)
[]
"""
return [int(sqrt(n)) for n in s if sqrt(n)==round(sqrt(n))]注意返回的是开方后的数，是整数
Double Eights
使用递归判断给定的数中是否存在相邻的两个8
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def double_eights(n):
""" Returns whether or not n has two digits in row that
are the number 8. Assume n has at least two digits in it.
>>> double_eights(1288)
True
>>> double_eights(880)
True
>>> double_eights(538835)
True
>>> double_eights(284682)
False
>>> double_eights(588138)
True
>>> double_eights(78)
False
>>> from construct_check import check
>>> # ban iteration
>>> check(LAB_SOURCE_FILE, 'double_eights', ['While', 'For'])
True
"""
if n==0:
return False
if n%100==88:
return True
else:
return double_eights(n//10)Making Onions
判断是否能通过f函数与g函数进行至多limit次操作实现将x转化为y
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def make_onion(f, g):
"""Return a function can_reach(x, y, limit) that returns
whether some call expression containing only f, g, and x with
up to limit calls will give the result y.
>>> up = lambda x: x + 1
>>> double = lambda y: y * 2
>>> can_reach = make_onion(up, double)
>>> can_reach(5, 25, 4) # 25 = up(double(double(up(5))))
True
>>> can_reach(5, 25, 3) # Not possible
False
>>> can_reach(1, 1, 0) # 1 = 1
True
>>> add_ing = lambda x: x + "ing"
>>> add_end = lambda y: y + "end"
>>> can_reach_string = make_onion(add_ing, add_end)
>>> can_reach_string("cry", "crying", 1) # "crying" = add_ing("cry")
True
>>> can_reach_string("un", "unending", 3) # "unending" = add_ing(add_end("un"))
True
>>> can_reach_string("peach", "folding", 4) # Not possible
False
"""
def can_reach(x, y, limit):
if limit < 0:
return False
elif x == y:
return True
else:
return can_reach(f(x), y, limit - 1) or can_reach(g(x), y, limit - 1)
return can_reach