冬雪雪冬 发表于 2020-2-28 17:09
用集合的 issubset 方法判断所有元素是否在一个集合里，这种方法我还没想到{:10_275:}
def func(list1):
row1 = "qwertyuiop"
row2 = "asdfghjkl"
row3 = "zxcvbnm"
in1, in2, in3 = False, False, False
result = []
for word in list1:
word = word.lower()
for letter in word:
if letter in row1:
in1 = True
if letter in row2:
in2 = True
if letter in row3:
in3 = True
if (in1 and not in2 and not in3) or (not in1 and in2 and not in3) or (not in1 and not in2 and in3):
result.append(word)
in1, in2, in3 = False, False, False
return result
一个账号 发表于 2020-3-1 13:15
用集合的 issubset 方法判断所有元素是否在一个集合里，这种方法我还没想到
一开始就想到了用集合运算，但忘了如何求子集，还是现查的
def function(strs):
tp=({'t', 'o', 'i', 'w', 'u', 'y', 'r', 'p', 'q', 'e'},{'g', 'h', 'k', 'd', 'a', 'l', 'f', 's', 'j'},{'x', 'n', 'c', 'm', 'v', 'z', 'b'})
rst=[]
for i in strs:
s=set(i.lower())
if not all((s-k for k in tp)):
rst.append(i)
return rst
def f_338(lst: list) -> list:
line_one = 'qwertyuiop'
line_two = 'asdfghjkl'
line_three = 'zxcvbnm'
return list(filter(lambda x: set(x.lower()).issubset(line_one) or
set(x.lower()).issubset(line_two) or
set(x.lower()).issubset(line_three), lst))
print(f_338(["Hello", "Alaska", "Dad", "Peace"]))
def fun338(x):
result = []
line = ['QWERTYUIOP','ASDFGHJKL','ZXCVBNM']
upper_x =
for i in range(len(x)):
for j in line:
for n in upper_x:
if n not in j:
break
else:
result.append(x)
break
return result