1.利用map()函数,把用户输入的不规范的英文名字,变为首字母大写,其他小写的规范名字。输入:[‘adam’, ‘LISA’, ‘barT’],输出:[‘Adam’, ‘Lisa’, ‘Bart’]:
def normalize(name):
lower = name.lower()
return lower.title()
2.Python提供的sum()函数可以接受一个list并求和,请编写一个prod()函数,可以接受一个list并利用reduce()求积:
from functools import reduce
def prod(L):
return reduce(mul,L)
def mul(x,y):
return x * y
3.利用map和reduce编写一个str2float函数,把字符串’123.456’转换成浮点数123.456:
from functools import reduce
def str2float(s):
def number_int(s):
numbers = {'1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8,
'9': 9}
return numbers[s]
def flt(x, y):
return x * 10 + y
return reduce(flt, map(number_int, s.replace('.', ''))) / (10 ** (len(s) -
s.index('.') - 1))
测试:
print('str2float(\'123.456\') =', str2float('123.456'))
if abs(str2float('123.456') - 123.456) < 0.00001:
print('测试成功!')
else:
print('测试失败!')
4.回数是指从左向右读和从右向左读都是一样的数,例如12321,909。请利用filter()筛选出回数:
方法一:
from functools import reduce
def is_palindrome(n):
return n == reverse_number(n)
def reverse_number(n):
n_r = list(str(n))
n_r.reverse()
return reduce(int_number, map(get_int, n_r))
def get_int(n):
ints = {'1':1, '2':2, '3':3, '4':4, '5':5, '6':6, '7':7, '8':8, '9':9, '0':0}
return ints[n]
def int_number(x, y):
return x * 10 + y
方法二:
def is_palindrome(n):
t = list(str(n))[:]
t.reverse()
if list(str(n)) == t:
return True
方法三:
def is_palindrome(n):
if str(n)[:] == str(n)[::-1]:
return True
# 测试:
output = filter(is_palindrome, range(1, 1000))
print('1~1000:', list(output))
if list(filter(is_palindrome, range(1, 200))) == [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191]:
print('测试成功!')
else:
print('测试失败!')