1.map()函數(shù)
接收兩個(gè)參數(shù),一個(gè)是函數(shù)套啤,一個(gè)是Iterable,map將傳入的函數(shù)依次作用到序列的每個(gè)元素随常,并把結(jié)果作為新的Iterator返回
map()傳入的第一個(gè)參數(shù)是f潜沦,即函數(shù)對(duì)象本身。由于結(jié)果r是一個(gè)Iterator绪氛,Iterator是惰性序列唆鸡,因此通過(guò)list()函數(shù)讓它把整個(gè)序列都計(jì)算出來(lái)并返回一個(gè)list
def normalize(name):
return str.capitalize(name)
L1 = ['adam', 'LISA', 'barT']
L2 = list(map(normalize, L1))
print(L2) #['Adam', 'Lisa', 'Bart']
2.reduce()函數(shù)
reduce把一個(gè)函數(shù)作用在一個(gè)序列[x1, x2, x3, ...]上,這個(gè)函數(shù)必須接收兩個(gè)參數(shù)枣察,reduce把結(jié)果繼續(xù)和序列的下一個(gè)元素做累積計(jì)算
reduce(f, [x1, x2, x3, x4]) = f(f(f(x1, x2), x3), x4)
# 請(qǐng)編寫(xiě)一個(gè)prod()函數(shù)争占,可以接受一個(gè)list并利用reduce()求積
from functools import reduce
def prod(L):
return reduce(lambda x,y:x*y,L)
print('3 * 5 * 7 * 9 =', prod([3, 5, 7, 9]))
if prod([3, 5, 7, 9]) == 945:
print('測(cè)試成功!')
else:
print('測(cè)試失敗!')
# 利用map和reduce編寫(xiě)一個(gè)str2float函數(shù),把字符串'123.456'轉(zhuǎn)換成浮點(diǎn)數(shù)123.456:
from functools import reduce
DIGITS = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
def char2num(s):
return DIGITS[s]
def str2float(s):
L1 = s[:s.index('.')]
L2 = s[s.index('.')+1:]
a=reduce(lambda x,y:x*10+y,map(char2num,L1))
b=reduce(lambda x,y:x*10+y,map(char2num,L2))/10**len(L2)
return a+b
print(str2float('123.456'))