Python3之Zip

Posted 2020-06-17

from collections import defaultdict
from collections import OrderedDict

d = defaultdict(list)
d[‘a‘].append(1)
d[‘a‘].append(2)
d[‘b‘].append(5)

print(d)

#di = dict(list)
#di[‘b‘].append(11)
#di[‘b‘].append(22)
#di[‘c‘].append(55)
#print(di)

ds = defaultdict(set)
ds[‘b‘].add(11)
ds[‘b‘].add(22)
ds[‘c‘].add(33)
print(ds)

do = OrderedDict()
do[‘foo‘] = 1
do[‘far‘] = 2
do[‘coo‘] = 5

for key in do:
    print(key, do[key])

# 字典的运算
prices = {
    ‘IBM‘: 205.5,
    ‘APPLE‘: 500,
    ‘HP‘: 43,
    ‘FB‘: 88.3
}

min_prices = min(zip(prices.values(), prices.keys()))
print(min_prices)

max_prices = max(zip(prices.values(), prices.keys()))
print(max_prices)

prices_sorted = sorted(zip(prices.values(), prices.keys()))
print(prices_sorted)

z1 = [1,2,3]
z2 = [‘A‘,‘B‘,‘C‘]
z3 = zip(z1, z2)
print(list(z3))
z4 = zip(*zip(z1, z2))
print(list(z4))

x = [1,2]
y = [‘a‘,‘b‘,‘c‘]
print(list(zip(x,y)))

#zip()配合*号操作符,可以将已经zip过的列表对象解压, 并且返回的是元祖


#使用zip和iterators生成滑动窗口 (n -grams) 
from itertools import islice
def n_grams(a, n):
    z = (islice(a, i, None) for i in range(n))
    return zip(*z)

a = [1,2,3,4,5,6]
print(‘(a,3):‘,list(n_grams(a,3)))
print(‘(a,2):‘,list(n_grams(a,2)))
print(‘(a,4):‘,list(n_grams(a,4)))

# 使用zip反转字典
m = {‘a‘: 1, ‘b‘: 2, ‘c‘: 3, ‘d‘: 4}
print(‘m.items = ‘, m.items())
print(list(zip(m.values(), m.keys())))
md = dict(zip(m.values(), m.keys()))
print(md)