小练习

1 寻找第n次出现位置
def search_n(s, c, n):
    size = 0
    for i, x in enumerate(s):
        if x == c:
            size += 1
        if size == n:
            return i
    return -1



print(search_n("fdasadfadf", "a", 3))# 结果为7，正确
print(search_n("fdasadfadf", "a", 30))# 结果为-1，正确

2 斐波那契数列前n项
def fibonacci(n):
    a, b = 1, 1
    for _ in range(n):
        yield a
        a, b = b, a + b


list(fibonacci(5))  # [1, 1, 2, 3, 5]

3 找出所有重复元素
from collections import Counter


def find_all_duplicates(lst):
    c = Counter(lst)
    return list(filter(lambda k: c[k] > 1, c))


find_all_duplicates([1, 2, 2, 3, 3, 3])  # [2,3]

4 联合统计次数
from collections import Counter
a = ['apple', 'orange', 'computer', 'orange']
b = ['computer', 'orange']

ca = Counter(a)
cb = Counter(b)
#Counter对象间可以做数学运算
ca + cb  # Counter({'orange': 3, 'computer': 2, 'apple': 1})


# 进一步抽象，实现多个列表内元素的个数统计


def sumc(*c):
    if (len(c) < 1):
        return
    mapc = map(Counter, c)
    s = Counter([])
    for ic in mapc: # ic 是一个Counter对象
        s += ic
    return s


#Counter({'orange': 3, 'computer': 3, 'apple': 1, 'abc': 1, 'face': 1})
sumc(a, b, ['abc'], ['face', 'computer'])


5 groupby单字段分组
天气记录：

a = [{'date': '2019-12-15', 'weather': 'cloud'},
 {'date': '2019-12-13', 'weather': 'sunny'},
 {'date': '2019-12-14', 'weather': 'cloud'}]
按照天气字段weather分组汇总：

from itertools import groupby
for k, items in  groupby(a,key=lambda x:x['weather']):
     print(k)
输出结果看出，分组失败！原因：分组前必须按照分组字段排序，这个很坑~

cloud
sunny
cloud
修改代码：

a.sort(key=lambda x: x['weather'])
for k, items in  groupby(a,key=lambda x:x['weather']):
     print(k)
     for i in items:
         print(i)
输出结果：

cloud
{'date': '2019-12-15', 'weather': 'cloud'}
{'date': '2019-12-14', 'weather': 'cloud'}
sunny
{'date': '2019-12-13', 'weather': 'sunny'}


6 itemgetter和key函数
注意到sort和groupby所用的key函数，除了lambda写法外，还有一种简写，就是使用itemgetter：

a = [{'date': '2019-12-15', 'weather': 'cloud'},
 {'date': '2019-12-13', 'weather': 'sunny'},
 {'date': '2019-12-14', 'weather': 'cloud'}]
from operator import itemgetter
from itertools import groupby

a.sort(key=itemgetter('weather'))
for k, items in groupby(a, key=itemgetter('weather')):
     print(k)
     for i in items:
         print(i)
结果：

cloud
{'date': '2019-12-15', 'weather': 'cloud'}
{'date': '2019-12-14', 'weather': 'cloud'}
sunny
{'date': '2019-12-13', 'weather': 'sunny'}


7 groupby多字段分组
from operator import itemgetter
from itertools import groupby

a = [{'date': '2019-12-15', 'weather': 'cloud'},
 {'date': '2019-12-13', 'weather': 'sunny'},
 {'date': '2019-12-14', 'weather': 'cloud'}]
a.sort(key=itemgetter('weather', 'date'))
for k, items in groupby(a, key=itemgetter('weather','date')):
     print(k)
     for i in items:
         print(i)

#result：
('cloud', '2019-12-14')
{'weather': 'cloud', 'date': '2019-12-14'}
('cloud', '2019-12-15')
{'weather': 'cloud', 'date': '2019-12-15'}
('sunny', '2019-12-13')
{'weather': 'sunny', 'date': '2019-12-13'}


8 sum函数计算和聚合同时做
Python中的聚合类函数sum,min,max第一个参数是iterable类型，一般使用方法如下：

a = [4,2,5,1]
sum([i+1 for i in a]) # 16
使用列表生成式[i+1 for i in a]创建一个长度与a一行的临时列表，这步完成后，再做sum聚合。

试想如果你的数组a长度十百万级，再创建一个这样的临时列表就很不划算，最好是一边算一边聚合，稍改动为如下：

a = [4,2,5,1]
sum(i+1 for i in a) # 16
此时i+1 for i in a是(i+1 for i in a)的简写，得到一个生成器(generator)对象，如下所示：

In [8]:(i+1 for i in a)
OUT [8]:<generator object <genexpr> at 0x000002AC7FFA8CF0>
生成器每迭代一步吐出(yield)一个元素并计算和聚合后，进入下一次迭代，直到终点。

9 list分组(生成器版)
from math import ceil

def divide_iter(lst, n):
    if n <= 0:
        yield lst
        return
    i, div = 0, ceil(len(lst) / n)
    while i < n:
        yield lst[i * div: (i + 1) * div]
        i += 1

list(divide_iter([1, 2, 3, 4, 5], 0))  # [[1, 2, 3, 4, 5]]
list(divide_iter([1, 2, 3, 4, 5], 2))  # [[1, 2, 3], [4, 5]]


10 列表全展开（生成器版）
#多层列表展开成单层列表
a=[1,2,[3,4,[5,6],7],8,["python",6],9]
def function(lst):
    for i in lst:
        if type(i)==list:
            yield from function(i)
        else:
            yield i
print(list(function(a))) # [1, 2, 3, 4, 5, 6, 7, 8, 'python', 6, 9]