python basis of the first class (white piao share)

python data structure (and algorithm analysis)

1.1 decomposed into individual variable sequence:

1.1.1 Solution:

Any sequence can be decomposed into a plurality of sequences of simple assignment, the only requirement is that the structure and the total number of variables to be consistent with the sequence of
example:

tp = (1,1,2)
var1,var2,var3 = tp
print(var1,var2,var3)

data = ['hello','a','b','c','d']
var1,var2,var3,var4,var5 = data
print(var1,var2,var3,var4,var5)

#若变量个数和元素个不统一则会报错:
#ValueError:need more than 2 values to unpack / too many values to unpack

1.2 iterables elements from the decomposition of any length

1.2.1 Solution:

than the one of the most cattle x python very flexible "* expression " the
example:

num1,*kargs,num2 = [1,2,3,4,5,6]
print(num1)# 1
print(num2)# 2
print(*kargs)# [2,3,4,5]
avg = sum(*args)/len(*args)
print(avg)# 2+3+4+5 / 4 == 3.5

A more complex example:

record = [
		('foo',1,2),
		('bar','hello'),
		('foo',2,3),
		('foo',4,4)
		]
		
def do_foo(x,y):
	print('foo',x,y)
def do_bars(s):
	print('bar',s)

for tag,*args in record:
	if tag == 'foo':
		do_foo(*args)
	elif tag == 'bar':
		do_bars(*args)

#有些时候想要丢弃某些值:
record = ('白piao老师', 50, 123.45, (12, 9, 2012)]
name,*_,(*_,year) = record #常常使用_或者ign来表示丢弃值
print(name,year)
#输出结果: 小白piao老师 2012		

#有的时候依托于*可以做出更加精妙的算法
#例子:
def func(items):
	head_var, *tail_var = items
	return var_head + func(tail_var) if tail_var else head_var 

>>>func([1,2,3,4,5])
>>>15
#其实python做这个真心鸡肋....

1.3 Save the last N elements:

If we want iteration or other forms of processing to do a limited historical records of the last few records:

1.3.1 Solution

from collections import deque
def search(lines,pattern,history):
	previous_lines = deque(maxlen = history)
	for line in lines:
		if pattern == line :
			yield line,privious_lines
		privious_lines.append(line)

if __name__ == '__main__':
	with open('test.txt') as f:
		for line,prevlines in search(f,'whitepiao',5):
			for line1 in prevlines:
				print(line1)
			print(line,end = '')
			print('-'*20)

1.3.2 About deque

In fact, essentially deque a queue length may be fixed, if a fixed length, in the case where the queue is full, a new value is added thereto, the oldest value will be automatically ejected,

>>>q = deque(maxlen = 3)
>>>q
>>>deque([],maxlen=3)
>>>q.append(1)
>>>q.append(2)
>>>q.append(3)
>>>q
>>>deque([1,2,3],maxlen = 3)
>>>q.append(4)
>>>q
>>>deque([2,3,4],maxlen = 3)
>>>q.appendleft(77)
>>>q
>>>deque([77,2,3],maxlen = 3)

1.4 find the maximum or minimum of N elements

1.4.1 find the maximum or minimum of N elements in a collection

1.4.2 Solution:

import heapq as h

nums = [1,8, 2, 23, -4, 18, 23, 42, 37, 2]
print(h.nlargest(3,nums))# [42,37,23]
print(h.nsmallest(3,nums)# [-4,1,2]

More used nlargest and nsmallest can accept a parameter key:

proInfo = [
	{'name':'IB','shares':100,'prices':91.1},
	{'name':'APP','shares':99,'prices':107.2},
	{'name':'LEN','shares':80,'prices':99.0},
	{'name':'ASU','shares':80,'prices':91.0},
	{'name':'DEL','shares':95,'prices':89.7},
]
cheap = h.nsmallest(3,proInfo,lambda x:x['price'])
expensive = h.nlargest(3,proInfo,lambda x:x['price'])
>>>cheap
>>>[89.7,91.0,91.1]