From廖雪峰的官方網(wǎng)站----Python3_學(xué)習(xí)記錄
17\04\2017
Some useful tips
|字符| ASCII| Unicode |UTF-8|
| ------------- |:-------------:|: -----:|:----|
|A |01000001 |00000000 01000001 |01000001
|中 |x |01001110 00101101 |11100100 10111000 10101101
Usually we use these at the front of our python programme
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
常見的占位符有：
|%d |整數(shù)| int|
| ---- |:-------------:|: -----:|
|%f |浮點數(shù)|float
|%s |字符串|string
|%x |十六進(jìn)制整數(shù)|decimal
>>> 'Age: %s. Gender: %s' % (25, True)
'Age: 25. Gender: True'
About ''
>>> print('I\'m ok.')
I'm ok.
>>> print('I\'m learning\nPython.')
I'm learning
Python.
>>> print('\\\n\\')
\
\
>>> print('\\\t\\')
\ \
>>> print(r'\\\t\\')
\\\t\\
>>> print('''line1
... line2
... line3''')
line1
line2
line3
Tower of Hanoi
The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:
- Only one disk can be moved at a time.
- Each move consists of taking the upper disk from one of the stacks and placing it
on top of another stack i.e. a disk can only be moved if it is the uppermost disk
on a stack. - No disk may be placed on top of a smaller disk.
Tower of Hanoi
To solve this problem in python, we need to use recursion.
#where n is the number of disk on A
#Generally it's like mathematical induction, find n=1 and n=n-1, then get the answer
def move(n, a, b, c):
if n = 1:
print(a, '-->', c)
else:
move(n-1, a, c, b)#n-1 disks move from A to B as buffer
move(1, a, b, c)#move the biggest disk from A to C
move(n-1, b, a, c)#move the rest n-1 disks left on B
move(3, 'A', 'B', 'C')
expected output is
A --> C
A --> B
C --> B
A --> C
B --> A
B --> C
A --> C
***
#####Pascal's triangle (楊輝三角）
1
1 1
1 2 1
1 3 3 1
To solve this problem in python, we can use a **generator**.
-- coding: utf-8 --
def triangles():
f = [1]
while True:
yield f
f = [f[0]] + [f[i-1] + f[i] for i in range(len(f)) if i>0] + [f[-1]]
# Or in another way
# f = [1] + [f[i-1] +f[i] for i in range(len(f)) if i>0] + [1]
n = 0
for t in triangles():
print(t)
n = n + 1
if n == 10:
break
>
The expect output is:
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
***
#####Iterable and Iterator
可以直接作用于for循環(huán)的對象統(tǒng)稱為可迭代對象：Iterable。
可以使用isinstance()判斷一個對象是否是Iterable對象：
from collections import Iterable
isinstance([], Iterable)
True
isinstance({}, Iterable)
True
isinstance('abc', Iterable)
True
isinstance((x for x in range(10)), Iterable)
True
isinstance(100, Iterable)
False
可以被next()函數(shù)調(diào)用并不斷返回下一個值的對象稱為迭代器：Iterator。
可以使用isinstance()判斷一個對象是否是Iterator對象：
from collections import Iterator
isinstance((x for x in range(10)), Iterator)
True
isinstance([], Iterator)
False
isinstance({}, Iterator)
False
isinstance('abc', Iterator)
False
把list、dict、str等Iterable變成Iterator可以使用iter()函數(shù)：
isinstance(iter([]), Iterator)
True
isinstance(iter('abc'), Iterator)
True
***
#####Map/Reduce
Simple example:
In Python code:
def f(x):
... return x * x
...
r = map(f, [1, 2, 3, 4, 5, 6, 7, 8, 9])
list(r)
[1, 4, 9, 16, 25, 36, 49, 64, 81]
Simple reduce example:
>Combine with **map()** and **lambda function**, change str into int.
from functools import reduce
def char2num(s):
return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
def str2int(s):
return reduce(lambda x, y: x * 10 + y, map(char2num, s))
Transform String '123.456' into Float 123.456:
-- coding: utf-8 --
from functools import reduce
def str2float(s):
def str2f(s):
return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
return reduce(lambda x, y : x*10+y, map(str2f, s[:s.find('.')])) + reduce(lambda x, y : x/10+y, map(str2f, s[:s.find('.'):-1]))/10
print('str2float('123.456') =', str2float('123.456'))
***
end of 17/04/2017
***
19/04/2017
***
#####enumerate
一般情況下對一個列表或數(shù)組既要遍歷索引又要遍歷元素時，會這樣寫：
for i in range (0,len(list)):
print i ,list[i]
使用enumerate后：
for index，text in enumerate(list)):
print index ,text
