04. tuple, list, dict, map(), reduce(), filter(), set()

由 datahunter 在日, 09/09/2012 - 00:45 發表

最後更新: 2020-12-11

list

Usage:

mylist = ["Linux", "Mac OS" , "Windows"]

String to List:

list('Hello')

['H', 'e', 'l', 'l', 'o']

for 與 if

for var in list:

if 'curly' in list:

range

range(n) 輸出 0, 1, ... n-1

range(a, b) 輸出 a, a+1, ... b-1

max 與 min

num=range(10)
max (num)
min (num)

del

var = 6

del var

----

list = ['a', 'b', 'c', 'd']

del list[0] # Delete first element

del list[-2:] # Delete last two elements

----

dict = {'a':1, 'b':2, 'c':3}

del dict['b'] # Delete 'b' entry

count:

>>> x = [[1, 2], 1, 1, [2, 1, [1, 2]]]
>>> x.count(1)
2

結合:

c=a+b

a.extend(b) # 相當於 +=

組合 zip:

names = ['apple', 'orange', 'banana', 'pear']

color = ['red', 'orange', 'yellow', 'green']

zip(names, color)

result:

[('apple', 'red'), ('orange', 'orange'), ('banana', 'yellow'), ('pear', 'green')]

再把 list 轉化為 dict

dict(zip(names, color))

remove

list.remove(elem)

>>> x = ['a', 'b', 'c', 'a']
>>> x.remove('a')
>>> x # ['b', 'c', 'a'] 只移除第一個

# Raises ValueError if the value is not present.

Code:

mylist=["A","B","C","A"]

for myitem in mylist:
        if mylist.count(myitem) > 1:
                mylist.remove(myitem)
                print myitem

./test.py

append 與 pop:

list.append(elem)

沒有 return 的 !

list.pop(index)

return 被 pop 那個 item, 沒有指定 index 時, Default: index=-1

-1 = 由頭開始

0 = 順序開始

N = 第幾個.

list.insert(index, elem) #

list.index(elem)

sorting

list.sort()

list.sorted() # 有 reture

ist.reverse()

sorted(strs, key=MyFn, cmp=cmpFn)
Default cmp: cmp(a, b)

strs = ['ccc', 'aaaa', 'd', 'bb']

print sorted(strs, key=len) ## ['d', 'bb', 'ccc', 'aaaa']

print sorted(strs, key=str.lower) ## ['aa', 'BB', 'CC', 'zz']

join:

# 用空格將 a, b, c join 埋

" ".join(["a", "b"," c"])

in:

mylist = [1,2,3,4,'a','b','c']
'a' in mylist        # True

2-Level

arr = []
arr.append([])
arr[0].append('aa1')
arr[0].append('aa2')

arr = []
arr.append(['aa1', 'aa2'])

Example

>>> foo = ['bar' for _ in range(10)]
>>> print foo
['bar', 'bar', 'bar', 'bar', 'bar', 'bar', 'bar', 'bar', 'bar', 'bar']

tuple

特性:

fixed size
immutable

mytuple = (1, 2, 'hi')
mytuple[2]="bye" ## 唔得
mytuple = (1, 2, 'bye') ## this works
(x, y, z) = (42, 13, "hike")

tuple([1, 2, 3])

dict

Group values into a structure
Refer to each value by name - key
Arbitrary order

定義 dict:

dict = {key1:value1, key2:value2, ... }

兩個 list 合成 dict

names = ["Peter", "Tom", "May"]
foods = ["apple", "orange", "pear"]
mydict = dict(zip(names, foods))

# Test
print(mydict["Peter"])                 # apple

建立 Dict

空白的 dict

mydict = {}

fromkeys

x = ('key1', 'key2', 'key3')
y = 0
mydict = {}.fromkeys(x, y)
print mydict                       # {'key3': 0, 'key2': 0, 'key1': 0}

key = value

dict['a'] = 'alpha'
dict['g'] = 'gamma'
dict['o'] = 'omega'

把 seq 轉成 dict

mydict = dict(seq)

dict 中 dict

people = {
    'Peter': {
        'id': '001',
        'addr': 'Here'
    },
    'Tom': {
        'id': '002',
        'addr': 'There'
}

keys(), values(), items()

dict.keys()

dict.values()

dict.items()

{"a":"alpha", "o":"omega", "g":"gamma"} 轉成了 [('a', 'alpha'), ('o', 'omega'), ('g', 'gamma')]

應用

mydict = {"a":"alpha", "o":"omega", "g":"gamma"}
for i in mydict:
    print(i)             # 一行一個 a o g

for i in mydict.items():
    print(i)             # 一行一個 ('a', 'alpha') ('o', 'omega') ('g', 'gamma')

clear:

# {} 與 .clear()

a = {"x", "y"}

b = a

b={} # 只影響自己 "b"

b.clear() # 連 a 也會影響, print a 只有 None

copy 與 deepcopy

x = {'username': 'admin', 'machines': ['foo', 'bar', 'baz']}

y = x.copy()

y['machines'].remove('bar') # 雖然只對 y 操作, 但仍會影響 x

解決方法:

from copy import deepcopy

y = x.deepcopy()

Output:

>>> d.has_key('name') # 如果直接 print d['name'] 就會出 Error

dict.get 的好處:

>>> print d.get('name') # 如果直接 print d['name'] 就會出 Error

None

Default Value

#!/usr/bin/python

myDict={'A':1, 'B':2, 'C':3}
print(myDict['D'])

output

Traceback (most recent call last):
  File "./test.py", line 5, in <module>
    print(myDict['D'])
KeyError: 'D'

# Set default value

print(myDict.get('D', 'Null'))

%

"Peter 's user_id is %(Peter)s." % mydict

>>> d = {'x': 1, 'y': 2}

>>> d.pop('x')

>>> d

{'y': 2}

用 dict 來 update dict

>>> x = {'name': 'peter'}

>>> d.update(x)

map(function, iterable, ...)

Apply function to every item of iterable and return a list of the results

Syntax: map(fun, iter ...)

the result is always a List(py2), Iterator(py3)

If additional iterable arguments are passed,

function must take that many arguments and is applied to the items from all iterables in parallel.

If one iterable is shorter than another it is assumed to be extended with None items.

Example:

[1]

strip_list = map(str.strip, lines)

str.strip # unbound method

[2]

lines1=(1,2,3)
lines2=(4,5,6)

def add(x,y):
    return x + y

print map(add,lines1,lines2)

Remark

lists comprehensions:

strip_list = [item.strip() for item in lines]

reduce()

Applies function of two arguments cumulatively to the items of iterable, from left to right,
so as to reduce the iterable to a single value.

Syntax: reduce (function, iterable)

i.e.

reduce(lambda x, y: x+y, [1, 2, 3, 4])

10 # ((1+2)+3)+4)

filter()

Returns a sequence from those elements of iterable for which function returns True.

Syntax: filter (function, iterable)

i.e.

>>> def f(x):
...     if x > 0:
...         return x
>>> filter(f, [-1, 0, 1])
[1]

set()

介紹:

Set 是無序及元素不重複的集合

e.g.

myset = set("[email protected]")
print myset

set(['@', 'c', 'e', 'm', 'o', '.', 's', 't'])

運算:

in 元素是否在集合中
& 交集
| 聯集
^ XOR
> 左是否為右的父集
- 差

{} 不是空的 set 來. 如果要建立空的 set, 那要用 set()

測試:

>>> set1 = {'AAA', 'BBB'}
>>> set2 = {'AAA', 'CCC', 'DDD'}
>>> 'AAA' in set1
True
>>> set1 & set2
{'AAA'}
>>> set1 | set2
{'AAA', 'BBB', 'CCC', 'DDD'}
>>> set1 - set2
{'caterpillar'}
>>> set1 ^ set2
{'BBB', 'CCC', 'DDD'}
>>> set1 > set2
False
>>> set1 < set2
False
>>>

# 加元素

s.add('e')

# 當set與list混在一起時用"&","|"

# &

admins.intersection(users)

# |

admins.union(users)

# 詳見

>>> dir(set()

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