What are the python sequence types?
What are the sequence types in python? What is a sequence type in Python? Learn more about it in this article.
Sequence: character, list, tuple
All sequences support iteration
Sequence Represents an ordered collection of objects whose index is a non-negative integer
Characters and tuples belong to immutable sequences, and lists are mutable
1)Characters
String literal: Put the text in single quotes, double quotes or triple quotes;
' '' ''' >>> str1 = ' hello, fanison ' >>> type(str1) str
If you want to use unicode encoding, use the character u before the character to identify it
>>> str2 = u'你好,fanison' >>> type(str2) unicode
Documentation string: If the first statement of a module, class or function is one character, the string becomes the documentation string and can be referenced using the __doc__ attribute;
Example:
>>> def printName(): "the function is print hello" print 'hello' >>> printName.__doc__
operator:
# S [i] Return to a sequence element I
slicing computing s [i: j] Return to one Slicing
Extended slicing operator s[i:j:stride]
Example:
>>> str3 = 'hello,fanison' >>> str2[0:] 'hello,fanison' 返回所有元素 >>> str2[0:7] 'hello,f' 返回索引7之前的所有元素 >>> str2[0:7:2] 'hlof' 返回从索引0到6内步径为2的元素,即隔一个取一个 >>> str2[7:0:-2] 'a,le' 从索引7处倒着隔一个取一个取到索引1处 >>> str2[-4:-1] 'iso' 从索引-4处取到-2处 >>> str2[-4::-1] 'inaf,olleh' 从-4处到开始处倒着取
Note:
Step If it is positive, it means that the index is taken from small to large i < j
The step path is negative, it means that it is taken backward, and the index is from large to small i > j
Supported Operations:
Indexing, slicing, min(), max(), len(), etc.
len(s) The number of elements in s
min (s) s minimum value
Max (s) s) The maximum value of
## Support method: S.index (SUB [,start [,end]]) Find the position where the specified string sub first appears S.upper() Convert a string to uppercase form S.lower() one String into a lowercase form # S.Join (T) uses S as a string in the separatist connection sequence T
>>> l1 = list(str1) >>> l1 ['h', 'e', 'l', 'l', 'o', ',', 'f', 'a', 'n', 'i', 's', 'o', 'n'] >>> ''.join(l1) 'hello,fanison' 使用空字符作为分隔符连接列表l1 S.replace(old, new[, count]) 替换一个字符串 >>> str1.replace('fan','FAN') 'hello,FANison'
## Use help () to get Its help
>>> help(str.join)
2) List
List: Container Type
. Modify
to modify the specified index element, modify the specified shard, delete the statement, the built -in method
##
>>> list1 = [ 1,2,3,'x','n' ] >>> list1[1]=56 >>> print list1 [1, 56, 3, 'x', 'n'] >>> list1[1:3]=[] 会删除索引1到索引3之前的元素 >>> print list1 [1, 'x', 'n'] >>> del(list1[1]) 使用del函数删除list索引为1的元素 >>> print list1 [1, 'n']
## Because support the original modification of the original place , Will not change the memory location, you can use ID () to view its location change
Built
## L.append(object) Append a new element to the end of L ’s . Add a merged list (the contents of the second list will be appended to the end as a single element)>>> l1 = [ 1,2,3 ] >>> l2 = [ 'x','y','z'] >>> l1.append(l2) >>> l1 [1, 2, 3, ['x', 'y', 'z']] 使用append方法会以其原有存在形式追加 >>> l1 = [ 1,2,3 ] >>> l1.extend(l2) >>> l1 [1, 2, 3, 'x', 'y', 'z'] 注意两种增加的区别
>>> l1 = [ 'x',2,'abc',16,75 ] >>> l1.pop(2) pop方法是按索引移除 'abc' >>> l1 ['x', 2, 16, 75] >>> l1.remove(16) remove方法是按值移除 >>> l1 ['x', 2, 75]
L.insert(index, object) 在索引index处插入值
>>> l1.insert(1,'abc') >>> l1 ['x', 'abc', 2, 75]
L.sort() 排序
L.reverse() 逆序
>>> l1.sort() [2, 75, 'abc', 'x'] >>> l1.reverse() ['x', 'abc', 75, 2]
l1 + l2: 合并两个列表,返回一个新的列表;不会修改原列表;
>>> l1 = [ 1,2,3] >>> l2 = [ 'x','y','z'] >>> l1 + l2 [1, 2, 3, 'x', 'y', 'z']
l1 * N: 把l1重复N次,返回一个新列表;
>>> l1 * 3 [1, 2, 3, 1, 2, 3, 1, 2, 3] 使用id()查看是否生成新列表
成员关系判断字符:
in 用法: item in container
not in item not in container
>>> l1 = [ 'x','y',3 ] >>> 'y' in l1 True >>> 'x' not in l1 False
列表解析:[]
列表复制方式:
浅复制:两者指向同一内存对象
>>> l1 = [ 1,2,3,4 ] >>> l2 = l1 >>> id(l1) == id(l1) True 可以看出两者内存地址相同 >>> l1.append('x') >>> print l1 [ 1,2,3,4,'x' ] >>> print l2 [ 1,2,3,4,'x' ]
深复制:两者指向不同内存对象
1)导入copy模块,使用deepcoop方法
>>> import copy >>> l3 = copy.deepcopy(l1) >>> id(l3) == id(l1) False 地址不同
2)复制列表的所有元素,生成一个新列表
>>> l4 = l1[:] >>> print l4 [ 1,2,3,4,'x' ] >>> l1.append(6) >>> print l1 [ 1,2,3,4,'x',6 ] l1改变 >>> print l4 [ 1,2,3,4,'x' ] l4不变
3)元组
表达式符号:()
容器类型
任意对象的有序集合,通过索引访问其中的元素,不可变对象,长度固定,异构,嵌套
常见操作:
>>> t1 = ( 1,2,3,'xyz','abc') >>> type(t1) tuple >>> len(t1) 5 >>> t2 = () 定义一个空元组 >>> t3 = ( , ) SyntaxError: invalid syntax 报错:使用逗号分隔的条件是最少要有一个元素
(1,)
>>> t1[:] ( 1,2,3,'xyz','abc' ) >>> t1[1:] (2, 3, 'xyz', 'abc')
(1,2)
>>> t1[1:4] (2, 3, 'xyz') >>> t4 = 'x',1,'yz',45,[2,4,6] 注意!!!这样也可以生成元组 >>> t4 ('x', 1, 'yz', 45, [2, 4, 6])
t1 + t4: 合并两个元组,返回一个新的元组;不会修改原元组;
>>> t1 + t4 (1, 2, 3, 'xyz', 'abc', 'x', 1, 'yz', 45, [2, 4, 6])
t1 * N: 把l1重复N次,返回一个新元组;
>>> t1 * 3 (1, 2, 3, 'xyz', 'abc', 1, 2, 3, 'xyz', 'abc', 1, 2, 3, 'xyz', 'abc')
成员关系判断
in
not in
注意:
虽然元组本身不可变,但如果元组内嵌套了可变类型的元素,那么此类元素的修改不会返回新元组;
例:
>>> t4 = ('x', 1, 'yz', 45, [2, 4, 6]) >>> id(t4) 44058448 >>> t4[4] [2, 4, 6] >>> t4[4].pop() 弹出列表内一个元素 6 >>> print t4[4] [2, 4] >>> print t4 ('x', 1, 'yz', 45, [2, 4]) >>> id(t4) 44058448 由此可见,对元组内列表内的修改也会使元组发生改变,没有返回新元组
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