1、列表listpython
a.基礎app
1.spa
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
中括號括起來;,分割每一個元素;列表中的元素能夠是數字,字符串,列表,布爾值......;列表能夠嵌套orm
(他就是個集合,內部放置任何東西)對象
2.能夠進行索引,切片取值blog
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
print(li[2])
print(li[2:5])
3.利用for、while循環 continue和break也能夠用排序
4.不一樣於字符串,列表元素是能夠被修改的 索引、切片也能夠修改索引
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
li[2] = 1200
print(li)
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
li[2:3] = [90, 100]
print(li)
5.刪除 del 列表[a] /del 列表[a:b]ip
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
del li[2:6]
print(li)
6.支持in操做 in 、not in 進行列表判斷rem
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
v=12 in li
print(v)
7.能夠一直往裏邊找
li = [1, 12, 9, "age", ["石振文", ["19", 10], "龐麥郎"], "alex", True]
print(li[4][2][0])
8.字符串直接轉成list格式,他默認每個字符成爲列表元素(內部使用for循環)
數字不能迭代(循環),不能轉換爲list類型
9.列表轉換爲字符串的時候,須要本身調用for循環,不然系統直接加一個" " 既有數字又有字符串,用下面的方法
li = [11, 22, 33, 44, "123", "alex"]
s = ""
for i in li:
s = s + str(i)
print(s)
列表中的元素若是隻有字符串的話,用join
li = [ "123", "alex"]
v="".join(li)
print(v)
b.list類中提供的方法
1.append 原來值最後追加
li = [11, 22, 33, 22, 44] v=li.append([1234,2323]) print(li)
print(v)
2.clear 清空列表
li = [11, 22, 33, 22, 44]
li.clear()
print(li)
3.拷貝,淺拷貝cope
v = li.copy()
print(v)
4.count 計算元素出現的次數
li = [11, 22, 33, 22, 44]
v = li.count(22)
print(v)
5.extend 擴展原列表,參數:可迭代對象 區分一下extend appendli = [11, 22, 33, 22, 44]li.append([9898,"不得了"])
[11, 22, 33, 22, 44, [9898, '不得了']]
li.extend([9898,"不得了"])
for i in [9898,"不得了"]:
li.append(i)
[11, 22, 33, 22, 44, 9898, '不得了']
li.extend("不得了")
print(li)
[11,22,33,44,9898,「不」,"得」,"了"]
6.index 根據值獲取當前值索引位置(左邊優先)
li = [11, 22, 33, 22, 44]
v= li.index(22)
print(v)
7. insert 在指定索引位置插入元素
li = [11, 22, 33, 22, 44]
li.insert(0,99)
print(li)
[]
八、pop 刪除某個值(1.指定索引;2. 默認最後一個),並獲取刪除的值
li = [11, 22, 33, 22, 44]
v = li.pop()
print(li)
print(v)
li = [11, 22, 33, 22, 44]
v = li.pop(1)
print(li)
print(v)
9.remove 刪除列表中的指定值,左邊優先
li = [11, 22, 33, 22, 44]
li.remove(22)
print(li)
PS:刪除 pop remove del li[0] del li[7:9] clear
10. reverse 將當前列表進行翻轉
li = [11, 22, 33, 22, 44]
li.reverse()
print(li)
11. sort 列表的排序
li = [11,44, 22, 33, 22]
li.sort() 從小到大排序
li.sort(reverse=True) 從大到小排序
print(li)
2、元祖
元組,元素不可被修改,不能被增長或者刪除
tuple
tu = (11,22,33,44)
tu.count(22),獲取指定元素在元組中出現的次數
tu.index(22)
1. 書寫格式
tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
通常寫元組的時候,推薦在最後加入 ,
元素不可被修改,不能被增長或者刪除
2. 索引
v = tu[0]
print(v)
3. 切片
v = tu[0:2]
print(v)
4. 能夠被for循環,元祖可迭代對象
for item in tu:
print(item)
5. 轉換
s = "asdfasdf0"
li = ["asdf","asdfasdf"]
tu = ("asdf","asdf")
v = tuple(s)
print(v)
v = tuple(li)
print(v)
v = list(tu)
print(v)
v = "_".join(tu)
print(v)
li = ["asdf","asdfasdf"]
li.extend((11,22,33,))
print(li)
6.元組的一級元素不可修改/刪除/增長
tu = (111,"alex",(11,22),[(33,44)],True,33,44,)
元組,有序。
v = tu[3][0][0]
print(v)
v=tu[3]
print(v)
tu[3]=123 報錯:元祖的一級元素不能修改
tu[3][0] = 567 進入了列表,列表內部能夠修改
print(tu)
7.tu.count(2),獲取指定元素在元祖中出現的次數
tu.index(22),獲取某個值的索引位置
3、字典
dict
dict 鍵值對 字典的key:數字,字符串,元祖 字典的value能夠是任意類型的值 字典是無序的
基本結構
dic = {
"k1": 'v1',
"k2": 'v2'
} 兩個鍵值對
1 根據序列,建立字典,並指定統一的值
v = dict.fromkeys(["k1",123,"999"],123)
print(v)
v=v[key] 能夠取裏邊的值
2 根據Key獲取值,key不存在時,能夠指定默認值(None)
v = dic['k11111']
print(v)
v = dic.get('k1',111111)
print(v)
3 刪除del v[key][index]
刪除並獲取值
dic = {
"k1": 'v1',
"k2": 'v2'
}
v = dic.pop('k1',90)
print(dic,v)
k,v = dic.popitem()
print(dic,k,v)
4 設置值,
已存在,不設置,獲取當前key對應的值
不存在,設置,獲取當前key對應的值
dic = {
"k1": 'v1',
"k2": 'v2'
}
v = dic.setdefault('k1111','123')
print(dic,v)
5 更新
dic = {
"k1": 'v1',
"k2": 'v2'
}
dic.update({'k1': '111111','k3': 123})
print(dic)
dic.update(k1=123,k3=345,k5="asdf")
print(dic)
6 keys() 7 values() 8 items() get update
#########
一、基本機構
info = {
"k1": "v1", # 鍵值對
"k2": "v2"
}
### 2 字典的value能夠是任何值
info = {
"k1": 18,
"k2": True,
"k3": [
11,
[],
(),
22,
33,
{
'kk1': 'vv1',
'kk2': 'vv2',
'kk3': (11,22),
}
],
"k4": (11,22,33,44)
}
print(info)
### 3 列表、字典不能做爲字典的key
info ={
1: 'asdf',
"k1": 'asdf',
True: "123",
# [11,22]: 123
(11,22): 123,
# {'k1':'v1'}: 123
}
print(info)
4 字典無序
info = {
"k1": 18,
"k2": True,
"k3": [
11,
[],
(),
22,
33,
{
'kk1': 'vv1',
'kk2': 'vv2',
'kk3': (11,22),
}
],
"k4": (11,22,33,44)
}
print(info)
五、索引方式找到指定元素
info = {
"k1": 18,
2: True,
"k3": [
11,
[],
(),
22,
33,
{
'kk1': 'vv1',
'kk2': 'vv2',
'kk3': (11,22),
}
],
"k4": (11,22,33,44)
}
# v = info['k1']
# print(v)
# v = info[2]
# print(v)
v = info['k3'][5]['kk3'][0]
print(v)
6 字典支持 del 刪除
info = {
"k1": 18,
2: True,
"k3": [
11,
[],
(),
22,
33,
{
'kk1': 'vv1',
'kk2': 'vv2',
'kk3': (11,22),
}
],
"k4": (11,22,33,44)
}
del info['k1']
del info['k3'][5]['kk1']
print(info)
7 for循環
dict
info = {
"k1": 18,
2: True,
"k3": [
11,
[],
(),
22,
33,
{
'kk1': 'vv1',
'kk2': 'vv2',
'kk3': (11,22),
}
],
"k4": (11,22,33,44)
}
對字典for循環的時候,默認循環它的全部的key
for item in info:
print(item)
for item in info.keys():
print(item)
for item in info.values():
print(item)
for item in info.keys():
print(item,info[item])
for k,v in info.items(): (items獲取鍵值對)
print(k,v)
True 1 False 0
info ={
"k1": 'asdf',
True: "123",
# [11,22]: 123
(11,22): 123,
# {'k1':' v1'}: 123
}
print(info)
4、小結
1、數字int(..)2、字符串replace/find/join/strip/startswith/split/upper/lower/formattempalte = "i am {name}, age : {age}"# v = tempalte.format(name='alex',age=19)v = tempalte.format(**{"name": 'alex','age': 19})print(v)3、列表append、extend、insert索引、切片、循環4、元組忽略索引、切片、循環 一級元素不能被修改5、字典get/update/keys/values/itemsfor,索引dic = { "k1": 'v1'}v = "k1" in dicprint(v)v = "v1" in dic.values()print(v)6、布爾值0 1bool(...)None "" () [] {} 0 ==> False