【列表推导式-以及练习题】

通过一行循环判断,遍历出一系列数据的方式是推导式
语法: val for val in Iterable (把想要的值写在 for的左侧)
里面是一行循环判断!根据套在推导式外层的符号判断具体是什么类型的推导式

推导式种类三种:
    [val for val in Iterable]  列表推导式
    {val for val in Iterable}  集合推导式
    {a:b for a,b in iterable}  字典推导式

lst=[1,2,3,4,5,3]
lst_new=[]
for i in range(1,51):
    lst_new.append(i)
print(lst)

print(lst_new)
#基本写法
lst=[i for i in range(2,44)]
print(lst)
#普通推导式
lst=[1,24,2]
lst_new=[]
for i in lst:
    if i %2==0:
        lst_new.append(i)
print(lst_new)
lst1=[i for i in lst if i%2==0]
print(lst1)
#关于推导式的练习


#(1).{'x': 'A', 'y': 'B', 'z': 'C' } 把字典写成x=A,y=B,z=C的列表推导式
dic={"x":"A","y":"B","z":"C"}
lst=[]
# for k,v in dic.items():
#     # print(k,v)
#     res=k+"="+v
#     lst.append(res)
# print(lst)

lst=[k+"="+v for k,v in dic.items()]
print(lst)
# (2).把列表中所有字符变成小写  ["ADDD","dddDD","DDaa","sss"]
lst=["ADDD","dddDD","DDaa","sss"]
# lst_new=[]
# for i in lst:
#     res=i.lower()
#     lst_new.append(res)
# print(lst_new)

lst=[i.lower() for i in lst]
print(lst)

# (3).x是0-5之间的偶数,y是0-5之间的奇数 把x,y组成一起变成元组,放到列表当中
lst=[]
# for i in range(0,6):
#     for j in range(0,6):
#         if i%2==0 and  j%2==1:
#             res=i,j
#             lst.append(res)
# print(lst)
lst = [ (x,y) for x in range(6) if x % 2 == 0 for y in range(6) if y % 2 == 1 ]
print(lst)

# (4).使用列表推导式 制作所有99乘法表中的运算
print("==============")
# for i in range(1,10):
#     for j in range(1,i+1):
#         #:d 整数占位符，可以写也可以不写
#         print("{:d}*{:d}={:2d}".format(i,j,i*j),end=" ")
#     print()
lst=["{:d}*{:d}={:2d}".format(i,j,i*j)  for i in range(1, 10) for j in range(1,i+1)]
print(lst)

# (5)#求M,N中矩阵和元素的乘积
# M = [ [1,2,3],
#       [4,5,6],
#       [7,8,9]  ]

# N = [ [2,2,2],
#       [3,3,3],
#       [4,4,4]  ]
# =>实现效果1   [2, 4, 6, 12, 15, 18, 28, 32, 36]
# =>实现效果2   [[2, 4, 6], [12, 15, 18], [28, 32, 36]]
M=[ [1,2,3],[4,5,6],[7,8,9] ]
N=[ [2,2,2],[3,3,3],[4,4,4] ]
#控制下标就等于控制了最后的结果
"""
M[0][0] * N[0][0] = 2
M[0][1] * N[0][1] = 4
M[0][2] * N[0][2] = 6

M[1][0] * N[1][0] = 12
M[1][1] * N[1][1] = 15
M[1][2] * N[1][2] = 18

M[2][0] * N[2][0] = 28
M[2][1] * N[2][1] = 32
M[2][2] * N[2][2] = 36"""

# lst=[M[i][j]*N[i][j] for i in range(0,3) for j in range(0,3)]
# print(lst)

# 第二个效果
# [ [] , [] , [] ] 通过推导式遍历出三个新列表
lst=[[i] for i in range(3)]
print(lst)

"""
外层i动的慢的,里层的j动的快的,所以下标M[i][j]
在拿出i的时候, 里面的for 循环了三遍 是在一个新的列表当中实现的;
"""
lst=[[M[i][j]*N[i][j] for j in range(3)]for i in range(3)]
print(lst)