python 之 匿名函数

5.14 匿名函数

lambda x , y : x+y

1 匿名的目的就是要没有名字，给匿名函数赋给一个名字是没有意义的

2 匿名函数的参数规则、作用域关系与有名函数是一样的

3 匿名函数的函数体通常应该是 一个表达式,该表达式必须要有一个返回值

f=lambda x,n:x ** n
print(f(2,3))

lambda匿名函数的应用：**max,min,sorted,map,reduce,filter**

求工资最高的人：max

salaries={
    'egon':3000,
    'alex':100000000,
    'wupeiqi':10000,
    'yuanhao':2000
}
def get(k):
    return salaries[k]
print(max(salaries,key=get)) #'alex' 
print(max(salaries,key=lambda x:salaries[x]))
info = [
    {'name': 'egon', 'age': '18', 'salary': '3000'},
    {'name': 'wxx', 'age': '28', 'salary': '1000'},
    {'name': 'lxx', 'age': '38', 'salary': '2000'}
]
max(info, key=lambda dic: int(dic['salary']))
max([11, 22, 33, 44, 55])

求工资最低的人：min

salaries={
    'egon':3000,
    'alex':100000000,
    'wupeiqi':10000,
    'yuanhao':2000
}
print(min(salaries,key=lambda x:salaries[x]))   # 'yuanhao' 
 info=[
            {'name':'egon','age':'18','salary':'3000'},
            {'name':'wxx','age':'28','salary':'1000'},
            {'name':'lxx','age':'38','salary':'2000'}
        ]
min(info,key=lambda dic:int(dic['salary']))

把薪资字典，按照薪资的高低排序sort

salaries={
    'egon':3000,
    'alex':100000000,
    'wupeiqi':10000,
    'yuanhao':2000
}
alaries=sorted(salaries) # 默认按照字典的键排序
print(salaries)
​
# salaries=sorted(salaries,key=lambda x:salaries[x])  #默认是升序排
alaries=sorted(salaries,key=lambda x:salaries[x],reverse=True) #降序
print(salaries)
info=[
            {'name':'egon','age':'18','salary':'3000'},
            {'name':'wxx','age':'28','salary':'1000'},
            {'name':'lxx','age':'38','salary':'2000'}
        ]
l=sorted(info,key=lambda dic:int(dic['salary']))

map 映射, 循环让每个元素执行函数，将每个函数执行的结果保存到新的列表中

v1 = [11,22,33,44]
result = map(lambda x:x+100,v1) # 第一个参数为执行的函数,第二个参数为可迭代元素.
print(list(result)) # [111,122,133,144]
names=['alex','wupeiqi','yuanhao','egon']
res=map(lambda x:x+'_NB' if x == 'egon' else x + '_SB',names)
print(list(res))

reduce , 对参数序列中元素进行累积.

import functools
v1 = ['wo','hao','e']
​
def func(x,y):
    return x+y
result = functools.reduce(func,v1) 
print(result)   # wohaoe
​
result = functools.reduce(lambda x,y:x+y,v1)
print(result)   # wohaoe
​
from functools import reduce
l=['my','name','is','alex','alex','is','sb']
res=reduce(lambda x,y:x+' '+y+' ',l)
print(res)
#my name  is  alex  alex  is  sb 

filter , 按条件筛选.

result=filter(lambda x:x > 2,[1,2,3,4])
print(list(result))
v1 = [11,22,33,'asd',44,'xf']
​
# 一般做法
def func(x):
    if type(x) == int:
        return True
    return False
result = filter(func,v1)
print(list(result))     # [11,22,33,44]
​
# 简化做法
result = filter(lambda x: True if type(x) == int else False ,v1)
print(list(result))
​
# 极简做法
result = filter(lambda x: type(x) == int ,v1)
print(list(result))
names=['alex_sb','wxx_sb','yxx_sb','egon']
res=filter(lambda x:True if x.endswith('sb') else False,names)
res=filter(lambda x:x.endswith('sb'),names)
print(list(res))        #['alex_sb', 'wxx_sb', 'yxx_sb']
ages=[18,19,10,23,99,30]
res=filter(lambda n:n >= 30,ages)
print(list(res))        #[99, 30]
salaries={
    'egon':3000,
    'alex':100000000,
    'wupeiqi':10000,
    'yuanhao':2000
}
res=filter(lambda k:salaries[k] >= 10000,salaries)
print(list(res))            #['alex', 'wupeiqi']