集合生成式
print({i ** 2 for i in {1,2,3,3}})
print({i ** 2 for i in {1,2,3,9,12,9} if i % 3 == 0 })
结果
{1, 4, 9}
{81, 9, 144}
列表生成式
#生成一个列表,列表元素分别为[12,22,32,…n2]
li = []
for i in range(1,9):
li.append(i ** 2)
print(li)
print([i ** 2 for i in range(1,9)])
[expression for i in 序列 if …]
#1.找出1-10之间的所有偶数,
#并返回一个列表(包含以这个偶数为半径的圆的面积)
a = math.pi
print(a)
print([math.pi * r * r for r in range(2,11,2)])
测试
3.141592653589793
[12.566370614359172, 50.26548245743669, 113.09733552923255, 201.06192982974676, 314.1592653589793]
#2.找出1~100之间所有的质数
def isPrime(num):
for i in range(2,num):
if num % i == 0:
return False
else:
return True
print([i for i in range(2,101) if isPrime(i)])
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
(二)
def zhishu(num):
for i in range(2,num):
if num % i == 0:
return False
else: #循环完之后再执行else
return True
n = int(input("Num:"))
print([i for i in range(2,n) if zhishu(i)])
#找出.log结尾的文件
import os
print(os.listdir('/var/log'))
print([i for i in os.listdir('/var/log') if i.endswith('.log')])
(2018-腾讯-在线编程题)
-题目描述:
给定一个正整数,编写程序计算有多少对质数的和等于输入的这个正整
数,并输出结果。输
入值小于1000。
如,输入为10, 程序应该输出结果为2。(共有两对质数的和为10,分别
为(5,5),(3,7))
#[2,3,5,7]
-输入描述:
输入包括一个整数n,(3 ≤ n < 1000)
-输出描述:
输出对数
-示例1 :
``
输入:
10
输出:
2
num = int(input())
def isPrime(num):
for i in range(2,num):
if num % i == 0:
return False
else:
return True
primeli = [i for i in range(2,num) if isPrime(i)]
print(primeli)
"""
1.先从列表中拿出两个数
2.判断两个数之和是否等于num
"""
#[2,3,5,7] # 2 10 -2 =8
primecount = 0
for item in primeli:
if (num - item) in primeli and item <= num - item:
primecount += 1
print(primecount)
def zhishu(num):
for i in range(2,num):
if num % i == 0:
return False
else:
return True
n = int(input("Num:"))
prime = [i for i in range(2,n) if zhishu(i)]
print(prime)
count = 0
for i in prime:
for j in prime:
if n == i + j and i <= j:
print('(%s,%s)' %(i,j))
count +=1
print('%s' %(count))
字典生成式
d = dict(a=1,b=2)
print(d)
new_d = {}
for i in d:
new_d[i.upper()] = d[i]
print(new_d)
#print('key转化为大写的字典:',new_d)
print({k.upper():v for k,v in d.items()})
结果
{'a': 1, 'b': 2}
{'A': 1, 'B': 2}
{'A': 1, 'B': 2}
d = dict(a=2,b=1,c=2,B=9,A=10)
#输出{'a':12,'b':10,'c':2}
new_d = {}
for k,v in d.items():
low_k = k.lower()
if low_k not in new_d:
new_d[low_k] = v
else:
new_d[low_k] += v
print(new_d)
print({k.lower(): d.get(k.lower(),0) + d.get(k.upper(),0) for k in d})
结果
{'a': 12, 'b': 10, 'c': 2}
{'a': 12, 'b': 10, 'c': 2}