z1446773686 发表于 2018-2-6 11:19
count = 0
grade = []
for x in range(1,5):
count = 0
grade = []
for x in range (1,5):
for y in range (1,5):
for d in range (1,5):
if ((x != y) and ((y != d)and (x != d))):
count += 1
n = str(x) ; b = str(y) ; a = str(d)
grade.append(list(zip(n,b,a)))
print('内容如下：')
for each in grade:
print(each)
print ('排列个数为:%d' % count )
print(grade)
运行内容：
内容如下：
[('1', '2', '3')]
[('1', '2', '4')]
[('1', '3', '2')]
[('1', '3', '4')]
[('1', '4', '2')]
[('1', '4', '3')]
[('2', '1', '3')]
[('2', '1', '4')]
[('2', '3', '1')]
[('2', '3', '4')]
[('2', '4', '1')]
[('2', '4', '3')]
[('3', '1', '2')]
[('3', '1', '4')]
[('3', '2', '1')]
[('3', '2', '4')]
[('3', '4', '1')]
[('3', '4', '2')]
[('4', '1', '2')]
[('4', '1', '3')]
[('4', '2', '1')]
[('4', '2', '3')]
[('4', '3', '1')]
[('4', '3', '2')]
排列个数为:24
[[('1', '2', '3')], [('1', '2', '4')], [('1', '3', '2')], [('1', '3', '4')], [('1', '4', '2')], [('1', '4', '3')], [('2', '1', '3')], [('2', '1', '4')], [('2', '3', '1')], [('2', '3', '4')], [('2', '4', '1')], [('2', '4', '3')], [('3', '1', '2')], [('3', '1', '4')], [('3', '2', '1')], [('3', '2', '4')], [('3', '4', '1')], [('3', '4', '2')], [('4', '1', '2')], [('4', '1', '3')], [('4', '2', '1')], [('4', '2', '3')], [('4', '3', '1')], [('4', '3', '2')]]
nums = []
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != j and i != k and j != k:
num = i * 100 + j * 10 + k
nums.append(num)
print(nums)
print(len(nums))
count = 0
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != k and i != j and j != k:
count +=1
print(count,i*100+j*10+k,'\n')
def test1():
for a in range(1,5):
for b in range(1, 5):
for c in range(1, 5):
if a != b and a != c and b != c:
print (a, b, c)
都乱写代码我也来说说，其实很简单嵌套三个for循环，下面我就帮助一些刚学习for的整理一下思路
for i in range(1,4): #这是控制百位1,4循环
for ii in range(1,4): #控制十位的1,4循环 只有等嵌套里面循环都循环完毕才可开启新一轮
for iii in range(1,4): #控制个位l1,4循环
if i != !! and ii != iii and i != !!!: #条件控制 百位不等于十位，十位不等于个位，百位不等于各位
print (i,ii,iii) #拼接每次循环结果输出
就像我们打字111,112,113,114,121,122,123,124,循环的规律可见而知{:10_249:}
加油加油
本帖最后由 凌九霄 于 2018-2-13 09:22 编辑
import itertools
for i in list(itertools.permutations(range(1,5),3)):
print("{0}{1}{2}".format(i,i,i))
from itertools import combinations
print(list(combinations(range(1,5),3)))
a=0
for i in :
for j in :
for k in :
number=100*i+10*j+k
if i!=j and i!=k and j!=k:
a+=1
print(number)
else:
continue
print(a)
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != k and i != j and j != k:
print(i,j,k)
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != j and i != k and j !=k:
print(i,j,k)
m = 0
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != k and i != j and j != k:
print(i,end = '')
print(j,end = '')
print(k)
m = m+1
print('\n',m)
m = 0
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i != k and i != j and j != k:
print(i,end = '')
print(j,end = '')
print(k)
m = m+1
print('\n','有%d种组合'%m)
lis = []
for i in range(1, 5):
for j in range(1, 5):
for k in range(1, 5):
if i != j and i != k and j != k:
print(i * 100 + j * 10 + k)
lis.append(i * 100 + j * 10 + k)
print("共组成数字：")
print(len(lis))
lis = []
for i in range(1, 5):
for j in range(1, 5):
for k in range(1, 5):
if i != j and i != k and j != k:
print(i * 100 + j * 10 + k)
lis.append(i * 100 + j * 10 + k)
print("共组成数字：")
print(len(lis))
a=set()
for i in range(1,5):
for ii in range(1,5):
for iii in range(1,5):
a.add(i*100+ii*10+iii)
print(a)
def number():
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if (i == j) or (j == k) or (i == k):
continue
print(100*i + 10*j + k)
number()
l =
v = []
for i in l:
for n in l:
if i==n:
continue
for x in l:
if x==n or x==i:
continue
v.append(i*100+n*10+x)
print (v)
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
if i !=j!=k:
print (i,j,k)
看完评论长见识了,原来还有itertool模块
本帖最后由 tianqi85 于 2018-3-13 09:48 编辑
num=list(range(1,5))
new=[]
for a in num:
for b in num:
for c in num:
new.append(int(str(a)+str(b)+str(c)))
print (new)
print (len(new))