Python：每日一题 121,Python交流,编程语言专区,鱼C论坛

冬雪雪冬 发表于 2017-11-5 17:03:37

Python：每日一题 121

本帖最后由冬雪雪冬于 2017-11-10 15:19 编辑

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题目：
这是一个排序的练习。
有如下的一个列表：
list1 =
要求先求出各位之和，如果结果不是一位数，继续此步骤，直到成为1~9的数字，例如684各位之和为18，再次运算为9，然后对其排序，排序不是从1到9，也不是从9到1，是按照下面列表的次序。
list2 =
最后给出list1排序后的列表。

SixPy 发表于 2017-11-5 18:05:00

本帖最后由 SixPy 于 2017-11-5 18:18 编辑

list1 =
list2 =
order = {n:i for i,n in enumerate(list2)}

def x(n):
while n>9:
n = sum(int(i)for i in str(n))
return order

list1.sort(key=x)
print(list1)
#

wc365 发表于 2017-11-6 00:11:16

#数字n先求出各位之和，如果结果不是一位数，继续此步骤，直到成为1~9的数字
def sumbit(n):
while n > 9:
   total = 0
   for each in list(str(n)):
         total += eval(each)
   n = total
return n

#序列list1按照序列list2的顺序排列
def sortlist(list1,list2):
list3 = [ for x in list1]
list4 = []
for i in list2:
   for j in range(len(list3)):
         if i == list3:
            list4.append(list3)
return list4

list1 =
list2 =
print(sortlist(list1,list2))

BngThea 发表于 2017-11-6 09:49:21

list1 =
list2 =

tar = []

for each in list1:
while each > 9:
each = sum(map(int, str(each)))
tar.append(each)

tar = sorted(tar, key = lambda x: list2.index(x))
print(tar)

wyp02033 发表于 2017-11-6 11:03:03

本帖最后由 wyp02033 于 2017-11-6 11:16 编辑

def get_index(num):
if num < 10:
   return num

else:
   new_num = 0
   for each in str(num):
         new_num += int(each)
   return get_index(new_num)

list1 =
index_list = []
for each in list1:
index_list.append(get_index(each))

the_order = {1:0, 6:1, 7:2, 4:3, 9:4, 3:5, 2:6, 5:7, 8:8}
num_order = []
for each in index_list:
num_order.append(the_order)

sorted_order = sorted(num_order)

new_list = []
for each in sorted_order:
new_list.append(list1)
list1.remove(list1)
num_order.remove(each)

print(new_list)

排序结果：

bush牛 发表于 2017-11-6 14:15:02

def len_sum(n:int):
if n < 10:
   return n
else:
   n = str(n)
   s_sum = sum(int(i) for i in n)
   if len(str(s_sum)) > 1:
         s_sum = len_sum(s_sum)
   return s_sum

list1 =
list2 =
list3 = [(len_sum(i), i) for i in list1]

from collections import defaultdict
d = defaultdict(list)
for k, v in list3:
d.append(v)

list4 = for i in list2 if i in d.keys()]
from itertools import chain
print(list(chain(*list4)))

无边落木 发表于 2017-11-6 14:57:10

看看答案

ytbihpl2017 发表于 2017-11-6 16:03:43

真的想不出来

飞非妃飞飞 发表于 2017-11-6 16:27:17

比较菜，代码很陋，就做个简单的
---------------------------
def toList(numbers):
List =
return List

if __name__ == "__main__":
L = toList();
a=[]
t=[for I in L]
print(t)
print(type(len(t)))
for i in t:
   print(sum(i))
   if sum(i)>10:
   a.append(sum(i))
print(a)
K=toList(a)
if __name__ == "__main__":
print(K)
ba = []
bt = [ for I in K]
print(bt)
for i in bt:
      print(sum(i))
      ba.append(sum(i))
print(ba)

renze 发表于 2017-11-6 17:18:39

list1 =
list2 = []

def doit(i):
add = 0
sum1 = list(str(i))
for j in sum1:
   add = add + int(j)
return add

def sort(in_list):
for j in range(len(in_list)) :
   if in_list == 1:
         in_list = 1
   elif in_list == 6:
         in_list = 2
   elif in_list == 7:
         in_list = 3
   elif in_list == 4:
         in_list = 4
   elif in_list == 9:
         in_list = 5
   elif in_list == 3:
         in_list = 6
   elif in_list == 2:
         in_list = 7
   elif in_list == 5:
         in_list = 8
   elif in_list == 8:
         in_list = 9
return in_list


for i in list1:
add = doit(i)
while add > 9 :
   add = doit(add)
list2.append(add)

list2 = sort(list2)

#list2.sort()

print(list2)

p5494846 发表于 2017-11-6 17:55:41

本帖最后由 p5494846 于 2017-11-6 17:58 编辑

list1 =
def fun_sum(x): #定义求位和的函数
while x > 9:
x = str(x)
y = 0
for n in x:
y += int(n)
x = y
return x
list = map(fun_sum,list1) #得到初始的列表

list_tmp = []
dict = {'1':'1','6':'2','7':'3','4':'4','9':'5','3':'6','2':'7','5':'8','8':'9'}
for num in list:
tmp = dict
list_tmp.append(tmp)
list_tmp.sort()#根据映射关系得到值的列表

#以下步骤是利用索引根据dict的值查找键
list_fin = []
key_list = []
value_list = []
for key,value in dict.items():
key_list.append(key)
value_list.append(value)
for x in list_tmp:
index = value_list.index(x)
y = key_list
list_fin.append(y)#根据映射关系反过来得到最终列表
print(list_fin)

solomonxian 发表于 2017-11-6 18:25:42

递归，各位数相加直到只剩一位数，按要求排序
list1 =
list2 =

new_sum = lambda n: n if n<10 else new_sum(sum(int(i) for i in str(n)))
new_list1 =
print(sorted(new_list1, key=lambda x: list2.index(x)))

ooxx7788 发表于 2017-11-6 20:37:23

def foo(num):
while num >=10:
   num = sum(map(int,list(str(num))))
return num

list1 =
dict1 = {v:k for k,v in zip(list(map(int,(list('123456789')))),)}
d = {v:k for k,v in dict1.items()}
x = for s in sorted( for r in map(foo,list1)])]
y = list(map(foo,list1))
temp = list(zip(y,list1))
result = []
for i in x:
for each in range(len(temp)):
   if i == temp:
         result.append(temp.pop(each))
         break

result

感觉写的有点麻烦

编程新血 发表于 2017-11-6 21:34:16

list1 =
list2 =
print(list1)
print(list2)
'''各位加到9'''
def addnum9(x):
while x>9:
   u=0
   for i in str(x):
         u+=int(i)
   x=u
return x
'''list1的数字的排序'''
listcc=[]
for i in list1:
listcc.append(addnum9(i))
'''组合在一起'''
list4=list(zip(listcc,list1))

list5=[]
for i in list2:
for i2 in list4:
   if i2==i:
         list5.append(i2)
print(list5)

lihw 发表于 2017-11-6 23:38:21

list1 =
list2=[]
list3=[]
list4=
he=0
def fun(list1):

for i in list1:
   he=sum(map(int,str(i)))
   if he <=9:
         list2.append(he)

   else:
         list1=[]
         list1.append(he)
         fun(list1)

return (sorted(list2,key=list4.index))

xindong 发表于 2017-11-7 08:35:28

数位相加用递归倒是没有什么问题。

按照规定的序列排序倒是有点烧脑子，代码看着有点繁琐。不过结果在这里：
排序前：

排序后：

# -*- coding: utf-8 -*-
"""
Created on Mon Nov6 15:59:02 2017

"""
import numpy as np

def numSum(Num):
NumOfDigitals = int(np.log10(Num))
numAdd=0;
for i in range(NumOfDigitals, 0, -1):
   if ( i == NumOfDigitals):
         Digital = int(Num / pow(10, i))
         numAdd=Digital
         remain = Num - Digital*pow(10,i)
   else:
         Digital = int(remain / pow(10, i))
         numAdd += Digital
         remain = remain - Digital*pow(10,i)

return numAdd + remain

def digitalSum(Num):
dig_sum = Num
while(dig_sum > 9):
   dig_sum = numSum(dig_sum)
return dig_sum

list1 =

list2 =
nlist1=[]
for i in list1:
nlist1.append(digitalSum(i))

print(nlist1)

nlist1=[]

for i in list1:
dig_sum = digitalSum(i)
if(len(nlist1)==0):
   nlist1.append(dig_sum)
else:
   index_j = 0
   flag=0
   order_old = list2.index(nlist1)
   order_new = list2.index(dig_sum)
   while (order_new > order_old and flag==0): #and index_j<len(nlist1)-1):
         index_j = index_j + 1
         if (index_j < len(nlist1)):
            order_old = list2.index(nlist1)
         else:
            flag=1

   if (flag==1):
            nlist1.append(dig_sum)
   else:
         nlist1.insert(index_j,dig_sum)

print(nlist1)

colinshi 发表于 2017-11-7 10:56:30

本帖最后由 colinshi 于 2017-11-7 10:58 编辑

from functools import reduce
list1 =

def add2(x, y):
return int(x) + int(y)
def CaiFen(element):
a = element
while a > 9:
   a = int(reduce(add2,list(str(a))))
return a
g=[]
for i in list1:
g.append(CaiFen(i))
z=zip(g,list1)
dic={}
for k,v in z:
dic=k
l=[]
list2 =
for i in list2:
for v,k in dic.items():
   if i == k:
         l.append(v)
print(l)
代码感觉写的有点长，没有精简。

cyz1993 发表于 2017-11-7 11:39:10

list1=
list2 =
num=0
d={}
list3=[]
jud=False
while not jud:
jud=True
for each in list1:
   jud=jud and (each < 10)

for each in list1:
   if each >= 10:
         for i in str(each):
            num +=int(i)
   else:
         num=each
   list3.append(num)
   num=0
list1=list3[:]
list3=[]
for i in range(9):
d]=i
list1.sort(key=lambda x:d)

天使骑魔鬼 发表于 2017-11-7 19:06:41

list1 =
list2 =
list3 = []
while not(set(list1) < set(list2) or set(list1) == set(list2)):
for i in range(9):
   list1 = sum(map(int,list(str(list1))))
for each in list2:
for i in range(len(list1)):
   if list1 == each:
         list3.append(list1)
print(list3)

WD2017 发表于 2017-11-8 00:32:20

需要什么基础才能写出，老师？

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Python：每日一题 121