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冬雪雪冬 发表于 2018-1-5 10:50:17

Python：每日一题 133

本帖最后由 冬雪雪冬 于 2018-1-10 19:46 编辑

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题目：
已有一个列表如下
list1 =
由此生成新的列表，要求为一个2维列表，包括list1的每个不重复元素和在list1中出现的次数。例如12在list1中出现11次，就记为，最后按元素由小到大排序，生成如下的列表
[, , , , , , , , , , , , , , , , , , , , ]

jerryxjr1220 发表于 2018-1-5 11:03:51

print(sorted([ for i in set(list1)]))

千雪寒蝉 发表于 2018-1-5 14:17:14

list1 = [12, 1, 3, 19, 0, 1, 16, 8, 19, 10, 6, 10, 0, 4, 3, 18, 2, 18, 17, 6, 16, 19, 12, 18, 4, 4,
         16, 12, 13, 11, 0, 0, 12, 7, 10, 2, 7, 12, 4, 18, 13, 18, 16, 17, 14, 12, 1, 12, 4, 14, 8,
         18, 4, 12, 10, 1, 18, 7, 2, 9, 3, 11, 12, 2, 9, 13, 12, 18, 15, 1, 9, 5, 11, 6, 12, 9, 16,
         14, 18, 0, 15, 4, 17, 5, 1, 13, 0, 10, 0, 6, 16, 15, 3, 16, 20, 18, 4, 18, 16, 3]
new_list = set(list1)
final_list = []
for i in new_list:
    num_group =
    final_list.append(num_group)
print(final_list)

woduhaitao 发表于 2018-1-5 15:10:28

# 查找列表中每个元素出现的次数
# list_original原始列表
# list_target    目标输出
# list_temp      转换用列表，用于存放原始列表中不重复的元素
# list_cell      转换用列表，用于存放原始列表中出现的元素和次数
list_original = [12, 1, 3, 19, 0, 1, 16, 8, 19, 10, 6, 10, 0, 4, 3, 18, 2, 18,
                17, 6, 16, 19, 12, 18, 4, 4, 16, 12, 13, 11, 0, 0, 12, 7, 10,
                2, 7, 12, 4, 18, 13, 18, 16, 17, 14, 12, 1, 12, 4, 14, 8, 18,
                4, 12, 10, 1, 18, 7, 2, 9, 3, 11, 12, 2, 9, 13, 12, 18, 15, 1,
                9, 5, 11, 6, 12, 9, 16, 14, 18, 0, 15, 4, 17,5, 1, 13, 0, 10,
                0, 6, 16, 15, 3, 16, 20, 18, 4, 18, 16, 3]
list_target = []
list_temp = []
list_cell = []
#将原始列表从小到大排序
list_original.sort()
#将元素列表中不重复的元素找出来，存放起来
for i in list_original:   
    if i not in list_temp:
      list_temp.append(i)
#print(list_temp)
#循环查找每个元素出现的次数      
for i in list_temp:
    list_cell =
    #将查找结果存放到列表中
    list_target.append(list_cell)
print(list_target)

冰封雪舞 发表于 2018-1-5 15:40:23

list1 = [12, 1, 3, 19, 0, 1, 16, 8, 19, 10, 6, 10, 0, 4, 3, 18, 2, 18, 17, 6,
         16, 19, 12, 18, 4, 4, 16, 12, 13, 11, 0, 0, 12, 7, 10, 2, 7, 12, 4,
         18, 13, 18, 16, 17, 14, 12, 1, 12, 4, 14, 8, 18, 4, 12, 10, 1, 18,
         7, 2, 9, 3, 11, 12, 2, 9, 13, 12, 18, 15, 1, 9, 5, 11, 6, 12, 9,
         16, 14, 18, 0, 15, 4, 17, 5, 1, 13, 0, 10, 0, 6, 16, 15, 3, 16,
         20, 18, 4, 18, 16, 3]
set1 = set(list1)
TT = []
for i in set1:
    d= list1.count(i)
    TT.append()
print(TT)

shigure_takimi 发表于 2018-1-5 15:46:15

list1 =
print([ for i in set(list1)])


# set好像是自动按从小到大排序的？

776667 发表于 2018-1-5 17:02:17

def fun133(list_x):
    result = []
    for i in range(min(list_x),max(list_x)+1):
      result.append()
    return result

if __name__ == '__main__':
    print(fun133())

浪子无名1517 发表于 2018-1-5 17:46:10

set1 = set(list1)
    list2 = []
    for i inset1:
      list2 .append()
    print(list2)

小黑大人 发表于 2018-1-5 18:33:53

list1 =
list1.sort()
list2 = []
for i in range(max(list1)+1):
    n = list1.count(i)
    list2.append()
print(list2)

solomonxian 发表于 2018-1-5 18:37:23

好像很久不见题目了，多写几个{:10_333:}
from collections import Counter

def fun(lst):
    return sorted(Counter(lst).most_common())

def fun2(lst):
    counter = Counter(lst)
    return [] for i in counter]

def fun3(lst):
    return [ for i in set(lst)]

bush牛 发表于 2018-1-5 19:15:00

from collections import Counter
print(sorted([ for i,j in Counter(list1).most_common()]))

wyp02033 发表于 2018-1-5 22:31:04

list1 = [12, 1, 3, 19, 0, 1, 16, 8, 19, 10, 6, 10, 0, 4, 3, 18, 2, 18, 17, 6, 16, 19, 12, 18, 4, 4, 16, 12, 13, 11, 0,
         0, 12, 7, 10, 2, 7, 12, 4, 18, 13, 18, 16, 17, 14, 12, 1, 12, 4, 14, 8, 18, 4, 12, 10, 1, 18, 7, 2, 9, 3, 11,
         12, 2, 9, 13, 12, 18, 15, 1, 9, 5, 11, 6, 12, 9, 16, 14, 18, 0, 15, 4, 17, 5, 1, 13, 0, 10, 0, 6, 16, 15, 3,
         16, 20, 18, 4, 18, 16, 3]

list1.sort()

set1 = set(list1)
result = []
for each in set1:
    result.append()
print(result)

结果：[, , , , , , , , , , , , , , , , , , , , ]

tuple小白 发表于 2018-1-6 10:58:57

先排序列表 然后去重 在找出出现的位置

万事屋 发表于 2018-1-6 13:56:08

本帖最后由 万事屋 于 2018-1-6 13:57 编辑

用堆写的。。也不知道名称和写法正不正规
def dsort(ll):
    d=[]
    ld=1
    for i in ll:
      ss=0
      while 1:
            if d==i:
                d+=1
                break
            elif d>i:
                if d==-1:
                  d.append()
                  d=ld
                  ld+=1
                  break
                else:
                  ss=d
            elif d<i:
                if d==-1:
                  d.append()                  
                  d=ld
                  ld+=1
                  break
                else:
                  ss=d
    return d
def showlist(d,ss):
    if ss==-1:
      return []
    else:
      return showlist(d,d)+]+showlist(d,d)
   
ll=
ff=dsort(ll)
ans=showlist(ff,0)
print(ans)

无郁 发表于 2018-1-6 15:16:07

list2=
#创建集合，存放不重复数字
j=set()
#遍历list2
for x in list2:

    j.add(x)
#将集合转换成列表
list3=list(j)
#排序
list3.sort()
print (list2)
print(list3)
#创建列表5存放结果
list5=[]
#print (list4)
for y in list3:
    #统计数字出现的次数
    a=list2.count(y)
    #print (a)
    #创建新列表4存放数字和出现的次数
    list4 = []
    list4.append(y)
    list4.append(a)
    #将列表4添加到列表5中
    list5.append(list4)
print (list5)

aegis1417 发表于 2018-1-6 23:05:43

list1 = [12, 1, 3, 19, 0, 1, 16, 8, 19, 10, 6, 10, 0, 4, 3, 18, 2, 18, 17, 6, 16, 19, 12, 18, 4, 4, 16, 12, 13,
         11, 0, 0, 12, 7, 10, 2, 7, 12, 4, 18, 13, 18, 16, 17, 14, 12, 1, 12, 4, 14, 8, 18, 4, 12, 10, 1, 18, 7,
         2, 9, 3, 11, 12, 2, 9, 13, 12, 18, 15, 1, 9, 5, 11, 6, 12, 9, 16, 14, 18, 0, 15, 4, 17, 5, 1, 13, 0, 10,
         0, 6, 16, 15, 3, 16, 20, 18, 4, 18, 16, 3]
answer=[]
list2=list(set(list1))
list2.sort()
for i in list2:
      answer.append()
print(answer)

timeislife 发表于 2018-1-7 09:58:39

本帖最后由 timeislife 于 2018-1-7 09:59 编辑

list1 = [12, 1, 3, 19, 0, 1, 16, 8, 19, 10, 6, 10, 0, 4, 3, 18, 2, 18, 17, 6,16,
         19, 12, 18, 4, 4, 16, 12, 13, 11, 0, 0, 12, 7, 10,2, 7, 12, 4, 18, 13,
         18, 16, 17, 14,12, 1, 12, 4, 14, 8, 18, 4, 12, 10, 1, 18, 7, 2, 9, 3,
         11, 12, 2, 9, 13,12, 18, 15, 1, 9, 5, 11, 6, 12, 9, 16, 14,18, 0, 15,
         4, 17, 5, 1, 13, 0, 10, 0, 6, 16, 15, 3, 16, 20, 18, 4, 18,16, 3]
temp = sorted(list(set(list1)))
returnedlist = [ for each in temp]
print(returnedlist)

321123 发表于 2018-1-7 11:34:44

list1 =
>>> temp=[]
>>> for each in list1:
        temp.append(list1.count(each))

       
>>> list2=sorted(zip(list1,temp))
>>> print(list2)

321123 发表于 2018-1-7 11:35:36

321123 发表于 2018-1-7 11:34
list1 =

但是我的输出是[(0, 7), (0, 7), (0, 7), (0, 7), (0, 7), (0, 7), (0, 7), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (2, 4), (2, 4), (2, 4), (2, 4), (3, 5), (3, 5), (3, 5), (3, 5), (3, 5), (4, 8), (4, 8), (4, 8), (4, 8), (4, 8), (4, 8), (4, 8), (4, 8), (5, 2), (5, 2), (6, 4), (6, 4), (6, 4), (6, 4), (7, 3), (7, 3), (7, 3), (8, 2), (8, 2), (9, 4), (9, 4), (9, 4), (9, 4), (10, 5), (10, 5), (10, 5), (10, 5), (10, 5), (11, 3), (11, 3), (11, 3), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (12, 11), (13, 4), (13, 4), (13, 4), (13, 4), (14, 3), (14, 3), (14, 3), (15, 3), (15, 3), (15, 3), (16, 8), (16, 8), (16, 8), (16, 8), (16, 8), (16, 8), (16, 8), (16, 8), (17, 3), (17, 3), (17, 3), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (18, 11), (19, 3), (19, 3), (19, 3), (20, 1)] 怎么解决能让他只出现一次{:5_94:}

321123 发表于 2018-1-7 12:01:16

temp=[]
>>> for each in list1:
      temp.append(list1.count(each))

      
>>> list2=sorted(zip(list1,temp))
>>> list3=[]
>>> for each in list2:
        if each not in list3:
                list3.append(each)

               
>>> list3

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