python学习习题总结(6)——统计重复数和非重复数,列表，字典

#随机生成若干数字 统计重复数和非重复数，按顺序输出 ||例50个[100,200]的数
import random
#lst = [randome.randint(100,200) for i in range(50)]    同append，效率比乘法*一次开辟空间慢
lst = [0]*50
for i in range(50):
lst[i] = random.randint(100,200)
print('Random numbers:{}'.format(','.join(map(str,lst))))
print('Method one:')
##局限性大,数组索引记录值
lst1 = [0]*201
for i in lst:
lst1[i] += 1
repenum = []
no_repenum = []
for i in range(len(lst1)):
if lst1[i] == 1:
no_repenum.append(i)
if lst1[i] >1:
repenum.append((i,lst1[i]))
print('The no-repetition numbers is :{}'.format(','.join(map(str,no_repenum))))
print('The repetition numbers is :{}'.format(','.join(map(str,repenum))))
print(no_repenum,repenum)
print('~~~~~~~~~~~~~~~~~~~~~~~~'*5)
print('Method two:')
##新建一个计数列表，一一对应,重复的数打标记,在第一个不重复的数上计数
lst2 = [0]*50
repenum = []
no_repenum = []
for i in range(len(lst)):
if lst2[i] != 0:
continue
else:
lst2[i] = 1
for j in range(i+1,len(lst)):
if lst2[j] != 0:
continue
if lst[i] == lst[j]:
lst2[i] += 1
lst2[j] = 1
if lst2[i] == 1:
no_repenum.append(lst[i])
else:
repenum.append((lst[i],lst2[i]))
print(sorted(no_repenum),sorted(repenum))
print('~~~~~~~~~~~~~~~~~~~~~~~~'*5)
print('Method three:')
#遍历每一个数，相同的数放入一个数组
lst3 = []
repenum = []
no_repenum = []
for i in range(len(lst)):
flag = False
for j in range(len(lst3)):
if lst[i] in lst3[j]:
lst3[j].append(lst[i])
flag = True
if flag == False:
lst3.append([lst[i]])
for i in range(len(lst3)):
if len(lst3[i]) == 1:
no_repenum.append(lst3[i][0])
else:
repenum.append((lst3[i][0],len(lst3[i])))
print(sorted(no_repenum),sorted(repenum))
print('~~~~~~~~~~~~~~~~~~~~~~~'*5)
print('Method four:')
#先把列表排序，重复数在一起，分别计数,统计过的变为None
lst4 = sorted(lst)
no_repenum = []
repenum = []
for i in range(len(lst4)):
if lst4[i] == None:
continue
for j in range(i+1,len(lst4)):
if lst4[j] != lst4[i]:
if j - i == 1:
no_repenum.append(lst4[i])
lst4[i] = None
else:
repenum.append((lst4[i],j-i))
for x in range(i,j):
lst4[x] = None
break
else:
if i == len(lst4)-1:
no_repenum.append(lst4[i])
else:
repenum.append((lst4[i],len(lst4)-i))
break
print(no_repenum,repenum)
print('~~~~~~~~~~~~~~~~~~~~~~'*5)
print('Method five')
#字典
dic = {}
no_repenum = []
repenum = []
for i in sorted(lst):
dic[i] = dic.get(i,0)+1
for i in dic:                         #python 3.6 字典有序输出
if dic[i] == 1:
no_repenum.append(i)
else:
repenum.append((i,dic[i]))
print(no_repenum,repenum)
print('~~~~~~~~~~~~~~~~~~~~~'*5)
print('Method six')
#count方法，效率最低，抛弃