Python：每日一题 115,Python交流,编程语言专区,鱼C论坛

冬雪雪冬 发表于 2017-10-22 19:42:57

Python：每日一题 115

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题目：
我们知道用add方法可以给集合增加元素，如：
>>> set1 = {'qwe', 'asd', 'z'}
>>> set1.add('123')
>>> set1
{'123', 'qwe', 'asd', 'z'}

当add已有元素时，则不会有任何变化，这也是集合唯一性的表现，现在有一个全是字符串的集合，你设计程序，当加入一个字符串是集合中已有元素时，会自动在字符串后面加上“_1”再加入元素中，如
set1 = {'qwe', 'asd', 'z'}
#加入'qwe',实际加入qwe_1，集合变成
{'qwe_1', 'qwe', 'asd', 'z'}
进一步，如果已有'qwe'和'qwe_1'，再加入'qwe'时，实际加入'qwe_2'，以此类推。

BngThea 发表于 2017-10-22 21:33:36

set1 = {'qwe_1','qwe', 'asd', 'z'}

adds = input("请输入想要加入的元素：")

count = 0
for each in set1:
if adds in each:
count += 1
if count:
adds += '_%d' % count

print("原set为：",set1)

set1.add(adds)

print("原set变为：",set1)

ooxx7788 发表于 2017-10-22 21:35:14

本帖最后由 ooxx7788 于 2017-10-23 09:04 编辑

def foo(se,strng):
import re
if strng not in se:
   se.add(strng)
else:
   if re.search(r'(.*?)_\d*', strng):
         s,i = re.findall(r'(.*?)_(\d*)',strng)
         strng = '{0}_{1}'.format(s,int(i)+1)
         return foo(se, strng)
   else:
         strng = strng + '_1'
         return foo(se, strng)

不过如果出现本身就是_\d*的文本好像有点问题，应该要增加一个是不是从1开始的判断，有点小复杂呢，不改了。

SixPy 发表于 2017-10-23 00:09:57

本帖最后由 SixPy 于 2017-10-23 00:14 编辑

class SET(set):
def add(self,x):
   n,x1 = 0,x
   while x in self:
            n += 1
            x = '%s_%d'%(x1,n)
   super().add(x)

s=SET('abc')
print(s)
s.add('c')
print(s)
s.add('c')
print(s)

wyp02033 发表于 2017-10-23 10:09:45

def my_add(the_set, st):
if st not in the_set:
   the_set.add(st)

else:
   i = 1
   st += "_%d" % i
   while st in the_set:
         i += 1
         st = st.split('_') + '_%d' % i

   the_set.add(st)

return the_set

solomonxian 发表于 2017-10-27 20:39:29

很少碰到用到类的题目，不熟，重写方法不用super就感觉要无限循环{:10_269:}
class New_set(set):
def add(self, value):
   if value not in self:
         super().add(value)
   else:
         length = len(self)
         i = 1
         while len(self) == length:
            super().add("".join())
            i += 1

shigure_takimi 发表于 2017-12-2 03:08:33

def addSet(aSet, something):
if isinstance(something, str):
   if something not in aSet:
         aSet.add(something)
   else:
         n = 1
         while something+'_'+str(n) in aSet:
            n += 1
         aSet.add(something+'_'+str(n))
return aSet

set1 = {'qwe', 'asd', 'z'}
for i in range(10):
print(addSet(set1, 'qwe'))

>>>
{'z', 'qwe', 'qwe_1', 'asd'}
{'qwe_2', 'z', 'qwe', 'qwe_1', 'asd'}
{'z', 'qwe_2', 'qwe', 'qwe_1', 'qwe_3', 'asd'}
{'z', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd'}
{'z', 'qwe_5', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd'}
{'z', 'qwe_6', 'qwe_5', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd'}
{'z', 'qwe_6', 'qwe_5', 'qwe_7', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd'}
{'z', 'qwe_6', 'qwe_5', 'qwe_7', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd', 'qwe_8'}
{'qwe_9', 'z', 'qwe_6', 'qwe_5', 'qwe_7', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd', 'qwe_8'}
{'qwe_9', 'z', 'qwe_10', 'qwe_6', 'qwe_5', 'qwe_7', 'qwe_2', 'qwe', 'qwe_1', 'qwe_4', 'qwe_3', 'asd', 'qwe_8'}

PYTHON90小菜鸟 发表于 2018-1-1 15:47:01

def Add(set1,element):
i=0
length_0=len(set1)
set1.add(element)
length_1=len(set1)
while length_0==length_1:
   i+=1
   set1.add(element+'_'+str(i))
   length_1=len(set1)

return set1

set1={'qwe','asd','z'}
y_n=None

while y_n!='no':
element=input('Please enter the element whilch you want to join:')
print(Add(set1,element))
y_n=input('Do you want to enter again :')
y_n.lower()

yjsx86 发表于 2018-2-3 11:18:07

text = {'qwe_1', 'qwe', 'asd', 'z','qwe_2'}
def setAdd(string):
before = len(text)
text.add(string)
last = len(text)
temp = []
if last == before:
   for x in text:
         if string in x:
            n = x
            temp.append(int(n if n else 0))
   text.add('%s_%d' % (string,max(temp)+1))

setAdd('qwe')
setAdd('asd')
setAdd('3721')

#result
{'z', 'asd_1', '3721', 'asd', 'qwe_1', 'qwe_3', 'qwe_2', 'qwe'}

majia1015 发表于 2018-2-25 19:03:35

set1 = {'majia','weiwei'}
x = 1

while True:

element = str(input('> '))
'''判断元素是否属于这个集合成员，如果属于就添加后缀'''
if element in set1:
   element = element + '_' + str(x)
   set1.add(element)
   x += 1

else:
   set1.add(element)
   break

print(set1)

print(set1)

victor.xu 发表于 2018-4-1 11:00:40

def add_element(sett,elem):
counter = 0
for i in sett:
   if i.count(elem) == 1:
         counter += 1
if counter:
   suffix = '%d' % (counter)
   elem = elem + '_' + suffix
   sett.add(elem)
else:
   sett.add(elem)
return sett
set1 = {'qwe', 'asd', 'z', 'qwe_1', 'qwe_2'}
lis = ['123', 'qwe', 'qwe_1', 'qwe_2']
for j in lis:
result = add_element(set1,j)
print(result)

Wangzy1025 发表于 2018-5-29 15:45:58

def iter():
a = 0
while True:
   a += 1
   yield a

def insert_set(str_set, str):
if str in str_set:
   temp = str
   a = iter()
   while temp in str_set:
         temp = str + '_%d' % next(a)
   str_set.add(temp)
else:
   str_set.add(str)

set1 = {'qwe', 'asd', 'zxc'}
insert_set(set1, 'qwe')
insert_set(set1, 'qwe')
insert_set(set1, 'qwe')
insert_set(set1, 'qwe')
insert_set(set1, 'asd')
insert_set(set1, 'tyu')
insert_set(set1, 'qwe')
print(set1)

hj5818 发表于 2019-4-6 21:59:35

set1={"qwe","asd","z"}
dict1={"qwe":0,"asd":0,"z":0}
while True:
n=str(input("输入集合元素："))
if n not in set1:
   set1.add(n)
   dict1=0
else:
   for each in set1:
         if each==n:
            dict1+=1
   set1.add(n+"_"+str(dict1))
print (set1)

kinkon 发表于 2022-3-7 13:14:33

from collections import Counter

set1 = {'qwe', 'asd', 'z'}

s = input("请输入想要加入的元素：")

set2 = Counter(set1)

if s in set1:
set1.add(s + '_' + str(set2))
else:
set1.add(s)

set2 += 1

print(set1)

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Python：每日一题 115