Python：每日一题 201

冬雪雪冬 · 发表于 2018-8-21 10:35:43

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本帖最后由冬雪雪冬于 2018-8-24 20:26 编辑

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题目：
zip函数大家都很熟悉，我抄录一下网上的说明：

zip() 函数用于将可迭代对象作为参数，将对象中对应的元素打包成一个个元组，然后返回由这些元组组成的对象。

如果各个可迭代对象的元素个数不一致，则返回的对象长度与最短的可迭代对象相同。

例如：

>>> a = [1,2,3] #此处可迭代对象为列表
>>> b = [4,5,6]
>>> c = [4,5,6,7,8]
>>> zipped = zip(a,b)
>>> zipped
<zip object at 0x02B01B48> #返回的是一个对象
>>> list(zipped)
[(1, 4), (2, 5), (3, 6)] #使用list()函数转换为列表
>>> list(zip(a,c))
[(1, 4), (2, 5), (3, 6)]

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这次我们自己编写一个类似的函数，与zip不同的是：如果各个可迭代对象的元素个数不一致，则返回的对象长度与最长的可迭代对象相同，不足的用None补齐。
例如：

str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
print(newzip(str1, list1, tuple1))

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返回：

[('A', 1, 1.1), ('B', 2, 2.2), ('C', 3, None), (None, [4, 5], None)]

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注，你的程序可以返回一个列表，也可是用yield返回迭代器对象。

凌九霄 · 发表于 2018-8-21 11:10:27

本帖最后由凌九霄于 2018-8-21 15:37 编辑

def newzip(*args):
nargs = [ ]
lmax = sorted([ len(i) for i in args ], reverse=True)
for i in args:
i = list(i)
while len(i) < lmax[ 0 ]:
i.append(None)
nargs.append(i)
return zip(*nargs)

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graceasyi · 发表于 2018-8-21 11:22:15

本帖最后由 graceasyi 于 2018-8-21 11:27 编辑

# 方法一：list

def newzip(*args):
ml = max([len(a) for a in args]) # 参数中最长的长度
result = []
for i in range(0, ml):
tup = ()
for a in range(0, len(args)):
try:
tup += (args[a][i],)
except IndexError:
tup += (None, )
result.append(tup)
return result
str1 = 'ABC'
list1 = [1, 2, 3, [4, 5]]
tuple1 = (1.1, 2.2)
print(newzip(str1, list1, tuple1))

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结果：
[('A', 1, 1.1), ('B', 2, 2.2), ('C', 3, None), (None, [4, 5], None)]

# 方法二：yield

def newzip1(*args):
ml = max([len(a) for a in args]) # 参数中最长的长度
for i in range(0, ml):
tup = ()
for a in range(0, len(args)):
try:
tup += (args[a][i],)
except IndexError:
tup += (None, )
yield tup
str1 = 'ABC'
list1 = [1, 2, 3, [4, 5]]
tuple1 = (1.1, 2.2)
for r in newzip1(str1, list1, tuple1):
print(r)

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结果：
('A', 1, 1.1)
('B', 2, 2.2)
('C', 3, None)
(None, [4, 5], None)

塔利班 · 发表于 2018-8-21 11:44:07

本帖最后由塔利班于 2018-8-21 14:18 编辑

def newzip(*p):
return [tuple((each[i] if i<len(each) else None for each in p)) for i in range(max([len(t) for t in p]))]

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def newzip(*p):
for i in range(max([len(t) for t in p])):
yield tuple((each[i] if i<len(each) else None for each in p))

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JessiFly · 发表于 2018-8-21 11:46:27

def newzip(*iterator):
lengths = [] #存储每个可迭代对象的长度
for each_iter in iterator:
lengths.append(len(each_iter))
max_len = max(lengths)
result = []
for i in range(max_len):
each_list = []
for iter_num in range(len(iterator)):
#若索引值不小于可迭代对象长度，用None补齐
if i < lengths[iter_num]:
each_list.append(iterator[iter_num][i])
else:
each_list.append(None)
result.append(tuple(each_list))
return result
if __name__ == '__main__':
str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
print(newzip(str1, list1, tuple1))

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天圆突破 · 发表于 2018-8-21 11:53:46

本帖最后由天圆突破于 2018-8-21 12:22 编辑

def func201(*args):
return list(zip(*map(lambda x:list(x)+[None for i in range(max(map(lambda x:len(x), args))-len(x))] , args)))

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if __name__ == '__main__':
res = func201([1,2,3], [1], [])
print(res)

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qwertyuiop1822 · 发表于 2018-8-21 12:49:33

def newzip(*arg):
#取元祖最长的可迭代对象的长度
n = len(sorted(arg,key=len)[-1])

for i in range(n):
      a = ()
      for j in range(len(arg)):
         try:
            a += (arg[j][i],)
         except IndexError:
            a += (None,)
      yield a

拉了盏灯 · 发表于 2018-8-21 13:23:07

def newzip(*args):
lenth = 0
for i in args:
if len(i) > lenth:
lenth = len(i)
l = []
for i in range(lenth):
t = []
for each in args:
try:
t.append(each[i])
except IndexError:
t.append(None)
l.append(tuple(t))
return l

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晓屁屁 · 发表于 2018-8-21 13:59:59

本帖最后由晓屁屁于 2018-8-21 14:03 编辑

格式有误

晓屁屁 · 发表于 2018-8-21 14:03:23

str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
def new_zip(str1,list1,tuple1):
ziped = []
for i in range(max(len(str1),len(list1),len(tuple1))):
ziped.append(list())
try:
ziped[i].append(str1[i])
except:
ziped[i].append(None)
try:
ziped[i].append(list1[i])
except:
ziped[i].append(None)
try:
ziped[i].append(tuple1[i])
except:
ziped[i].append(None)
for i in range(len(ziped)):
yield tuple(ziped[i])
for i in new_zip(str1,list1,tuple1):
print(i)

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ErTian · 发表于 2018-8-21 14:12:54

def newzip(*args):
temp = []
args=list(args)
length_args=len(args)
length = len(sorted(args,key=lambda k:len(k),reverse=True)[0])
for i in range(length):
      empty = []
      for j in range(length_args):
         try:
            empty.append(args[j][i])
         except IndexError:
            empty.append(None)
      temp.append(tuple(empty))
return temp

newzip([0,1,2],[5,1,5],[3,2,1,5])

chongchuigu · 发表于 2018-8-21 16:03:40

本帖最后由 chongchuigu 于 2018-8-21 16:05 编辑

str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
def a201(*nums):
length=[]
for i in nums:
length.append(len(i))
maxnums=max(length)
newlist=[]
for i in range(maxnums):
for j in nums:
try:
newlist.append(j[i])
except IndexError:
newlist.append(None)
result=[(tuple(newlist[i:i+len(nums)])) for i in range(0,len(newlist),len(nums))]
return result
print(a201(str1,list1,tuple1))

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Legend丶Hu · 发表于 2018-8-21 16:20:38

def newzip(it_p1,it_p2,it_p3):
len1 = len(it_p1)
len2 = len(it_p2)
len3 = len(it_p3)
len_max = max(len1,len2,len3)
res = []
for i in range(len_max):
temp_tuple = []
if len1 > i:
temp_tuple.append(it_p1[i])
else:
temp_tuple.append(None)
if len2 > i:
temp_tuple.append(it_p2[i])
else:
temp_tuple.append(None)
if len3 > i:
temp_tuple.append(it_p3[i])
else:
temp_tuple.append(None)
res.append(tuple(temp_tuple))
return res
str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
print(newzip(str1, list1, tuple1))

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a641743376 · 发表于 2018-8-21 16:36:58

def newzip(*args):
maxlen=0
minlen=len(args[0])
for i in args:
      if len(i)>maxlen:
         maxlen=len(i)
      elif len(i)<minlen:
         minlen=len(i)
      list1=[]

for i in range(minlen):
      list2=[]
      for j in args:
         list2.append(j[i])
      list3=tuple(list2)
      list1.append(list3)

for i in range(minlen,maxlen):
      list2=[]
      for j in args:
         try:
            list2.append(j[i])
         except IndexError as f:
            list2.append('None')
      list3=tuple(list2)
      list1.append(list3)
return list1

Wangzy1025 · 发表于 2018-8-21 16:41:04

本帖最后由 Wangzy1025 于 2018-8-23 09:48 编辑

def fun201(*iter):
max_len, list_temp = max([len(x) for x in iter]), [list(x) for x in iter]
for each in list_temp:
if len(each) < max_len:
for i in range(max_len - len(each)):
each.append(None)
for i in range(max_len):
temp = []
for each in list_temp:
temp.append(each[i])
yield tuple(temp)
str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
print(list(fun201(str1, list1, tuple1)))

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a641743376 · 发表于 2018-8-21 16:42:29

精简一下：
def newzip(*args):
maxlen=0
for i in args:
      if len(i)>maxlen:
         maxlen=len(i)
list1=[]
for i in range(0,maxlen):
      list2=[]
      for j in args:
         try:
            list2.append(j[i])
         except IndexError as f:
            list2.append('None')
      list3=tuple(list2)
      list1.append(list3)
return list1
str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
print(newzip(str1, list1, tuple1))

colinshi · 发表于 2018-8-21 17:17:56

def newzip(*args):
num = len(args)
count = 0
for i in range(num):
if len(args[i]) > count:
count = len(args[i])
for i in range(count):
tmp = []
for j in range(num):
try:
tmp.append(args[j][i])
except:
tmp.append(None)
yield tmp
str1 = 'ABC'
list1 = [1, 2, 3,[4, 5]]
tuple1 = (1.1, 2.2)
for i in newzip(str1, list1, tuple1):
print(i)

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久疤K · 发表于 2018-8-21 22:57:18

>>> def newzip( *ls ):
m = max(len(x) for x in ls)
r = []
for i in range(m):
r.append(tuple(x[i] if i < len(x) else None for x in ls))
return r
>>> str1 = 'ABC'
>>> list1 = [1, 2, 3,[4, 5]]
>>> tuple1 = (1.1, 2.2)
>>> r = newzip(str1, list1, tuple1)
>>> r
[('A', 1, 1.1), ('B', 2, 2.2), ('C', 3, None), (None, [4, 5], None)]
>>> newzip(str1)
[('A',), ('B',), ('C',)]
>>>

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mongoole · 发表于 2018-8-21 23:03:07

#coding:utf-8
#自己编写一个类似的函数，与zip不同的是：如果各个可迭代对象的元素个数不一致
# 则返回的对象长度与最长的可迭代对象相同，不足的用None补齐。
def newzip(*args):
c = []
for i in args:
c.append(len(i))
num = max(c)
# print(num)
e = []
for m in range(num):
# print(m)
d = []
for n in args:
try:
if n[m] not in n:
d.append(None)
else:
d.append(n[m])
except:
d.append(None)
tupple = tuple(d)
e.append(tupple)
# print(e)
return e
str1 = 'ABC'
list1 = [1, 2, 3, [4, 5]]
tuple1 = (1.1, 2.2)
str2 = '1'
print(newzip(str1,list1,tuple1,str2))

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787330410 · 发表于 2018-8-22 00:17:29

def newzip(*seq):
seq=list(seq)
max_value=max([len(each_item) for each_item in seq])
for x in range(len(seq)):
seq[x]=list(seq[x])
for each_item in seq:
for n in range(max_value-len(each_item)):
each_item.append(None)
return [tuple([each_item[each_value] for each_item in seq]) for each_value in range(max_value)]

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