Python：每日一题 195

冬雪雪冬 · 发表于 2018-8-6 19:53:36

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本帖最后由冬雪雪冬于 2018-8-9 19:47 编辑

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题目：
这是一个奇怪的字典{"A": "B", "B": "C", "C": "D"}，一个item的value是另一个item的key
编写一个函数，参数为一个字典和一个key，如果得到的value是另一个item的key则继续求value，直到value不是key为止，返回这个value。
例如：
dict1 = {"A": "B", "B": "C", "C": "D"}
fun(dict1, "A") 返回 "D"
fun(dict1, "C") 返回 "D"

注意，如果value都是其他的item的key，即形成循环则返回None
例如
dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
fun(dict1, "A") 返回 None

--------------------------------------------------------------------
对大家答案的测试如下：
d1 = {"A": "B", "B": "C", "C": "D", "E": "F"}
d2 = {"A":"B", "B":"C","C":"D","D":"B"}
def f.....
print(f(d1, "A")) #链状
print(f(d1, "E")) #单独
print(f(d2, "A")) #循环

塔利班 · 发表于 2018-8-6 20:25:08

def fun195(d,k):
t=dict()
while True:
t[k]=d[k]
d.pop(k)
if t[k] in d:
k=t[k]
elif t[k] in t:
return None
else:
return t[k]

复制代码

zxz-python · 发表于 2018-8-6 20:51:43

def loop(dict1, key):
new_key = key
while True:
      if dict1[new_key] in dict1.keys():    #判断该key对应的值是否在key中
         new_key = dict1[new_key]
         if new_key == key:
            return None
      else:
         return dict1[new_key]

结果
dict1 = {"A": "B", "B": "C", "C": "D"}
print(loop(dict1, "A"))
D

dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(loop(dict1, "A"))
None

祥哥威武 · 发表于 2018-8-6 21:16:19

def func(dict0, key1):
if key1 not in dict0:
      return "字典中不包含该key!!!"
key2 = dict0[key1]
while True:
      if key2 == key1:
         return None
      if key2 not in dict0:
         return key2
      key2 = dict0[key2]

dict0 = {"A": "B", "B": "C", "C": "D"}
print(func(dict0, "A"))

dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(func(dict1, "A"))

JessiFly · 发表于 2018-8-6 21:16:29

def fun195(dict1,key):
list1 = []
value = dict1[key]
while value in dict1:
list1.append(value)
value = dict1[value]
if value in list1:
return None
return value

复制代码

>>> dict1 = {"A": "B", "B": "C", "C": "D"}
>>> fun195(dict1,'A')
'D'
>>> fun195(dict1,'C')
'D'
>>> dict2 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
>>> fun195(dict2,'A')
>>>

复制代码

天圆突破 · 发表于 2018-8-6 21:23:28

def func195(dct: dict, key: str):
if dct[key] in dct:
value = dct[key]
del dct[key]
return func195(dct, value)
else:
return None if dct=={} else dct[key]
if __name__ == '__main__':
dict1 = {"A": "B", "B": "C", "C": "D"}
res = func195(dict1,"A")
print(res)

复制代码

chongchuigu · 发表于 2018-8-6 21:57:18

本帖最后由 chongchuigu 于 2018-8-7 14:11 编辑

def a195(x,y):
for i in range(len(x)):
y=x[y]
if y not in dict1:
return y
elif x[y]==y:
return None
dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(a195(dict1, "A"))
dict1 = {"A": "B", "B": "C", "C": "D"}
print(a195(dict1, "A"))

复制代码

zhou995287902 · 发表于 2018-8-6 22:27:23

def fun(dicts,key):
count=1
while dicts[key] in dicts.keys():
key=dicts[key]
count+=1
if count>len(dicts):
print(None)
break
if count<=len(dicts):
print(dicts[key])
dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
fun(dict1, "A")

复制代码

久疤K · 发表于 2018-8-6 22:46:03

>>> def fun( dicts, key, rund=None):
if rund == None:
rund = []
if key in rund:
return None
rund.append(key)
v = dicts.get(key)
if v in dicts:
return fun(dicts, v, rund)
return v
>>> d = {1:2,2:3,3:4}
>>> fun(d, 1)
4
>>> d = {1:2,2:3,3:4, 4:1}
>>> fun(d,1)
>>> d = {"A": "B", "B": "C", "C": "D"}
>>> fun(d,'A')
'D'
>>> fun(d,'C')
'D'
>>> d = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
>>> fun(d, 'A')
>>>

复制代码

Henriz · 发表于 2018-8-6 23:56:23

def fun(dic1, str_a):
str_b = str_a
str_a = dic1[str_a]
while(1):
if str_a in dic1:
if str_a != str_b:
str_a = dic1[str_a]
else:
str_b = None
return str_b
break
else:
str_b = str_a
return str_b
break
dic1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(fun(dic1,"A"))
print(fun(dic1,"E"))

复制代码

拉了盏灯 · 发表于 2018-8-7 08:45:08

def fun195(dicts,key):
try:
fun195(dicts,dicts[key])
except KeyError:
print(dicts[key])
except RuntimeError:
return None
fun195({'A':'B','B':'C'},'A')

复制代码

mongoole · 发表于 2018-8-7 10:39:42

#coding:utf-8
def result(dic,str):
try:
if dic[str] in dic.keys():
return result(dic, dic[str])
else:
return dic[str]
except:
return None
dict1 = {"A": "B", "B": "C", "C": "D"}
dict2 = {"A": "B", "B": "C", "C": "D", "D": "A"}
print(result(dict1,'A'))
print(result(dict1,'C'))
print(result(dict1,'E'))
print(result(dict2,'A'))

复制代码

popython · 发表于 2018-8-7 11:15:33

本帖最后由 popython 于 2018-8-7 11:16 编辑

def fun(di,di_key):
A = di.keys()
ai = len(di)
di_key = di[di_key]
while ai:
if di_key in A:
di_key = di[di_key]
else:
return di_key
ai -= 1
if ai == 0:
return None

复制代码

只想到这样的。。。

故梦L · 发表于 2018-8-7 11:49:50

本帖最后由故梦L 于 2018-8-7 11:58 编辑

def function(dict, _key, keys=list(), values=list()):
"""
:param count:搜索_key的位置
:param flag:从_key开始索引
:return:value or None
"""
count = 0
for key,value in dict.items():
values.append(value)
keys.append(key)
if key == _key:
flag = count
else:
count += 1
for i in range(flag, len(dict)-1):
if keys[i+1] != values[i]:
return values[i]
else:
return None
dict1 = {"A": "H", "B": "C", "C": "D", "D": "E", "E": "F"}
print(function(dict1,'A'))

复制代码

colinshi · 发表于 2018-8-7 13:12:30

gtmp = []
def func195(dic, str):
global gtmp
try:
if dic[str]:
if dic[str] in gtmp:
gtmp = []
return
else:
gtmp.append(dic[str])
return func195(dic, dic[str])
except:
gtmp = []
return dic[str]
if __name__=='__main__':
dict1 = {"A": "B", "B": "C", "C": "D"}
print(func195(dict1, "A"))
print(func195(dict1, "C"))
dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(func195(dict1, "A"))

复制代码

grf1973 · 发表于 2018-8-7 14:24:56

def fun195(dic,x):
usedkeys=[x]
while True:
x=dic.get(x)
if x in usedkeys:
return None
if x in dic:
usedkeys.append(x)
else:
return x

dict1 = {"A": "B", "B": "C", "C": "D"}
print(fun195(dict1,"A"))
print(fun195(dict1,"C"))

dict1 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(fun195(dict1, "A"))

787330410 · 发表于 2018-8-7 16:18:47

def fun195(dic,key):
if dic[key] in dic:
dic[key],key=None,dic[key]
return fun195(dic,key)
else:
return dic[key]

复制代码

凌九霄 · 发表于 2018-8-7 16:43:56

本帖最后由凌九霄于 2018-8-7 21:26 编辑

dict1 = {"A": "B", "B": "C", "C": "D"}
dict2 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
dict3 = {"A": "B", "B": "C", "C": "D", "D": "Q", "E": "F"}
def func(ldict, key):
value = None
cdict = ldict.copy()
for i in range(len(ldict)):
if ldict.__contains__(key):
value = ldict[key]
try:
del cdict[ key ]
except KeyError:
return None
else:
key = value
else:
break
return value
print(func(dict1, "A")) #D
print(func(dict1, "M")) #None
print(func(dict2, "B")) #None
print(func(dict3, "C")) #Q

复制代码

小木子c · 发表于 2018-8-7 16:53:57

def fun193(dict1,str1):
value = dict1[str1]
while value in dict1.keys():
value = dict1[value]
if value == str1:
return None
return value

复制代码

Wangzy1025 · 发表于 2018-8-7 17:18:38

def fun195(dict1, key):
key1 = dict1.get(key)
if key1:
while dict1.get(key1) and key1 != key:
key1 = dict1.get(key1)
if not dict1.get(key1):
return key1
else:
return
else:
return
dict1 = {"A": "B", "B": "C", "C": "D"}
dict2 = {"A": "B", "B": "C", "C": "D", "D": "A", "E": "F"}
print(fun195(dict1, 'A'))
print(fun195(dict1, 'C'))
print(fun195(dict1, 'D'))
print(fun195(dict2, 'A'))

复制代码

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