实施 group_by_owners 字典 [关闭]

Question

我对 TestDome.com Fileowners 问题进行了成功尝试，并想看看是否有人有建议来简化我的答案。在线 IDE 使用 Python 3.5.1。如果您想自己解决问题并且只是在寻找答案，这里有一个。众所周知，我是一名 Python 专家，所以这花了我很长一段时间来制作大量的修补程序。即使关于语法或一般清洁度，任何 cmets 都会有所帮助。谢谢！

实现一个 group_by_owners 函数：

接受包含每个文件名的文件所有者名称的字典。 以任意顺序返回包含每个所有者名称的文件名列表的字典。 例如，对于字典 {'Input.txt': 'Randy', 'Code.py': 'Stan', 'Output.txt': 'Randy'}，group_by_owners 函数应返回 {'Randy': ['Input. txt', 'Output.txt'], 'Stan': ['Code.py']}。

class FileOwners:

@staticmethod
def group_by_owners(files):
    val = (list(files.values()))                    #get values from dict
    val = set(val)                                  #make values a set to remove duplicates
    val = list(val)                                 #make set a list so we can work with it
    keyst = (list(files.keys()))                    #get keys from dict
    result = {}                                     #creat empty dict for output
    for i in range(len(val)):                       #loop over values(owners)
        for j in range(len(keyst)):                 #loop over keys(files)
            if val[i]==list(files.values())[j]:     #boolean to pick out files for current owner loop
                dummylist = [keyst[j]]              #make string pulled from dict a list so we can add it to the output in the correct format
                if val[i] in result:                #if the owner is already in the output add the new file to the existing dictionary entry
                    result[val[i]].append(keyst[j]) #add the new file
                else:                               #if the owner is NOT already in the output make a new entry 
                    result[val[i]] = dummylist      #make a new entry
    return result

files = {
'Input.txt': 'Randy',
'Code.py': 'Stan',
'Output.txt': 'Randy'
}

print(FileOwners.group_by_owners(files))

输出：

{'Stan': ['Code.py'], 'Randy': ['Output.txt', 'Input.txt']} 

Answer 1

Holly molly，这么简单的东西有很多代码：

def group_by_owners(files):
    result = {}
    for file, owner in files.items():  # use files.iteritems() on Python 2.x
        result[owner] = result.get(owner, []) + [file]  # you can use setdefault(), too
    return result

files = {
    'Input.txt': 'Randy',
    'Code.py': 'Stan',
    'Output.txt': 'Randy'
}

print(group_by_owners(files))
# {'Stan': ['Code.py'], 'Randy': ['Output.txt', 'Input.txt']}

您可以通过将collections.defaultdict 用于result 并将其所有键初始化为list 来进一步简化它 - 如果它以前不存在，您甚至不需要执行创建新列表的杂技附加到它。

Answer 2

我个人发现投票的答案很难理解，而其他一些则有点笨重。这是我的版本：

def group_by_owners(files):

ownerdict = {}

for key, value in files.items():
    if value in ownerdict:
        ownerdict[value].append(key)
    else:
        ownerdict[value] = [key]
return ownerdict


files = {
    'Input.txt': 'Randy',
    'Code.py': 'Stan',
    'Output.txt': 'Randy'
}
print(group_by_owners(files))

Answer 3

这对我有用。我相信这会更高效、更简单。

# Input dictionary.
files = {
        'Input.txt': 'Randy',
        'Code.py': 'Stan',
        'Output.txt': 'Randy'
    }   
# Function to group the files.   
def group_by_owners(files):

    # Dictionary object to hold the result.
    result = {}

    for key, value in files.items():

        if value not in result.keys():

            # Insert the values into the resulting dictionary
            # by interchanging the key, values.
            result[value] = [key]
        else:

            # Append the othet file name if the owner is same.
            result[value].append(key)

    return result

print(group_by_owners(files))

这是输出：

{'Randy': ['Input.txt', 'Output.txt'], 'Stan': ['Code.py']}

Answer 4

我刚开始学习，我也是 python 的菜鸟，但使用的是 2.7。所以请省去打印的括号。

我喜欢你关于虚拟列表和附加的想法。类似的东西，但可能更干净一点：

files = {
    'Input.txt': 'Randy',
    'Code.py': 'Stan',
    'Output.txt': 'Randy'
}
owners=[]
filenames=files.keys()
for key in files:
    if files[key] not in owners:
        owners.append(files[key])
print owners
print filenames

result={}
for i in owners:
    resultvalues=[]
    for key in files:
        if files[key]==i: 
            resultvalues.append(key)
            print resultvalues
            result[i]=resultvalues
print result

Answer 5

我在这里提到易于理解的解决方案，

owner = []
for i in dict1.keys():
if j not in owner:
    owner.append(j)
results = {}
for i in owner:
    result = []
    for j in dict1.keys():
        if dict1.get(j) == i:
            result.append(j)
        results[i] = result
print results

Answer 6

def group_by_owners(files):    
    values = []
    dic = {}

    for k,v in files.items():
        values.append(v)

    s_values = set(values)
    l_values = list(s_values)

    for i in l_values:
        keys=[] 
        for k,v in files.items():
            if v == i:
             keys.append(k)
        dic[i]=keys            

    return dic

Answer 7

你们看起来好多了，我花了 3.5 个小时通过大量的尝试和错误得到答案，因为我是 python 新手，而且思维不集中，感谢 @zwer 教授新东西。

dic={'apple':['green','red'], 'power': ['voltage'],'banana': ['green','red'],'current':['voltage'],'grass':['green'],'tiger':['lion']}


i = 0
adic = {}
while i < len(dic):
    key = list(dic)[i]
    for key2 in list(dic):
        if key2 != key:
            j = 0
            while j < len(dic[key]):
                if dic[key][j] in dic[key2]:
                    color = dic[key][j]
                    if dic[key][j] in adic:
                        if key not in adic[color]:
                            adic[dic[key][j]].append(key)

                    else:
                        mdic = {dic[key][j]: [key]}
                        adic.update(mdic)
                else:
                    if dic[key][j] not in adic:
                        mdic = {dic[key][j]: [key]}
                        adic.update(mdic)
                j = j + 1
    i = i + 1
print(adic)

Answer 8

class FileOwners:
    def group_by_owners(files):
    temp=[]

    for i in files:
        if files[i] not in temp:
            temp.append (files[i])
        result={}

    for j in temp:
        temp1=[]
        for i in files:
            if files[i]==j: 
                temp1.append(i)
                result[j]=temp1
    return (result)

files = {'Input.txt': 'Randy',
        'Code.py': 'Stan',
        'Output.txt': 'Randy'}

print(FileOwners.group_by_owners(files))

Answer 9

您必须将 Dictionary 迭代到 Dictionary 并比较值并将键附加到新列表中。在新字典中获取输出。

class FileOwners:

    @staticmethod

    def group_by_owners(files):
        new_dic = {}

        for key,val in files.items():
            key_list = []
            for k,v in files.items():

                if v == val:
                    #print(v)
                    key_list.append(k)
                    new_dic[v]= key_list

        return new_dic
        #print(new_dic)
files = {
    'Input.txt': 'Randy',
    'Code.py': 'Stan',
    'Output.txt': 'Randy'
}
print(FileOwners.group_by_owners(files))

Answer 10

我制作了所有者及其相关文件名列表的字典。为此，我首先创建了字典，其中传递的字典 files 的值作为键，空列表作为值。然后循环遍历files 并在空列表中附加files 的键

def group_by_owners(files):
    return_dict = {}
    owner_files = files.keys()
    for f in owner_files:
        return_dict[files[f]] = []
    for f, n in files.items():
        return_dict[n].append(f)
    return return_dict

Answer 11

我昨天开始学习 Python，在 testdome 上遇到了这个问题。这是我的答案：

def group_by_owners(files):
    f = files.values()
    s = set(f)
    newdict={}
    fileOwners = list(s)
    for owner in fileOwners:
        l=[]
        for file in files:
            if owner == files[file]:
                l.append(file)
        newdict[owner]=l
    return newdict


files = {
    'Input.txt': 'Randy',
    'Code.py': 'Stan',
    'Output.txt': 'Randy'
}   
print(group_by_owners(files))

Answer 12

1. 这是一些需要理解的复杂代码。

2. 但是是解决方案之一。

       files={
         'input.txt':'randy',
         'output.txt':'randy',
         'code.py':'stan'
         }
     k=files.keys()
     newfiles={value:key for key,value in files.items()}
     emptylist=[]
     for key in k:
         if key not in newfiles.values():
             missfile=key
             valuetoadd=files[missfile]
     emptylist.append(newfiles[valuetoadd])
     emptylist.append(missfile)
     newfiles[valuetoadd]=emptylist
     print(newfiles)
   

Answer 13

这是我对 defaultdict 的回答：

    from collections import defaultdict
    
    d = defaultdict(list)

    def group_by_owners(files):
        for k,v in files.items():
            d[v].append(k)
        return dict(d.items())

    if __name__ == "__main__":    
        files = {
            'Input.txt': 'Randy',
            'Code.py': 'Stan',
            'Output.txt': 'Randy'
        }   
        print(group_by_owners(files))