如何將給定值和鍵作為元組的元組創建字典

Question

我正在嘗試編寫一個 function 代碼，它將創建一個字典給定的值和鍵作為元組的元組。 例如

long2wide ( ((" apple ","red "),(" banana "," yellow "),(" banana "," green "),
(" apple "," green "),(" cherry "," red "))
, (" fruit ", " colour ") )

返回

{'fruit': ['apple', 'banana', 'banana', 'apple', 'cherry'], 'colour': ['red', 'yellow', 'green', 'green', 'red']}

這是我現在擁有的（錯誤的）代碼：

def long2wide(data, headers):
    dictionary = {}
    for entry in headers:
        for i in range(len(data)): #Iterate through each row of data
            for j in range(len(data[0])): #Iterate through each column of data
                dictionary[(headers)] = data[i][j]
    return dictionary

我目前的 output 是 {'fruit': 'red', 'color': 'red'}

:(

如果有人可以幫助調試/解決這個問題，我將不勝感激。 謝謝！！

Answer 1

我是這樣做的：

def long2wide(data, headers):
    dictionary = {}
    for count, entry in enumerate(headers):
        dictionary[entry] = [i[count] for i in data]
    return dictionary

output是{' fruit ': [' apple ', ' banana ', ' banana ', ' apple ', ' cherry '], ' colour ': ['red ', ' yellow ', ' green ', ' green ', ' red ']}

當您有未定義數量的標頭時，也可以使用此代碼。

這是通過列表理解完成的。 我們為標題中的每個項目創建一個列表。 然后，我們遍歷數據並為每個數據元組添加一個元素，具體取決於標題的數量，這意味着取決於我們正在迭代的 header。

如果你真的喜歡列表推導，你可以進一步縮小范圍：

dictionary = {entry: [i[count] for i in data] for count, entry in enumerate(headers)}

但我不建議這樣做，因為它會變得凌亂且難以閱讀。

Answer 2

對於任意數量的字段，也許？

def long2wide(values, names):
    d = dict()
    for i in range(len(names)):
        d[names[i]] = [v[i] for v in values]
    return d

或者，如果您喜歡單線：

{n: [v[i] for v in values] for i,n in enumerate(names)}

其中values是您的第一個列表並names您的第二個列表。

Answer 3

您可以將元組加載到 dataframe 然后回到字典

data=(
    (" apple ","red "),
    (" banana "," yellow "),
    (" banana "," green "),
    (" apple "," green "),
    (" cherry "," red ")
)
headers= (" fruit ", " colour ") 

df=pd.DataFrame(data,columns=headers)
a_dict=df.to_dict()
for key,item in a_dict.items():
   print(key)
   for i in np.arange(len(item)):
       print(item[i])

output：

fruit 
  apple 
  banana 
  banana 
  apple 
  cherry 
colour 
  red 
  yellow 
  green 
  green 
  red 

或者您可以直接將數據鍵值對加載到字典中並使用 header 作為索引

  headers= ((" fruit ",0),(" colour ",1)) 

  dct = dict((y, x) for x, y in data)        
  print(dct)

output：

  {'red ': ' apple ', ' yellow ': ' banana ', ' green ': ' apple ', ' red ': ' cherry '}

Answer 4

這是一個示例：

def long2wide(data, headers):
    dictionary = {}
    dictionary[headers[0]] = []
    dictionary[headers[1]] = []
    for i in data:
        dictionary[headers[0]].append(i[0])
        dictionary[headers[1]].append(i[1])
    return dictionary

Answer 5

那這個呢？

def make_dic(list):
    d = {'fruit': [], 'color': [], }
    for fruit in long2wide:
        d['fruit'].append(fruit[0])
        d['color'].append(fruit[1])
    return d


long2wide = ((("apple", "red "), ("banana", "yellow"), ("banana", "green"),
              ("apple", "green"), ("cherry", "red")))
print(make_dic(long2wide))

Answer 6

首先，使用字典推導初始化空列表的字典。

init_dict = {k: [] for k in headers}

然后，遍歷您的元組列表，使每個 output 為 (, )。

for i in data:
    # i is the value (<fruit>, <color>)
    fruit = i[0]
    color = i[1]
    init_dict[headers[0]].append(fruit)
    init_dict[headers[1]].append(color)