如何将给定值和键作为元组的元组创建字典

Question

I'm trying to code a function which would create a dictionary given values and keys as tuples of tuples. Eg

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

returns

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

Here's the (erroneous) code I have now:

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

My current output is {'fruit': 'red', 'colour': 'red'}

:(

Would really appreciate if anyone could help debug/resolve this. Thanks!!

Answer 1

This is how I did it:

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

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

This code can also be used when you have an undefined number of headers.

This is done through list comprehension. We create a list for each item in the headers. Then, we loop through the data and add one element per tuple of the data, depending on the count of the headers, meaning depending on which header we are iterating through.

If you really liked list comprehensions, you could narrow it down further to:

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

But I wouldn't recommend this as it can get messy and hard to read.

Answer 2

For any number of fields, maybe?

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

Or if you like one-liners:

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

where values is your first list and names your second list.

Answer 3

you can load the tuple into a dataframe then back into a dictionary

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 

or you can load data key value pairs into a dictionary directly and use the header as an index

  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

here is a sample:

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

What about this?

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

First, initialize a dict of empty lists using a dictionary comprehension.

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

Then, iterate through your list of tuples so that each output is (, ).

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)