Matrix of Key-Value Pairs / Dict

Question

I have several objects from type dict with different keys. I want to create a table with all keys and foreach object one row. If one key isn't available, it should be empty.

For Example:

x1=dict( {"a":2, "b":3})
x2=dict( {"a":2, "b":3, "c":2})

and i want to get something like this:

"a","b","c"
2,3,
2,3,2

could you post a minimal input, the expected output and what you tried to achieve so far? — Colonel Beauvel, Aug 02 '16 at 08:25
@DeepSpace, yes I've tried to loop over x1.keys() but get failed — Roby, Aug 02 '16 at 08:36
Do you have a list of all the keys already, or do you need to extract them from all the `dict`s first? — PM 2Ring, Aug 02 '16 at 08:36

Julien Spronck · Accepted Answer · 2016-08-02T08:56:42.887

1

If you use pandas, you can do this:

import pandas as pd
df = pd.DataFrame({k: [v] for k, v in x1.iteritems()})
df2 = pd.DataFrame({k: [v] for k, v in x2.iteritems()})

df = pd.concat((df, df2), ignore_index=True)

#    a  b   c
# 0  2  3 NaN
# 1  2  3   2

Note: iteritems() only works in Python 2.x.

edited Aug 02 '16 at 08:56

answered Aug 02 '16 at 08:35

Julien Spronck

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1

You could simply use `DataFrame.from_dict` instead of iterating the dictionaries. – DeepSpace Aug 02 '16 at 08:37
1

@DeepSpace I think that does not work because the values are scalars instead of lists. – Julien Spronck Aug 02 '16 at 08:38
this worked for me with, after finding http://stackoverflow.com/a/10458567/722695 – Roby Aug 02 '16 at 08:48

score 0 · Answer 2 · answered Aug 02 '16 at 08:42

As a general approach (I am assuming you have a list of dicts, here, and you are fine with having the columns in "asciibetical" order):

def EmitDictsAsCSV(list_of_dicts):
  # First, accumulate the full set of dict keys
  key_set = set()
  for d in list_of_dicts:
     key_set.update(d.iterkeys())

  # make a sorted list out of them
  column_names = sorted(key_set)
  # print the header
  print ",".join(['"%s"' % c for c in column_names])

  # For each dict, loop over the columns and build a row, 
  # use the string representation of the value, if there's 
  # one, otherwise use an empty string, 
  # finish off by printing the row data, separated by commas
  for d in list_of_dicts:
    row_data = []
    for c in column_names:
      if c in d:
         row_data.append(str(d[c]))
      else:
        row_data.append("")
    print ",".join(row_data)

score 0 · Answer 3 · answered Aug 02 '16 at 08:49

Here is another simple solution that does not use pandas:

all_dics = [x1, x2]
keys = set(key for d in all_dics for key in d) # {'a', 'b', 'c'}
dic = {key: [None]*len(all_dics) for key in keys} # {'a': [None, None], 'b': [None, None], 'c': [None, None]}
for j, d in enumerate(all_dics):
    for key, val in d.iteritems():
        dic[key][j] = val

print dic
# {'a': [2, 2], 'b': [3, 3], 'c': [None, 2]}

score 0 · Answer 4 · answered Aug 02 '16 at 09:08

Here is a very crude and possibly inefficient solution

x1=dict( {"a":2, "b":3,"d":4,"e":5})
x2=dict( {"a":2, "b":3, "c":2})

z = dict(x1.items() + x2.items())
print(z.keys())

x1_vals = []
x2_vals = []
for keys in z.keys():
    if keys in x1.keys():
        x1_vals.append( x1[keys])
    else:
        x1_vals.append(None)

    if keys in x2.keys():
        x2_vals.append(x2[keys])
    else:
        x2_vals.append(None)

print (x1_vals)  
print (x2_vals)

Output

['a', 'c', 'b', 'e', 'd']
[2, None, 3, 5, 4] 
[2, 2, 3, None, None]

Matrix of Key-Value Pairs / Dict

4 Answers4