Print a dictionary of lists vertically

Question 1

Pretty-printing tables requires a considerable amount of code (table-recipe, pretty-table). It's no fun writing this kind of code on an ad-hoc basis; you might as well use a well-designed module.

If you have pandas, you could dump the dict directly into a DataFrame, and print it like this:

In [4]: import pandas as pd
In [5]: result = {'WeightedLevel': [388.850952, 716.718689, 1312.55957, 2405.087158, 4460.083984, 8543.792969, 18805.201172, 57438.140625, 1792.367554], 'Job': 'Desktop', 'LoadLevel': [0.212399, 0.393191, 0.727874, 1.347436, 2.494368, 4.617561, 8.548006, 15.824027, 1.0], 'Task': 'test', 'Failure': [2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0], 'Blocks': [7255.151855, 231.589661, 9.365415, 0.55364, 0.0504, 0.006408, 0.001204, 0.000842, 2.060041]}

In [6]: pd.DataFrame(result)
Out[6]: 
        Blocks  Failure      Job  LoadLevel  Task  WeightedLevel
0  7255.151855        2  Desktop   0.212399  test     388.850952
1   231.589661        2  Desktop   0.393191  test     716.718689
2     9.365415        2  Desktop   0.727874  test    1312.559570
3     0.553640        2  Desktop   1.347436  test    2405.087158
4     0.050400        2  Desktop   2.494368  test    4460.083984
5     0.006408        2  Desktop   4.617561  test    8543.792969
6     0.001204        2  Desktop   8.548006  test   18805.201172
7     0.000842        2  Desktop  15.824027  test   57438.140625
8     2.060041        2  Desktop   1.000000  test    1792.367554

[9 rows x 6 columns]

Here is a way to print the dict in a table-like format without using a third-party module:

import itertools as IT

result = {'WeightedLevel': [388.850952, 716.718689, 1312.55957, 2405.087158, 4460.083984, 8543.792969, 18805.201172, 57438.140625, 1792.367554], 'Job': 'Desktop', 'LoadLevel': [0.212399, 0.393191, 0.727874, 1.347436, 2.494368, 4.617561, 8.548006, 15.824027, 1.0], 'Task': 'test', 'Failure': [2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0], 'Blocks': [7255.151855, 231.589661, 9.365415, 0.55364, 0.0504, 0.006408, 0.001204, 0.000842, 2.060041]}

matrix = zip(*[value if isinstance(value, list) else IT.repeat(value) for key,value in result.items()])
print(''.join(['{:15}'.format(key) for key in result.keys()]))
for row in matrix:
    print(''.join(['{:15}'.format(str(item)) for item in row]))

yields

Task           Blocks         LoadLevel      Failure        Job            WeightedLevel  
test           7255.151855    0.212399       2.0            Desktop        388.850952     
test           231.589661     0.393191       2.0            Desktop        716.718689     
test           9.365415       0.727874       2.0            Desktop        1312.55957     
test           0.55364        1.347436       2.0            Desktop        2405.087158    
test           0.0504         2.494368       2.0            Desktop        4460.083984    
test           0.006408       4.617561       2.0            Desktop        8543.792969    
test           0.001204       8.548006       2.0            Desktop        18805.201172   
test           0.000842       15.824027      2.0            Desktop        57438.140625   
test           2.060041       1.0            2.0            Desktop        1792.367554

Question 2

If they were all the same length this would be easy. What you will want to do is split apart the dictionary then loop through printing with end kwarg in the print statement.

so something like below. If you make every item a list that is the same length it would be easier.

def print_dict_table(result):
    """Print a dictionary of lists like a table.

    Args:
        result (dict): Dictionary of lists to print as a table.
    """
    # Count the longest value
    keys = result.keys()
    count = 0
    for i in result.values():
        if isinstance(i, list) and len(i) > count:
            count = len(i)

    # print header
    for i in keys:
        print(i, end="\t")
    print()

    # print columns
    for i in range(count):
        for j in keys:
            if isinstance(result[j], list) and len(result[j]) >= i:
                    print(result[j][i], end="\t")
            else:
                print(result[j], end="\t")
        print()
# end print_dict_table


result = {'WeightedLevel': [388.850952, 716.718689, 1312.55957, 2405.087158, 4460.083984, 8543.792969, 18805.201172, 57438.140625, 1792.367554], 'Job': 'Desktop', 'LoadLevel': [0.212399, 0.393191, 0.727874, 1.347436, 2.494368, 4.617561, 8.548006, 15.824027, 1.0], 'Task': 'test', 'Failure': [2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0], 'Blocks': [7255.151855, 231.589661, 9.365415, 0.55364, 0.0504, 0.006408, 0.001204, 0.000842, 2.060041]}

print_dict_table(result)

results

Failure Job LoadLevel   WeightedLevel   Task    Blocks  
2.0 Desktop 0.212399    388.850952  test    7255.151855 
2.0 Desktop 0.393191    716.718689  test    231.589661  
2.0 Desktop 0.727874    1312.55957  test    9.365415    
2.0 Desktop 1.347436    2405.087158 test    0.55364 
2.0 Desktop 2.494368    4460.083984 test    0.0504  
2.0 Desktop 4.617561    8543.792969 test    0.006408    
2.0 Desktop 8.548006    18805.201172    test    0.001204    
2.0 Desktop 15.824027   57438.140625    test    0.000842    
2.0 Desktop 1.0 1792.367554 test    2.060041

Dictionaries are not ordered. Honestly, I would redefine your data structure. If it is table data then use a table.