Create a histogram in Python

Question

I'm trying to create a histogram in python

My fuction is the next one:

        def hist_hor(diccionario):
            dict={}
            
            for e in diccionario:
                if e  in dict:
                    dict[e] += '*'
                else: 
                    dict[e]= '*'
            for i in sorted(dict):
        
                print(f'{i}: {dict[i]}')
    
    histograma_horizontal(d)

It is supposed to look like this:

a: *****
b: **********
c: ************
d: ***********
e: ***************
f: ********************
g: ***************
h: *********
i: *******
j: **

but with my function it looks like this:

a: *
b: *
c: *
d: *
e: *
f: *
g: *
h: *
i: *
j: *

Besides, anyone who knows to represent it in that way too ?:

          *        
          *        
          *        
          *        
        * * *      
        * * *      
        * * *      
    *   * * *      
    * * * * *      
  * * * * * *      
  * * * * * * *    
  * * * * * * *    
  * * * * * * * *  
  * * * * * * * *  
* * * * * * * * *  
* * * * * * * * *  
* * * * * * * * *  
* * * * * * * * * *
* * * * * * * * * *

What is diccionario? Please share it

balderman
– balderman

2021-12-18 18:46:08 +00:00
Commented Dec 18, 2021 at 18:46 — balderman
– balderman, Commented Dec 18, 2021 at 18:46
This seems like a decent use for a defaultdict.

Chris
– Chris

2021-12-19 01:21:17 +00:00
Commented Dec 19, 2021 at 1:21 — Chris
– Chris, Commented Dec 19, 2021 at 1:21

Rob G · Accepted Answer · 2021-12-19 01:16:14Z

3

You can complete the first part without creating a new dictionary (remember to use end to prevent a new line):

NOTE - Tested in Ubuntu 20.04, using Python 3.8.

def hist_hor(diccionario):
    for key, value in diccionario.items():
        print(f"{key}: ", end="")
        for _ in range(value):
            print("*", end="")
        print()


diccionario = {"a": 5, "b": 10, "c": 12, "d": 11, "e": 15,
               "f": 20, "g": 15, "h": 9, "i": 7, "j": 2, }

hist_hor(diccionario)

Output:

a: *****
b: **********
c: ************
d: ***********
e: ***************
f: ********************
g: ***************
h: *********
i: *******
j: **

But if you need to create a new dictionary, use update instead of trying to manipulate indexes:

def hist_hor(diccionario):
    # Do not use dict as a variable name; it shadows a builtin type
    new_dicc = {}
    for key, value in diccionario.items():
        temp = ""
        for _ in range(value):
            temp += "*"
        new_dicc.update({key: temp})
    return new_dicc

d = hist_hor(diccionario)
for key, value in d.items():
    print(f"{key}: {value}")

The output will be the same.

For the second part, get the max value first and create a reverse loop. Iterate through the values, and if the value is greater than or equal to the loop index, print an asterisk:

def hist_hor(diccionario):
    m = max(diccionario.values())
    for i in range(m, 0, -1):
        for x in diccionario.values():
            if x >= i:
                print("* ", end="")
            else:
                print("  ", end="")
        print()

Output:

          *         
          *         
          *         
          *         
          *         
        * * *       
        * * *       
        * * *       
    *   * * *       
    * * * * *       
  * * * * * *       
  * * * * * * *     
  * * * * * * *     
  * * * * * * * *   
  * * * * * * * *   
* * * * * * * * *   
* * * * * * * * *   
* * * * * * * * *   
* * * * * * * * * * 
* * * * * * * * * *

edited Dec 19, 2021 at 1:16

answered Dec 18, 2021 at 18:55

Rob G

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2 Comments

Rob G Over a year ago

There are a few ways. You can use operator.itemgetter and collections.OrderedDict, or you can flip the values and the keys: new_dicc = dict(sorted((value, key) for (key, value) in new_dicc.items())) The easiest way is to use sorted with a lambda function: new_dicc = dict(sorted(new_dicc.items(), key=lambda item: item[1]))

Rob G Over a year ago

YW @maría-josé, but please review stackoverflow.com/help/someone-answers and meta.stackexchange.com/questions/5234/…

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