A simple Python graphing tool

Question

I've written a primitive Python script which "draws graphs" based on a list. It works, but it's a lot of lines for what I am trying to reach. The idea of the script is to function as a module that can be imported for a quick visual representation of data. Here's nextgraph.py:

from time import sleep
def draw_graph(
 x,
 y,
 previous=None, 
 initial=None,
 header_margin=1,
 footer_margin=1,
 anim_time=0.5,
 graph_symbol="*",
 indicator="|",
 graph_subtitle="My graph"
 ):
 if initial:
 graph_ = initial
 initial[x[y]][y] = graph_symbol
 # This may be a bit confusing, in this case,
 # x[y] means the current integer in a list (x)
 # and not "x-coordinate[y-coordinate]"
 else:
 graph_ = previous
 graph_[x[y]][y] = graph_symbol
 # Same thing here
 sleep(anim_time) 
 print(
 "\n"*header_margin, 
 indicator.join(graph_[0]), 
 "\n", indicator.join(graph_[1]),
 "\n", indicator.join(graph_[2]), 
 "\n", indicator.join(graph_[3]), 
 "\n", indicator.join(graph_[4]),
 "\n", indicator.join(graph_[5]), 
 "\n", indicator.join(graph_[6]), 
 "\n", indicator.join(graph_[7]),
 "\n", indicator.join(graph_[8]), 
 "\n", indicator.join(graph_[9]), 
 "\n", indicator.join(graph_[10]),
 "\n", indicator.join(graph_[11]), 
 "\n", indicator.join(graph_[12]), 
 "\n", indicator.join(graph_[13]),
 "\n", indicator.join(graph_[14]), 
 "\n", indicator.join(graph_[15]), 
 "\n", indicator.join(graph_[16]),
 "\n", indicator.join(graph_[17]),
 graph_subtitle,
 "033円[00m",
 "\n"*footer_margin)
 return graph_
def update_graph(data,
 top,
 bottom,
 sleep_,
 color="033円[00m",
 symbol="*",
 seperator="|",
 comment="My graph"
 ):
 graph = [
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "], 
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "], 
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "],
 [" ", " ", " ", " ", " ", " ", " ", " ", " ",
 " ", " ", " ", " ", " ", " ", " ", " ", " "]
 ]
 markers = 0
 while markers != 17:
 print("033円[00m")
 for _ in data:
 print(color)
 if markers == 0:
 old = draw_graph(data,
 markers, 
 initial=graph,
 header_margin=top,
 footer_margin=bottom,
 anim_time=sleep_,
 graph_symbol=symbol,
 indicator=seperator,
 graph_subtitle=comment
 )
 else:
 old = draw_graph(data, 
 markers, 
 old,
 header_margin=top,
 footer_margin=bottom,
 anim_time=sleep_,
 graph_symbol=symbol,
 indicator=seperator,
 graph_subtitle=comment
 )
 markers += 1
 print("033円[00m")

And here's a sample implementation (GraphTest.py):

from nextgraph import update_graph
from time import time
stream = [
 [3,3,3,3,4,4,4,3,3,3,3,4,3,2,1,2,2],
 [2,3,4,5,5,6,5,6,5,6,6,7,8,8,9,10,9],
 [10,11,12,13,13,14,14,15,14,14,13,12,11,10,9,8,7],
 [7,7,7,8,8,7,7,7,6,5,5,4,5,3,1,2,2],
 [3,3,4,3,4,3,4,5,6,6,5,4,3,2,1,1,2]
 ]
count = 0
for substream in stream:
 update_graph(data=substream,
 top=30,
 bottom=8,
 sleep_=0.2
 )
 count += 1

I am having trouble getting this to be more scalable. Adding a row/column would currently require adding another two lines to the code, so for a larger number (128 / 256 positions), this would become unmanageable. Also, for each added row, the print() function in draw_graph needs another line. If anybody has any suggestions on how to simplify the graph matrix or draw_graph / update_graph functions, or otherwise on how to improve the code, I'd love to hear.

Answer 1

• You may want to change update_graph so that it uses *args and **kwargs. update_graph(data, *args, color="033円[00m", **kwargs). This would allow you to then call draw_graph with less arguments, all on the same line:

  if markers == 0:
   old = draw_graph(data, markers, *args, initial=graph, **kwargs)
  else:
   old = draw_graph(data, markers, *args, previous=old, **kwargs)
  

• Rather than manually building the graph, you can instead use a comprehension and *. graph = [[" "] * 18 for _ in range(18)]. This is much more readable, and takes up much less space.

• Rather than manually joining all the lines in the graph, you could instead use join each line with " \n " as that's what , "\n", is doing. This assumes that graph_ is only 18 lines high.

  " \n ".join(indicator.join(g) for g in graph_)
  

• You could probably also use use iter and next, rather than using markers == 0, however that's may be less readable to you.

• Rather than using graph_ = initial and graph_ = previous you could just pass graph_ this would remove the if's in both update_graph and draw_graph. As they're doing the same thing.

• You may want to create an enum of colours, rather than using magic strings, such as "033円[00m".

• Daniel: Thanks for your answers, I'm currently implementing them, but I'm stuck with a question regarding your answer. You mention I should pass graph_. I'm not sure what you mean by that. Does this mean I call my old variable graph_, and then pass it to update_graph?
• Peilonrayz: @Coal_ Sorry I went out, yeah, remove initial and previous, and instead use graph_.