Python3 Print on Same Line - Numbers in 7-Segment-Device Format

早过忘川 提交于 2020-06-23 12:45:30

问题


I'm new to Python and having difficulty getting the output to print on one line.

This is pertaining to the online Python class Learning Python Essentials Lab 5.1.10.6 and printing to a 7-segment-device. If you are unfamiliar with a 7-segment-device, see Wikipedia.

I am NOT using any external device. I only need it to print to my own terminal. All the other StackOverflow solutions I found are related to using actual devices and didn't help.

  • Lab Link: https://edube.org/learn/programming-essentials-in-python-part-2/lab-a-led-display

  • Purpose: Prompt user for number; print number in 7-segment display format to your terminal.

  • Notes: Using Python3.9. I tried 3 alternate solutions (Option 1,2,3), but none do what I want it to.
  • INSTRUCTIONS: Un/Comment Option 1,2,or 3 to run just that option
  • I did find this alternate solution, which I mostly understand. However, it's a totally different approach and not one I would have come up with. I know there are many ways to skin a 7-segment-device, and if this is the most correct, then I'll learn it. But I feel like I'm so close and only a superfluous '\n' away from figuring it out with my own method and trying to understand what I'm missing.

Thank you for your help.

Desired Output

###   ##  ###  ###  # #  ###  ###  ###  ###  ###  
# #  ###    #    #  # #  #    #      #  # #  # #  
# #   ##  ###  ###  ###  ###  ###    #  ###  ###  
# #   ##  #      #    #    #  # #    #  # #    #  
###   ##  ###  ###    #  ###  ###    #  ###  ###

My Code

# clear screen each time you run the script
import os
clear = lambda: os.system('cls')
clear()
# 
# Dictionary of (number:7-segment-hash)
dict1 = {
    '0':('###','# #','# #','# #','###'),
    '1':('#####'),
    '2':('###','  #','###','#  ','###'),
    '3':('###','  #','###','  #','###'),
    '4':('# #','# #','###','  #','  #'),
    '5':('###','#  ','###','  #','###'),
    '6':('###','#  ','###','# #','###'),
    '7':('###','  #','  #','  #','  #'),
    '8':('###','# #','###','# #','###'),
    '9':('###','# #','###','  #','###')
}

# Function to print numbers in 7-segment-device format
def fun_PrintNums(num):
    if num < 0 or num % 1 > 0 or type(num)!=int:    # if num is NOT a positive whole integer
        return "Invalid entry, please try again"
    else: 
        display = [' ']
        for i in str(num):      # convert 'num' to STRING; for each "number" in string 'num'


#'''Option 1: works, but prints nums vertically instead of side-by-side; Return=None ''' #
            for char in dict1[i]:
                print(*char)
print(fun_PrintNums(int(input("Enter any string of whole numbers: "))))
#----------------------------------------------------------------#


#''' Option 2: Return works, but still vertical and not spaced out ''' #
#             for char in dict1[i]:
#                 display.append(char)
#     return display
# print('\n'.join(fun_PrintNums(int(input("Enter any string of whole numbers: ")))))
#---------------------------------------------------------------------#

#''' Option 3: 'display' row1 offset; spaced out as desired, but vertical; Return=None''' #
#             for char in dict1[i]:                
#                 display += char
#                 display += '\n'
#     a = print(*display,end='')
#     return a
# print(fun_PrintNums(int(input("Enter any string of whole numbers: "))))
#---------------------------------------------------------------#

Option 1 Output Works, but prints nums vertically instead of side-by-side; Return=None

# # #
    #
# # #
#    
# # #
# # #
    #
# # #
    #
# # #
None 

Option 2 Output Return works, but still vertical and not spaced out.


###
  #
###
#  
###
###
  #
###
  #
###

Option 3 Output 'display' row1 offset; spaced out as desired, but vertical; Return=None

  # # # 
     #  
 # # #  
 #      
 # # #  
 # # #  
     #  
 # # #  
     #  
 # # #  
None    

回答1:


Your problem is that you are printing each number before the next, but you need to print each row before the next. As a simplified example:

dict1 = {
    '0':('###','# #','# #','# #','###'),
    '1':(' ##','###',' ##',' ##',' ##'),
    '2':('###','  #','###','#  ','###'),
    '3':('###','  #','###','  #','###'),
    '4':('# #','# #','###','  #','  #'),
    '5':('###','#  ','###','  #','###'),
    '6':('###','#  ','###','# #','###'),
    '7':('###','  #','  #','  #','  #'),
    '8':('###','# #','###','# #','###'),
    '9':('###','# #','###','  #','###')
}

num = '0123456789'

for row in range(len(dict1['0'])):
    print(' '.join(dict1[i][row] for i in num))

Output:

###  ## ### ### # # ### ### ### ### ###
# # ###   #   # # # #   #     # # # # #
# #  ## ### ### ### ### ###   # ### ###
# #  ## #     #   #   # # #   # # #   #
###  ## ### ###   # ### ###   # ### ###

If you don't want to use a list comprehension inside join, you can unroll that like this:

for row in range(len(dict1['0'])):
    line = []
    for i in num:
        line.append(dict1[i][row])
    print(' '.join(line))


来源:https://stackoverflow.com/questions/61809719/python3-print-on-same-line-numbers-in-7-segment-device-format

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