Once again I'm asking for you advice. I'm trying to print a complex string block, it should look like this:
32 1 9999 523
+ 8 - 3801 + 9999 - 49
---- ------ ------ -----
40 -3800 19998 474
I wrote the function arrange_printer() for the characters arrangement in the correct format that could be reutilized for printing the list. This is how my code looks by now:
import operator
import sys
def arithmetic_arranger(problems, boolean: bool):
arranged_problems = []
if len(problems) <= 5:
for num in range(len(problems)):
arranged_problems += arrange_printer(problems[num], boolean)
else:
sys.exit("Error: Too many problems")
return print(*arranged_problems, end=' ')
def arrange_printer(oper: str, boolean: bool):
oper = oper.split()
ops = {"+": operator.add, "-": operator.sub}
a = int(oper[0])
b = int(oper[2])
if len(oper[0]) > len(oper[2]):
size = len(oper[0])
elif len(oper[0]) < len(oper[2]):
size = len(oper[2])
else:
size = len(oper[0])
line = '------'
ope = ' %*i\n%s %*i\n%s' % (size,a,oper[1],size,b,'------'[0:size+2])
try:
res = ops[oper[1]](a,b)
except:
sys.exit("Error: Operator must be '+' or '-'.")
if boolean == True:
ope = '%s\n%*i' % (ope,size+2, res)
return ope
arithmetic_arranger(['20 + 300', '1563 - 465 '], True)
#arrange_printer(' 20 + 334 ', True)
Sadly, I'm getting this format:
2 0
+ 3 0 0
- - - - -
3 2 0 1 5 6 3
- 4 6 5
- - - - - -
1 0 9 8
If you try printing the return of arrange_printer() as in the last commented line the format is the desired.
Any suggestion for improving my code or adopt good coding practices are well received, I'm starting to get a feel for programming in Python.
Thank you by your help!
The first problem I see is that you use += to add an item to the arranged_problems list. Strings are iterable. somelist += someiterable iterates over the someiterable, and appends each element to somelist. To append, use somelist.append()
Now once you fix this, it still won't work like you expect it to, because print() works by printing what you give it at the location of the cursor. Once you're on a new line, you can't go back to a previous line, because your cursor is already on the new line. Anything you print after that will go to the new line at the location of the cursor, so you need to arrange multiple problems such that their first lines all print first, then their second lines, and so on. Just fixing append(), you'd get this output:
20
+ 300
-----
320 1563
- 465
------
1098
You get a string with \n denoting the start of the new line from each call to arrange_printer(). You can split this output into lines, and then process each row separately.
For example:
def arithmetic_arranger(problems, boolean:bool):
arranged_problems = []
if len(problems) > 5:
print("Too many problems")
return
for problem in problems:
# Arrange and split into individual lines
lines = arrange_printer(problem, boolean).split('\n')
# Append the list of lines to our main list
arranged_problems.append(lines)
# Now, arranged_problems contains one list for each problem.
# Each list contains individual lines we want to print
# Use zip() to iterate over all the lists inside arranged_problems simultaneously
for problems_lines in zip(*arranged_problems):
# problems_lines is e.g.
# (' 20', ' 1563')
# ('+ 300', '- 465') etc
# Unpack this tuple and print it, separated by spaces.
print(*problems_lines, sep=" ")
Which gives the output:
20 1563
+ 300 - 465
----- ------
320 1098
If you expect each problem to have a different number of lines, then you can use the itertools.zip_longest() function instead of zip()
To collect all my other comments in one place:
return print(...) is pretty useless. print() doesn't return anything. return print(...) will always cause your function to return None.
Instead of iterating over range(len(problems)) and accessing problems[num], just do for problem in problems and then use problem instead of problems[num]
Debugging is an important skill, and the sooner into your programming career you learn it, the better off you will be.
Stepping through your program with a debugger allows you to see how each statement affects your program and is an invaluable debugging tool
Related
The following code is my view of checking whether the sum of a number and it's reverse is a palindrome or not.If the sum is a palindrome then sum will be displayed.Otherwise the process will be repeated until we get a palindrome. When it is set to execute, I am getting a time limit exceeded error.Where do I need to correct the code?
def pal(n1):
temp=n1
rev=0
while(temp>0):
rev=(rev*10)+(temp%10)
temp=temp/10
sum1=n1+rev
temp=sum1
rev=0
while(temp>0):
rev=(rev*10)+(temp%10)
temp=temp/10
if(rev==sum1):
print(sum1)
else:
pal(sum1)
n=int(input())
pal(n)
I expect the output of a number 453 to be 6666.
i.e.
453+354=807 (not a palindrome. So repeat the process)
807+708=1515
1515+5151=6666 (it is a palindrome)
Your problem is that you are checking for while temp > 0: but inside that loop you are using float division: temp=temp/10. So the condition will always hold. For example:
>>> 8/10
0.8
>>> 0.8/10
0.08
What you want is to change your divisions to int division:
>>> 8//10
0
Still you might consider working with strings which is much easier in that case:
def pal(n):
rev_n = str(n)[::-1]
sum_str = str(n + int(rev_n))
while sum_str != sum_str[::-1]:
# print(sum_str)
sum_rev = sum_str[::-1]
sum_str = str(int(sum_str) + int(sum_rev))
print(sum_str)
And with the commented print this gives:
>>> pal(453)
807
1515
6666
Here is one way of doing this using string manipulation, which goes a lot easier than trying to do this with numbers. It is also a more direct translation of what you describe afterwards. (I do not really see the link between your code and your description...)
def is_palindrome(text):
# : approach1, faster for large inputs
# mid_length = len(text) // 2
# offset = 0 if len(text) % 2 else 1
# return text[:mid_length] == text[:-mid_length - offset:-1]
# : approach2, faster for small inputs
return text == text[::-1]
def palindrome_sum(num):
while not is_palindrome(num):
num = str(int(num) + int(num[::-1]))
return num
num = input() # 453
palindrome = palindrome_sum(num)
print(palindrome)
# 6666
Whilst trying to create a piece of code in order to prove the Collatz Conjecture, I tried to, in a single line, format all of the results (as can be seen in the last line of the script). Even though the output looks as desired, the last line in particular is over complex and long. I was wondering if someone could write the last couple lines together and better. Thanks!
PD: The last line of the script prints each value of the history = [] list and then adds an index to it. The output of 10, in turn, looks like this:
$python collatz.py
In: 10
0 | 10
1 | 5
2 | 16
3 | 8
4 | 4
5 | 2
6 | 1
Done!
Here's my code: (The code has now been edited based on the answers :))
#!/usr/bin/env python3
import time
def formulate(number):
history = []
counter = 0
while True:
history.append(number)
if number == 1:
break
if number % 2 == 0:
number = number // 2
else:
number = number * 3 + 1
counter += 1
return counter, history
counter, history = formulate(int(input("In: ")))
for idx, x in enumerate(history):
print("{} | {}".format(idx, x))
print('Done!')
I would just keep it simple and perform one print per output line:
counter, history = formulate(int(input("In: ")))
for idx, x in enumerate(history):
print("{} | {}".format(idx, x))
print('Done!')
I know there are a couple of post about this question on S.O. but they have not helped me solve my problem. I am trying to use an accumulator to sum up the values in a text file. When there is a number on each line my code just prints each line that is in the file. When there is a blank space between I get an error message. I think it is a simple oversight but I am new to Python so I am not sure what I am doing wrong.
My code:
def main():
#Open a file named numbers.txt
numbers_file = open('numbers.txt','r')
#read the numbers on the file
number = numbers_file.readline()
while number != '':
#convert to integer
int_number = int(number)
#create accumulator
total = 0
#Accumulates a total number
total += int_number
#read the numbers on the file
number = numbers_file.readline()
#Print the data that was inside the file
print(total)
#Close the the numbers file
numbers_file.close()
#Call the main function
main()
Inputs in the text file:
100
200
300
400
500
Gives me error message:
ValueError: invalid literal for int() with base 10: '\n'
Inputs in the text file:
100
200
300
400
500
Prints:
100
200
300
400
500
You need to exclude empty lines because you can't convert them to an int(). One pythonic (EAFP) way to do this is to catch the exception and ignore (though this will silently ignore any non-number line):
with open('numbers.txt','r') as numbers_file:
total = 0
for line in numbers_file:
try:
total += int(line)
except ValueError:
pass
print(total)
Or you can explicitly test that you don't have an empty string after you .strip() all the whitespace (this would still error for a non-numeric line, e.g. 'hello'):
with open('numbers.txt','r') as numbers_file:
total = 0
for line in numbers_file:
if line.strip():
total += int(line)
print(total)
This second one can be written as a generator expression:
with open('numbers.txt','r') as numbers_file:
total = sum(int(line) for line in numbers_file if line.strip())
print(total)
You are assigning the value 0 to your accumulator each time you go through the loop, before you add the new value. This means you're adding the new value to 0 each time, which means you're just printing the new value.
If you move the line total = 0 to occur before the loop, then it should work as you were hoping.
If you want, you can clean this up a little:
numbers_file = open('numbers.txt','r')
total = 0
for number in numbers_file:
if number:
int_number = int(number)
total += int_number
print(total)
numbers_file.close()
would be a first pass. The check if number returns True if number contains a "truthy" value, which in this case would happen if you hit an empty line.
Hi you are missing to remove the 'new line symbol' which is \n.
To ensure you get only literals that can be converted to numbers you have to strip other characters.
With e.g.
a = '100\ntest'
print(a.isnumeric())
a = '103478'
print(a.isnumeric())
You can test if there is a character that prevents conversion to a number.
The regular expression package to manipulate string easily.
See this stack overflow threat.
import re
a = jkfads1000ki'
re.sub('\D','',a)
'1000'
See the Python docs on re.
I'm trying to figure out how to print a one line string while using a for loop. If there are other ways that you know of, I would appreciate the help. Thank you. Also, try edit off my code!
times = int(input("Enter a number: "))
print(times)
a = 0
for i in range(times+1):
print("*"*i)
a += i
print("Total stars: ")
print(a)
print("Equation: ")
for e in range(1,times+1):
print(e)
if e != times:
print("+")
else:
pass
Out:
Enter a number: 5
*
**
***
****
*****
Equation:
1
+
2
+
3
+
4
+
5
How do I make the equation in just one single line like this:
1+2+3+4+5
I don't think you can do a "backspace" after you've printed. At least erasing from the terminal isn't going to be done very easily. But you can build the string before you print it:
times = int(input("Enter a number: "))
print(times)
a = 0
for i in range(times+1):
print("*"*i)
a += i
print("Total stars: ")
print(a)
print("Equation: ")
equation_string = ""
for e in range(1,times+1):
equation_string += str(e)
if e != times:
equation_string += "+"
else:
pass
print(equation_string)
Basically, what happens is you store the temporary equation in equation_str so it's built like this:
1
1+
1+2
1+2+
...
And then you print equation_str once it's completely built. The output of the modified program is this
Enter a number: 5
5
*
**
***
****
*****
Total stars:
15
Equation:
1+2+3+4+5
Feel free to post a comment if anything is unclear.
Instead of your original for loop to print each number, try this:
output = '+'.join([str(i) for i in range(1, times + 1)])
print(output)
Explanation:
[str(i) for i in range(1, times + 1)] is a list comprehension that returns a list of all your numbers, converted to strings so that we can print them.
'+'.join(...) joins each element of your list, with a + in between each element.
Alternatively:
If you want a simple modification to your original code, you can simply suppress the newline from each print statement with the keyword paramater end, and set this to an empty string:
print(e, end='')
(Note that I am addressed the implied question, not the 'how do I do a backspace' question)
Too long for a comment, so I will post here.
The formatting options of python can come into good use, if you have a sequence you wish to format and print.
Consider the following...
>>> num = 5 # number of numbers to generate
>>> n = num-1 # one less used in generating format string
>>> times = [i for i in range(1,num+1)] # generate your numbers
>>> ("{}+"*n + "{}=").format(*times) # format your outputs
'1+2+3+4+5='
So although this doesn't answer your question, you can see that list comprehensions can be brought into play to generate your list of values, which can then be used in the format generation. The format string can also be produced with a l.c. but it gets pretty messy when you want to incorporate string elements like the + and = as shown in the above example.
I think you are looking for the end parameter for the print function - i.e. print(e, end='') which prints each value of e as it arrives followed by no space or newline.
I'm trying to use the str.find() and it keeps raising an error, what am I doing wrong?
import codecs
def countLOC(inFile):
""" Receives a file and then returns the amount
of actual lines of code by not counting commented
or blank lines """
LOC = 0
for line in inFile:
if line.isspace():
continue
comment = line.find('#')
if comment > 0:
for letter in range(comment):
if not letter.whitespace:
LOC += 1
break
return LOC
if __name__ == "__main__":
while True:
file_loc = input("Enter the file name: ").strip()
try:
source = codecs.open(file_loc)
except:
print ("**Invalid filename**")
else:
break
LOC_count = countLOC(source)
print ("\nThere were {0} lines of code in {1}".format(LOC_count,source.name))
Error
File "C:\Users\Justen-san\Documents\Eclipse Workspace\countLOC\src\root\nested\linesOfCode.py", line 12, in countLOC
comment = line.find('#')
TypeError: expected an object with the buffer interface
Use the built-in function open() instead of codecs.open().
You're running afoul of the difference between non-Unicode (Python 3 bytes, Python 2 str) and Unicode (Python 3 str, Python 2 unicode) string types. Python 3 won't convert automatically between non-Unicode and Unicode like Python 2 will. Using codecs.open() without an encoding parameter returns an object which yields bytes when you read from it.
Also, your countLOC function won't work:
for letter in range(comment):
if not letter.whitespace:
LOC += 1
break
That for loop will iterate over the numbers from zero to one less than the position of '#' in the string (letter = 0, 1, 2...); whitespace isn't a method of integers, and even if it were, you're not calling it.
Also, you're never incrementing LOC if the line doesn't contain #.
A "fixed" but otherwise faithful (and inefficient) version of your countLOC:
def countLOC(inFile):
LOC = 0
for line in inFile:
if line.isspace():
continue
comment = line.find('#')
if comment > 0:
for letter in line[:comment]:
if not letter.isspace():
LOC += 1
break
else:
LOC += 1
return LOC
How I might write the function:
def count_LOC(in_file):
loc = 0
for line in in_file:
line = line.lstrip()
if len(line) > 0 and not line.startswith('#'):
loc += 1
return loc
Are you actually passing an open file to the function? Maybe try printing type(file) and type(line), as there's something fishy here -- with an open file as the argument, I just can't reproduce your problem! (There are other bugs in your code but none that would cause that exception). Oh btw, as best practice, DON'T use names of builtins, such as file, for your own purposes -- that causes incredible amounts of confusion!