First question here.
I am picking up Python and have a question that may be quite basic.
I am trying to create this pattern with a nested loop:
x
x
x
xxx
With the code:
numbers = [1,1,1,3]
for count_x in numbers:
output = ""
for count in range(count_x):
output +=x
print(output)
My question is - why does my output change when I move the position of print(output).
The code above works but when I align print(output) with the for count_x in numbers:, I only get "xxx", when I align print(output) to output +=x, I get the following:
x
x
x
x
xx
xxx
which is very odd because there are only 4 items in the list and it shows me 6 lines of output.
Could someone please help? Really puzzled. Thank yall very much.
There's a difference between these two bits of code (fixing the x/"x" problem along the way - your code won't actually work as is unless you have a variable x):
# First:
numbers = [1,1,1,3]
for count_x in numbers:
output = ""
for count in range(count_x):
output += "x"
print(output)
# Second:
numbers = [1,1,1,3]
for count_x in numbers:
output = ""
for count in range(count_x):
output += "x"
print(output)
In the second, the print is done inside the loop that creates the string, meaning you print it out multiple times while building it. That's where your final three lines come from: *, ** and ***. This doesn't matter for all the previous lines since there's no functional difference. Printing a one character string at the end or after adding each of the one characters has the same effect.
In the first, you only print the string after fully constructing it.
You can see this effect by using a slightly modified version that outputs different things for each outer loop:
numbers = [1,1,1,3]
x = 1
for count_x in numbers:
output = ""
for count in range(count_x):
output += str(x)
print(output)
x += 1
This shows that the final three lines are part of a single string construction (comments added):
1 \
2 >- | Each is one outer/inner loop.
3 /
4 \ | A single outer loop, printing
44 >- | string under construction for
444 / | each inner loop.
In any case, there are better ways to do what you want, such as:
numbers = [1,1,1,3]
for count_x in numbers:
print('x' * count_x)
You could probably also do it on one line with a list comprehension but that's probably overkill.
Related
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
I have an assignment that has instructions as follows:
write a program that reads in 4 sets of 4 dashed lines and outputs the four binary symbols that each set of four lines represents.
input consists of 16 lines in total, consisting of any number of dashes and spaces.
the first four lines represents a symbol, the next four lines represents the next symbol and so on.
print out the four binary-encoded symbols represented by the 16 lines in total.
each binary symbol should be on its own line
This is based upon a previous program that I wrote where input is a single line of text consisting of any number of spaces and dashes. If there is an even number of dashes in the line, output 0. Otherwise, output 1.
This is the code for the above:
line = input()
num_dashes = line.count("-")
mod = num_dashes % 2
if mod == 0:
print("0")
else:
print("1")
Please may someone assist me?
Thank you.
The code you have for processing one line is fine, although you could replace the if...else with just:
print(mod)
Now to extend this to multiple lines, it might be better not to call print like that, but to collect the output in a variable, and only output that variable when all 16 lines have been processed. This way the output does not get mixed with the input from the console.
So for instance, it could happen like this:
output = []
for part in range(4): # loop 4 times
digits = ""
for line in range(4): # loop 4 times
line = input()
num_dashes = line.count("-")
mod = num_dashes % 2
digits += str(mod) # collect the digit
output.append(digits) # append 4 digits to a list
print("\n".join(output)) # print the list, separated by linebreaks
I am trying to convert a string of varchar to ascii. Then i'm trying to make it so any number that's not 3 digits has a 0 in front of it. then i'm trying to add a 1 to the very beginning of the string and then i'm trying to make it a large number that I can apply math to it.
I've tried a lot of different coding techniques. The closest I've gotten is below:
s = 'Ak'
for c in s:
mgk = (''.join(str(ord(c)) for c in s))
num = [mgk]
var = 1
num.insert(0, var)
mgc = lambda num: int(''.join(str(i) for i in num))
num = mgc(num)
print(num)
With this code I get the output: 165107
It's almost doing exactly what I need to do but it's taking out the 0 from the ord(A) which is 65. I want it to be 165. everything else seems to be working great. I'm using '%03d'% to insert the 0.
How I want it to work is:
Get the ord() value from a string of numbers and letters.
if the ord() value is less than 100 (ex: A = 65, add a 0 to make it a 3 digit number)
take the ord() values and combine them into 1 number. 0 needs to stay in from of 65. then add a one to the list. so basically the output will look like:
1065107
I want to make sure I can take that number and apply math to it.
I have this code too:
s = 'Ak'
for c in s:
s = ord(c)
s = '%03d'%s
mgk = (''.join(str(s)))
s = [mgk]
var = 1
s.insert(0, var)
mgc = lambda s: int(''.join(str(i) for i in s))
s = mgc(s)
print(s)
but then it counts each letter as its own element and it will not combine them and I only want the one in front of the very first number.
When the number is converted to an integer, it
Is this what you want? I am kinda confused:
a = 'Ak'
result = '1' + ''.join(str(f'{ord(char):03d}') for char in a)
print(result) # 1065107
# to make it a number just do:
my_int = int(result)
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 generate code to return the number of substrings within an input that are in sequential alphabetical order.
i.e. Input: 'abccbaabccba'
Output: 2
alphabet = 'abcdefghijklmnopqrstuvwxyz'
def cake(x):
for i in range(len(x)):
for j in range (len(x)+1):
s = x[i:j+1]
l = 0
if s in alphabet:
l += 1
return l
print (cake('abccbaabccba'))
So far my code will only return 1. Based on tests I've done on it, it seems it just returns a 1 if there are letters in the input. Does anyone see where I'm going wrong?
You are getting the output 1 every time because your code resets the count to l = 0 on every pass through the loop.
If you fix this, you will get the answer 96, because you are including a lot of redundant checks on empty strings ('' in alphabet returns True).
If you fix that, you will get 17, because your test string contains substrings of length 1 and 2, as well as 3+, that are also substrings of the alphabet. So, your code needs to take into account the minimum substring length you would like to consider—which I assume is 3:
alphabet = 'abcdefghijklmnopqrstuvwxyz'
def cake(x, minLength=3):
l = 0
for i in range(len(x)):
for j in range(i+minLength, len(x)): # carefully specify both the start and end values of the loop that determines where your substring will end
s = x[i:j]
if s in alphabet:
print(repr(s))
l += 1
return l
print (cake('abccbaabccba'))