for example in this code below the with the end the integers stay on the same line but without the it does not.
num = 5
for i in range(1, num +1):
for j in range(num, i-1, -1):
print(j, end="")
print()
The end statement for printing in Python allows the programmer to define a custom ending character for each print call other than the default \n
For instance, if you have a function that is to print all values within a list on the same line, you do:
def value(l):
for items in l:
print(l, end=' ')
So if the list argued to this function contains the values [1, 2, 3, 4], it will print them in this manner: 1 2 3 4. If this new ending character parameter was not defined they would be printed:
1
2
3
4
The same principle applies for ANY value you provide for the end option.
In Python 3, the default ist end='\n' for the print function which is a newline after the string is printed.
To suppress this newline, one can set end='' so that the next print starts in the same line.
This is unrelated to the for loop.
Related
I am creating a dice program that must roll the dice a certain amount of times and append those rolls to a list. I am trying to figure out how to add commas in between each item of the list. This is the code that I have so far:
listRolls = []
def listPrint():
for i, item in enumerate(listRolls):
if (i+1)%13 == 0:
print(item)
else:
print(item,end=' ')
print(', '.join(listRolls))
For future reference, it's more "pythonic" (not my word) to use lower case variable_names, meaning your listRolls would then be list_rolls. Your code will handle it JUST FINE, however!
change
print(item,end=' ')
into
print(item,end=',')
If you want to print your whole list in one line, komma seperated simply use
data = [2,3,4,5,6]
print( *data, sep=",")
The * before the list-variable will make single elements from your list (decomposing it), so the print command essentially sees:
print( 2,3,4,5,6 , sep=",")
The sep="," tells the print command to print all given elements with a seperator as specified instead of the default ' '.
If you need to print, say, only 4 consecutive elements from your list on one line, then you can slice your list accordingly:
data = [2,3,4,5,6,7,8,9,10,11]
# slice the list in parts of length 4 and print those:
for d in ( data[i:i+4] for i in range(0,len(data),4)):
print( *d, sep=",")
Output:
2,3,4,5
6,7,8,9
10,11
Doku:
unpacking lists
print(*objects, sep=' ', ...)
range(start, stop[, step])
I'm trying to print each element individually, which is fine but also repeat each element based on position eg. "abcd" = A-Bb-Ccc-Dddd etc
So my problems are making print statements print x times based off their position in the string. I've tried a few combinations using len and range but i often encounter errors because i'm using strings not ints.
Should i be using len and range here? I'd prefer if you guys didn't post finished code, just basically how to go about that specific problem (if possible) so i can still go about figuring it out myself.
user_string = input()
def accum(s):
for letter in s:
pos = s[0]
print(letter.title())
pos = s[0 + 1]
accum(user_string)
You can enumerate iterables (lists, strings, ranges, dictkeys, ...) - it provides the index and a value:
text = "abcdef"
for idx,c in enumerate(text):
print(idx,c)
Output:
(0, 'a')
(1, 'b')
(2, 'c')
(3, 'd')
(4, 'e')
(5, 'f')
You can use that to print something multiple times. The print command takes 2 optional parameters :
print("Bla","blubb", sep=" --->", end=" Kawumm\n")
Output:
Bla --->blubb Kawumm
that specify what is printed between outputs and on the end of output - you can specify an end="" - so you can continue printing on the same line.
Doku:
Print
Enumerate
Edit:
user_string = input()
def accum(s):
t = [] # list to store stuff into
for count, letter in enumerate(s):
total = letter.upper() + letter * (count) # 1st as Upper, rest as is
t.append(total) # add to list
print(*t, sep="-") # the * "unpacks" the list into its parts
accum(user_string)
Unpacking:
print( [1,2,3,4,5], sep=" +++ ") # its just 1 value to print, no sep needed
print(*[1,2,3,4,5], sep=" +++ ") # 5 values to print, sep needed
Output:
[1, 2, 3, 4, 5]
1 +++ 2 +++ 3 +++ 4 +++ 5
You could try having a counter that will increase by 1 as the loop traverses through the string. Then, within the loop that you currently have, have another for loop to loop the size of the counter. If you want it to print out the first letter capitalized then you will need to account for that along with the dashes.
So, I have this code:
print("%d" % a, end=(" "))
It works, but in the output, there's a whitespace after the last number. I need to get rid of the last whitespace. My output needs to be shown on the same line, separated by a blank space. There should be no space after the last value.
Here's an example: (n is input, so suppose n = 5)
0 1 1 2 3
I've tried .strip, .join, but none of them worked. What do i have to do to get the right output? I'm sorry if this is too much of a simple question, I'm new in python.
edit: edit2:
a, b, i = 0, 1, 0
n=int(input())
for j in range(0, n):
while i < n:
print("%d" % a)
a, b = b, a + b
i += 1
You are adding the trailing space yourself with the end argument.
print("%d" % a, end=(" "))
Means print a ending in a ' '. Remove the end argument and the trailing space will no longer be printed (the default '\n' will be printed instead). See the docs for print() for more details.
Note also, that the end argument does not affect the string you are printing, i.e., a is not affected by end. If there is a trailing space in a string a, then a.strip() will remove that space. The reason it doesn't get removed by strip() in your case is that the space is not in the string you are printing, but instead is added to the visual output by the print() function.
Update:
It is hard to say, because it is a mystery what happens before or after the code snippet in your edit, but it sounds like you want to do something like:
a, b, i = 0, 1, 0
n=int(input())
nums = []
for j in range(0, n):
while i < n:
nums.append(str(a))
...
# Based on your desired output,
# I assume you modify the value of `a` somwhere in here
...
print(' '.join(nums))
while index < len(diecount):
print(index)
for number in range(diecount[index]):
print('*')
index+=1
print("")
At the moment i am getting
1
**
2
3
**
i want the output to be
1 **
2
3 **
A more Pythonic way to write this:
for index, count in enumerate(diecount[1:], start=1):
print(index, '*' * count)
(Manually controlling loop indexes in a while is usually a sign you're trying to write C code in Python).
Each print function appends a newline at the end of whatever it currently prints.
You can append a space instead of newline after the first print using the 2nd argument:
print(index, end = " ")
How to put spaces between values in a print() statement
for example:
for i in range(5):
print(i, sep='', end='')
prints
012345
I would like it to print
0 1 2 3 4 5
While others have given an answer, a good option here is to avoid using a loop and multiple print statements at all, and simply use the * operator to unpack your iterable into the arguments for print:
>>> print(*range(5))
0 1 2 3 4
As print() adds spaces between arguments automatically, this makes for a really concise and readable way to do this, without a loop.
>>> for i in range(5):
... print(i, end=' ')
...
0 1 2 3 4
Explanation: the sep parameter only affects the seperation of multiple values in one print statement. But here, you have multiple print statements with one value each, so you have to specify end to be a space (per default it's newline).
Just add the space between quotes after the end. It should be:
end = " "
In Python 2.x, you could also do
for i in range(5):
print i,
or
print " ".join(["%s"%i for i in range(5)])