This question already has answers here:
How do I pad a string with zeroes?
(19 answers)
Closed 3 years ago.
I'm trying to make length = 001 in Python 3 but whenever I try to print it out it truncates the value without the leading zeros (length = 1). How would I stop this happening without having to cast length to a string before printing it out?
Make use of the zfill() helper method to left-pad any string, integer or float with zeros; it's valid for both Python 2.x and Python 3.x.
It important to note that Python 2 is no longer supported.
Sample usage:
print(str(1).zfill(3))
# Expected output: 001
Description:
When applied to a value, zfill() returns a value left-padded with zeros when the length of the initial string value less than that of the applied width value, otherwise, the initial string value as is.
Syntax:
str(string).zfill(width)
# Where string represents a string, an integer or a float, and
# width, the desired length to left-pad.
Since python 3.6 you can use fstring :
>>> length = 1
>>> print(f'length = {length:03}')
length = 001
There are many ways to achieve this but the easiest way in Python 3.6+, in my opinion, is this:
print(f"{1:03}")
Python integers don't have an inherent length or number of significant digits. If you want them to print a specific way, you need to convert them to a string. There are several ways you can do so that let you specify things like padding characters and minimum lengths.
To pad with zeros to a minimum of three characters, try:
length = 1
print(format(length, '03'))
I suggest this ugly method but it works:
length = 1
lenghtafterpadding = 3
newlength = '0' * (lenghtafterpadding - len(str(length))) + str(length)
I came here to find a lighter solution than this one!
Related
This question already has answers here:
Dynamically calculated zero padding in format string in python
(2 answers)
How do I pad a string with zeroes?
(19 answers)
Closed 9 months ago.
Sorry if this is a bit of a noob question. But moving on..
Say at the beginning of my code I set a variable, like this:
TestVar = 'A6'
But I later want it to print out as 000000A6
Or say it was
TestVar = 'C30'
I'd want it to print out as 00000C30
Basically always returning it with a length of 8
The reasoning for this is I've made a general script for modding a game, (I can link if asked) and you need to put in certain values which I want to have format automatically for ease of use. For example on running it'll print
Item ID Here:
And if you put in 166 it would convert the decimal number to hex which would be A6, however in order to be usable for all values (not just ones that are 2 digits once converted) I'm trying to make it detect it's length and format it with the 0s before.
Sorry if this doesnt make sense, in a simpler way of saying this, is there a way for it to detect the length of a variable? So for example in pseudo
TestVar = 'C30'
If TestVar length = 3
print('00000'+TestVar)
Print Result: 00000C30
Basically always returning it with a length of 8
That's what format strings do:
>>> print(f"{'C30':>08s}")
00000C30
As a sidenote, to output any number as 8-digit hex:
>>> print(f"{100:>08X}")
00000064
>>> print(f"{1024:>08X}")
00000400
See the documentation:
for f-strings (the f'I am an f-string' syntax);
for formatting syntax (the >08s and >08X thing).
Use string function rjust():
print(test.rjust(8,'0'))
The .zfill string method can be used.
For example:
s = 'C30'
s.zfill(8)
>>> '00000C30'
Try this code
txt = "A6"
x = txt.zfill(8)
print(x)
You can use string.zfill method
for example.
code = '3C0'
filledCode = code.zfill(8)
this method filled with zero the number of digit that you pass like a parameter
try something like this str.rjust() function
i = 1111
pad = '0'
n = 8
x = str(i).rjust(n, pad)
print(x) # 00001111
This question already has answers here:
How to implement conditional string formatting? [duplicate]
(3 answers)
Closed 1 year ago.
I currently am trying to work with a number that has variable decimal place lengths. It can either be an integer, or have up to 10 decimals i.e. 33.3333333. I wanted to restrict the number to only have 2 decimals when it exceeds the length, or maintain the original if it's less.
I've tried using "{:0:.2f}".format, but the problem is that for integers, it also adds .00 to the end of the string.
When I tried using round(3) it'll return 3.0.
Is there a method, preferably a single line, that can convert 3.333333 to 3.33 but still allow for 3 to stay as an int?
Try choosing the format as a function of the values wholeness:
"{d}" if int(a) == a else "{:0:.2f}"
Can you finish from there?
You can use a conditional expression to choose the format based on the type of the variable:
for x in (33.3333333, 3):
print(("{:0}" if isinstance(x, int) else "{:.2f}").format(x))
You could also implement it using a dictionary to map types to format strings:
formats = {int: "{:0}", float: "{:.2f}"}
for x in (33.3333333, 3):
print(formats.get(type(x)).format(x))
This question already has answers here:
Given n, take tsum of the digits of n. If that value has more than one digit, continue reducing a single-digit number is produced
(4 answers)
Closed 1 year ago.
I have problem and trying to get next:
new_string = "35" #and this result must be like new_int = 3+5.
How im available to do this? I know the type conversion, but not a clue how i should do this.
As you are new to the python, i suggest you doing it using
int(new_string[0]) # 3
int(new_string[1]) # 5
So now you have 2 integers, you can to whatever you want
This question already has answers here:
How do I pad a string with zeroes?
(19 answers)
Closed 6 years ago.
i'd like to convert a decimal to a string, where zeros at the end are preserved.
Using str method erases the last zeros.
Example:
number=0.20
Goal: "0.20"
e.g. using: str(number)="0.2" doesn't seem to work.
If you want 2 decimal places use:
number = 0.20
str_number = '%.2f' % number
Number before f indicates the desired number of places.
This can be done using string formatting.
"{0:.2f}".format(number)
Will return 0.20.
Doing your chosen method won't work because upon declaring number = 0.20 it omits the last zero right away. If you put that into your idle:
number = 0.20
number
0.2
So declaring number as str(number) is doing str(0.2).
Use the % operator with an appropriate format string:
'%1.2f' % number
=> '0.20'
I need to record SerialNumber(s) on an object. We enter many objects. Most serial numbers are strings - the numbers aren't used numerically, just as unique identifiers - but they are often sequential. Further, leading zeros are important due to unique id status of serial number.
When doing data entry, it's nice to just enter the first "sequential" serial number (eg 000123) and then the number of items (eg 5) to get the desired output - that way we can enter data in bulk see below:
Obj1.serial = 000123
Obj2.serial = 000124
Obj3.serial = 000125
Obj4.serial = 000126
Obj5.serial = 000127
The problem is that when you take the first number-as-string, turn to integer and increment, you loose the leading zeros.
Not all serials are sequential - not all are even numbers (eg FDM-434\RRTASDVI908)
But those that are, I would like to automate entry.
In python, what is the most elegant way to check for leading zeros (*and, I guess, edge cases like 0009999) in a string before iterating, and then re-application of those zeros after increment?
I have a solution to this problem but it isn't elegant. In fact, it's the most boring and blunt alg possible.
Is there an elegant solution to this problem?
EDIT
To clarify the question, I want the serial to have the same number of digits after the increment.
So, in most cases, this will mean reapplying the same number of leading zeros. BUT in some edge cases the number of leading zeros will be decremented. eg: 009 -> 010; 0099 -> 0100
Try str.zfill():
>>> s = "000123"
>>> i = int(s)
>>> i
123
>>> n = 6
>>> str(i).zfill(n)
'000123'
I develop my comment here, Obj1.serial being a string:
Obj1.serial = "000123"
('%0'+str(len(Obj1.serial))+'d') % (1+int(Obj1.serial))
It's like #owen-s answer '%06d' % n: print the number and pad with leading 0.
Regarding '%d' % n, it's just one way of printing. From PEP3101:
In Python 3.0, the % operator is supplemented by a more powerful
string formatting method, format(). Support for the str.format()
method has been backported to Python 2.6.
So you may want to use format instead… Anyway, you have an integer at the right of the % sign, and it will replace the %d inside the left string.
'%06d' means print a minimum of 6 (6) digits (d) long, fill with 0 (0) if necessary.
As Obj1.serial is a string, you have to convert it to an integer before the increment: 1+int(Obj1.serial). And because the right side takes an integer, we can leave it like that.
Now, for the left part, as we can't hard code 6, we have to take the length of Obj1.serial. But this is an integer, so we have to convert it back to a string, and concatenate to the rest of the expression %0 6 d : '%0'+str(len(Obj1.serial))+'d'. Thus
('%0'+str(len(Obj1.serial))+'d') % (1+int(Obj1.serial))
Now, with format (format-specification):
'{0:06}'.format(n)
is replaced in the same way by
('{0:0'+str(len(Obj1.serial))+'}').format(1+int(Obj1.serial))
You could check the length of the string ahead of time, then use rjust to pad to the same length afterwards:
>>> s = "000123"
>>> len_s = len(s)
>>> i = int(s)
>>> i
123
>>> str(i).rjust(len_s, "0")
'000123'
You can check a serial number for all digits using:
if serial.isdigit():