CODEWITHSUNDEEP
regex
Avinash
Avinash

Reputation: 3291

Regular expression to match decimals with or without leading zeros

^([0-9]*[1-9][0-9]*(\.[0-9]+)?|[0]+\.[0-9]*[1-9][0-9]*)$

I don't know regular expression well. Above regular expression does not allow input .2 .but it allows all other decimals like 0.2 , 0.02 etc . I need to make this expression allow the number like .2 ,.06 , etc.....

Upvotes: 3

Views: 12597

Answers (6)

notfeyyyyy
notfeyyyyy

Reputation: 1

I have been working on this from quite a while I got a some good expression that works well

 [+-]?(?:\d+|\D{0}\.\d+)(?:\.\d+)?

It identifies numbers/decimals with with/without leading 0's. Here is the demo on regx101

Adding demo here in case link do not works

import re      
x = "string .001 or -.001 or 1.00 or -1.00 or-213 or 123.132s1 or 012"   
print(re.findall(r"[+-]?(?:\d+|\D{0}\.\d+)(?:\.\d+)?", x))

O/P

['.001', '-.001', '1.00', '-1.00', '-213', '123.132', '1', '012']

Upvotes: 0

Atur
Atur

Reputation: 1770

You may also use a simple expression like:

^[-+]?\d*(\.\d+)?$

Upvotes: 1

Vitaliy Ulantikov
Vitaliy Ulantikov

Reputation: 10514

Replace it with:

^([0-9]*[1-9][0-9]*(\.[0-9]+)?|[0]*\.[0-9]*[1-9][0-9]*)$

or even shorter:

^(\d*[1-9]\d*(\.\d+)?|[0]*\.\d*[1-9]\d*)$

Upvotes: 0

Gumbo
Gumbo

Reputation: 655209

I would use this:

^(?:(?:0|[1-9][0-9]*)(?:\.[0-9]*)?|\.[0-9]+)$

This allows number expressions starting with either

  • some digits, followed by optional fractional digits, or
  • just fractional digits.

Allowed:

123
123.
123.45
.12345

But not:

.
01234

Upvotes: 2

gnarf
gnarf

Reputation: 106332

I like this regexp for floating point numbers, its pretty smart in that it wont match 0.0 as a number. It requires at least one non-zero number on either side of the period. Figured I'd break it into its parts to provide a deeper understanding of it.

^             #Match at start of string
 (            #start capture group
  [0-9]*       # 0-9, zero or more times
  [1-9]        # 1-9
  [0-9]*       # 0-9, zero or more times
  (            #start capture group
   \.           # literal .
   [0-9]+       # 0-9, one or more times
  )?           #end group - make it optional
 |            #OR - If the first option didn't match, try alternate
  [0]+         # 0, one or more times ( change this to 0* for zero or more times )
  \.           # literal .
  [0-9]*       # 0-9, zero or more times
  [1-9]        # 1-9
  [0-9]*       # 0-9, zero or more times
 )            #end capture group
$             #match end of string

The regexp has two smaller patterns inside of it, the first matches cases where the number is >= 1 (having at least one non-zero character left of the .) optionally allowing for a period with one or more trailing numbers. The second matches <1.0 and ensures that there is at least one non-zero digit on the right side of the dot.

Johannes' answer already gives you the [0]* solution to the problem.

Couple of regexp shortcuts, you could replace any instance of [0-9] with \d in most regexp flavors. Also [0] only matches 0 so you might as well just use 0* instead of [0]*. The final regexp:

/^(\d*[1-9]\d*(\.\d+)?|0*\.\d*[1-9]\d*)$/

Upvotes: 8

Joey
Joey

Reputation: 354426

Just change the + after [0] into an ansterisk `*":

^([0-9]*[1-9][0-9]*(\.[0-9]+)?|[0]*\.[0-9]*[1-9][0-9]*)$

So instead of allowing one or more zeroes preceding the dot, just allow 0 or more.

Upvotes: 11

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