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Regex: find numbers greater than specific value (with varying decimal lengths)

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

I am trying to regex find all values in a list that are greater than or equal to .03. The tricky part is that my values have between 9 and 15 decimal places.

My current code works somewhat but is unwieldy - any advice much appreciated:

^(?:0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9])$

Thank you.

python regex

share|improve this question

asked Mar 27 at 22:05

user3682157user3682157

50322 gold badges1919 silver badges3838 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I see your regex the same as ^0?.[0-9][3-9]d9,11$
  
  – revo
  Mar 27 at 22:12
  

  
  
  
  

add a comment | 

3

I am trying to regex find all values in a list that are greater than or equal to .03. The tricky part is that my values have between 9 and 15 decimal places.

My current code works somewhat but is unwieldy - any advice much appreciated:

^(?:0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9])$

Thank you.

python regex

share|improve this question

asked Mar 27 at 22:05

user3682157user3682157

50322 gold badges1919 silver badges3838 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I see your regex the same as ^0?.[0-9][3-9]d9,11$
  
  – revo
  Mar 27 at 22:12
  

  
  
  
  

add a comment | 

I am trying to regex find all values in a list that are greater than or equal to .03. The tricky part is that my values have between 9 and 15 decimal places.

My current code works somewhat but is unwieldy - any advice much appreciated:

^(?:0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9])$

Thank you.

python regex

share|improve this question

asked Mar 27 at 22:05

user3682157user3682157

50322 gold badges1919 silver badges3838 bronze badges

^(?:0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]|0?.[0-9][3-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9])$

share|improve this question

asked Mar 27 at 22:05

user3682157user3682157

50322 gold badges1919 silver badges3838 bronze badges

share|improve this question

asked Mar 27 at 22:05

user3682157user3682157

50322 gold badges1919 silver badges3838 bronze badges

asked Mar 27 at 22:05

user3682157user3682157

50322 gold badges1919 silver badges3838 bronze badges

asked Mar 27 at 22:05

user3682157user3682157

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3 Answers
3

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oldest

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1

You can use an asterisk to denote zero or more of a digit:

^(?:0?.d[3-9]d*)$

This has the added benefit of matching exactly 0.03 or something with (say) 100 decimal places.

If you want to be strict about the decimal places, you can use brace syntax, which matches any repetitions between 9 and 15 digits:

^(?:0?.d[3-9]d9,15)$

Note that, as written, this regex doesn't match anything greater than or equal to 0.03--it matches anything greater than or equal to 0.03 and less than 1, and it doesn't match, say, 0.1, which is greater than 0.03. To match anything greater than 0.03, best to skip regex entirely and parse the number.

share|improve this answer

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for your response! Apologies, but I am confused at the last sentence - why would it not match .01 if it matches everything greater than .03 and less than 1?
  
  – user3682157
  Mar 27 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because .d[3-9] is not optional, so it won't match 0.1 because that string fails the [3-9] character.
  
  – ggorlen
  Mar 27 at 22:25
  
  

  
  
  
  

add a comment | 

0

You should simply parse your data as float:

From your pattern you seem to have the numbers already seperated very well - why use regex at all? You numbers fill the whole line or aren't matched at all ( r"^........$" so essentially you have:

t = """0.0000002
0.4
0.04
0.004
24544.23354
Also Bad"""

Simply split into lines and check each line as float

# splits the string into a list of lines
for line in (x.strip() for x in t.split("n")):
 try:
 if float(line) >= 0.03:
 print(line)
 else: 
 print("Not ok:",line)
 except:
 print("Not ok:",line)

Output:

Not ok: 0.0000002
0.4
0.04
Not ok: 0.004
24544.23354
Not ok: Also Bad

share|improve this answer

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

add a comment | 

0

Here you go: 0.030000000 - 0.999999999999999

Greater than or equal to .03 and less than 1.0

at 9-15 decimal places

0?.(?:03d7,13|0[4-9]d7,13|[1-9]d8,14)

Expanded

 # 0.030000000 - 0.999999999999999
 0?
 .
 (?:
 03 d7,13 
 | 0 [4-9] d7,13 
 | [1-9] d8,14 
 )

Note - this was machine generated, there could be some overlap.

Example: 0?.(?:0[3-9]d7,13|[1-9]d8,14)

share|improve this answer

edited Mar 27 at 23:44

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges

add a comment | 

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1

You can use an asterisk to denote zero or more of a digit:

^(?:0?.d[3-9]d*)$

This has the added benefit of matching exactly 0.03 or something with (say) 100 decimal places.

If you want to be strict about the decimal places, you can use brace syntax, which matches any repetitions between 9 and 15 digits:

^(?:0?.d[3-9]d9,15)$

Note that, as written, this regex doesn't match anything greater than or equal to 0.03--it matches anything greater than or equal to 0.03 and less than 1, and it doesn't match, say, 0.1, which is greater than 0.03. To match anything greater than 0.03, best to skip regex entirely and parse the number.

share|improve this answer

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for your response! Apologies, but I am confused at the last sentence - why would it not match .01 if it matches everything greater than .03 and less than 1?
  
  – user3682157
  Mar 27 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because .d[3-9] is not optional, so it won't match 0.1 because that string fails the [3-9] character.
  
  – ggorlen
  Mar 27 at 22:25
  
  

  
  
  
  

add a comment | 

1

You can use an asterisk to denote zero or more of a digit:

^(?:0?.d[3-9]d*)$

This has the added benefit of matching exactly 0.03 or something with (say) 100 decimal places.

If you want to be strict about the decimal places, you can use brace syntax, which matches any repetitions between 9 and 15 digits:

^(?:0?.d[3-9]d9,15)$

Note that, as written, this regex doesn't match anything greater than or equal to 0.03--it matches anything greater than or equal to 0.03 and less than 1, and it doesn't match, say, 0.1, which is greater than 0.03. To match anything greater than 0.03, best to skip regex entirely and parse the number.

share|improve this answer

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for your response! Apologies, but I am confused at the last sentence - why would it not match .01 if it matches everything greater than .03 and less than 1?
  
  – user3682157
  Mar 27 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Because .d[3-9] is not optional, so it won't match 0.1 because that string fails the [3-9] character.
  
  – ggorlen
  Mar 27 at 22:25
  
  

  
  
  
  

add a comment | 

You can use an asterisk to denote zero or more of a digit:

^(?:0?.d[3-9]d*)$

This has the added benefit of matching exactly 0.03 or something with (say) 100 decimal places.

If you want to be strict about the decimal places, you can use brace syntax, which matches any repetitions between 9 and 15 digits:

^(?:0?.d[3-9]d9,15)$

Note that, as written, this regex doesn't match anything greater than or equal to 0.03--it matches anything greater than or equal to 0.03 and less than 1, and it doesn't match, say, 0.1, which is greater than 0.03. To match anything greater than 0.03, best to skip regex entirely and parse the number.

share|improve this answer

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

^(?:0?.d[3-9]d*)$

^(?:0?.d[3-9]d9,15)$

share|improve this answer

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

answered Mar 27 at 22:15

ggorlenggorlen

11.8k44 gold badges1313 silver badges3030 bronze badges

0

You should simply parse your data as float:

From your pattern you seem to have the numbers already seperated very well - why use regex at all? You numbers fill the whole line or aren't matched at all ( r"^........$" so essentially you have:

t = """0.0000002
0.4
0.04
0.004
24544.23354
Also Bad"""

Simply split into lines and check each line as float

# splits the string into a list of lines
for line in (x.strip() for x in t.split("n")):
 try:
 if float(line) >= 0.03:
 print(line)
 else: 
 print("Not ok:",line)
 except:
 print("Not ok:",line)

Output:

Not ok: 0.0000002
0.4
0.04
Not ok: 0.004
24544.23354
Not ok: Also Bad

share|improve this answer

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

add a comment | 

0

You should simply parse your data as float:

From your pattern you seem to have the numbers already seperated very well - why use regex at all? You numbers fill the whole line or aren't matched at all ( r"^........$" so essentially you have:

t = """0.0000002
0.4
0.04
0.004
24544.23354
Also Bad"""

Simply split into lines and check each line as float

# splits the string into a list of lines
for line in (x.strip() for x in t.split("n")):
 try:
 if float(line) >= 0.03:
 print(line)
 else: 
 print("Not ok:",line)
 except:
 print("Not ok:",line)

Output:

Not ok: 0.0000002
0.4
0.04
Not ok: 0.004
24544.23354
Not ok: Also Bad

share|improve this answer

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

add a comment | 

You should simply parse your data as float:

From your pattern you seem to have the numbers already seperated very well - why use regex at all? You numbers fill the whole line or aren't matched at all ( r"^........$" so essentially you have:

t = """0.0000002
0.4
0.04
0.004
24544.23354
Also Bad"""

Simply split into lines and check each line as float

# splits the string into a list of lines
for line in (x.strip() for x in t.split("n")):
 try:
 if float(line) >= 0.03:
 print(line)
 else: 
 print("Not ok:",line)
 except:
 print("Not ok:",line)

Output:

Not ok: 0.0000002
0.4
0.04
Not ok: 0.004
24544.23354
Not ok: Also Bad

share|improve this answer

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

t = """0.0000002
0.4
0.04
0.004
24544.23354
Also Bad"""

# splits the string into a list of lines
for line in (x.strip() for x in t.split("n")):
 try:
 if float(line) >= 0.03:
 print(line)
 else: 
 print("Not ok:",line)
 except:
 print("Not ok:",line)

Not ok: 0.0000002
0.4
0.04
Not ok: 0.004
24544.23354
Not ok: Also Bad

share|improve this answer

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

answered Mar 27 at 22:17

Patrick ArtnerPatrick Artner

29.7k66 gold badges2626 silver badges4545 bronze badges

0

Here you go: 0.030000000 - 0.999999999999999

Greater than or equal to .03 and less than 1.0

at 9-15 decimal places

0?.(?:03d7,13|0[4-9]d7,13|[1-9]d8,14)

Expanded

 # 0.030000000 - 0.999999999999999
 0?
 .
 (?:
 03 d7,13 
 | 0 [4-9] d7,13 
 | [1-9] d8,14 
 )

Note - this was machine generated, there could be some overlap.

Example: 0?.(?:0[3-9]d7,13|[1-9]d8,14)

share|improve this answer

edited Mar 27 at 23:44

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges

add a comment | 

0

Here you go: 0.030000000 - 0.999999999999999

Greater than or equal to .03 and less than 1.0

at 9-15 decimal places

0?.(?:03d7,13|0[4-9]d7,13|[1-9]d8,14)

Expanded

 # 0.030000000 - 0.999999999999999
 0?
 .
 (?:
 03 d7,13 
 | 0 [4-9] d7,13 
 | [1-9] d8,14 
 )

Note - this was machine generated, there could be some overlap.

Example: 0?.(?:0[3-9]d7,13|[1-9]d8,14)

share|improve this answer

edited Mar 27 at 23:44

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges

add a comment | 

Here you go: 0.030000000 - 0.999999999999999

Greater than or equal to .03 and less than 1.0

at 9-15 decimal places

0?.(?:03d7,13|0[4-9]d7,13|[1-9]d8,14)

Expanded

 # 0.030000000 - 0.999999999999999
 0?
 .
 (?:
 03 d7,13 
 | 0 [4-9] d7,13 
 | [1-9] d8,14 
 )

Note - this was machine generated, there could be some overlap.

Example: 0?.(?:0[3-9]d7,13|[1-9]d8,14)

share|improve this answer

edited Mar 27 at 23:44

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges

0?.(?:03d7,13|0[4-9]d7,13|[1-9]d8,14)

 # 0.030000000 - 0.999999999999999
 0?
 .
 (?:
 03 d7,13 
 | 0 [4-9] d7,13 
 | [1-9] d8,14 
 )

share|improve this answer

edited Mar 27 at 23:44

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges

answered Mar 27 at 23:38

slnsln

31.3k33 gold badges1717 silver badges4545 bronze badges