Home Parse String to Float or Int

# Parse String to Float or Int

Tristan Havelick
1#
Tristan Havelick Published in 2008-12-19 01:52:26Z
 In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 542.2222? Or parse the string "31" to an integer, 31? I just want to know how to parse a float string to a float, and (separately) an int string to an int.
Harley Holcombe
2#
Harley Holcombe Reply to 2008-12-19 01:54:51Z
 >>> a = "545.2222" >>> float(a) 545.22220000000004 >>> int(float(a)) 545 
codelogic
3#
codelogic Reply to 2008-12-19 01:54:11Z
 float("545.2222") and int(float("545.2222"))
Peter Mortensen
4#
Peter Mortensen Reply to 2015-03-21 08:28:35Z
 Users codelogic and harley are correct, but keep in mind if you know the string is an integer (for example, 545) you can call int("545") without first casting to float. If your strings are in a list, you could use the map function as well. >>> x = ["545.0", "545.6", "999.2"] >>> map(float, x) [545.0, 545.60000000000002, 999.20000000000005] >>>  It is only good if they're all the same type.
Peter Mortensen
5#
Peter Mortensen Reply to 2015-03-21 08:30:04Z
 Here's another interpretation of your question (hint: it's vague). It's possible you're looking for something like this: def parseIntOrFloat( aString ): return eval( aString )  It works like this... >>> parseIntOrFloat("545.2222") 545.22220000000004 >>> parseIntOrFloat("545") 545  Theoretically, there's an injection vulnerability. The string could, for example be "import os; os.abort()". Without any background on where the string comes from, however, the possibility is theoretical speculation. Since the question is vague, it's not at all clear if this vulnerability actually exists or not.
David Jones
6#
David Jones Reply to 2014-04-08 12:20:41Z
 def num(s): try: return int(s) except ValueError: return float(s) 
S.Lott
7#
S.Lott Reply to 2008-12-19 02:33:15Z
 float(x) if '.' in x else int(x) 
Nick
8#
Nick Reply to 2010-10-21 14:10:01Z
 You need to take into account rounding to do this properly. I.e. int(5.1) => 5 int(5.6) => 5 -- wrong, should be 6 so we do int(5.6 + 0.5) => 6 def convert(n): try: return int(n) except ValueError: return float(n + 0.5) 
Peter Mortensen
9#
Peter Mortensen Reply to 2015-03-21 08:32:25Z
 The question seems a little bit old. But let me suggest a function, parseStr, which makes something similar, that is, returns integer or float and if a given ASCII string cannot be converted to none of them it returns it untouched. The code of course might be adjusted to do only what you want:  >>> import string >>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \ ... int(x) or x.isalnum() and x or \ ... len(set(string.punctuation).intersection(x)) == 1 and \ ... x.count('.') == 1 and float(x) or x >>> parseStr('123') 123 >>> parseStr('123.3') 123.3 >>> parseStr('3HC1') '3HC1' >>> parseStr('12.e5') 1200000.0 >>> parseStr('12$5') '12$5' >>> parseStr('12.2.2') '12.2.2' 
Stefano
10#
Stefano Reply to 2012-03-15 02:23:45Z
 This is another method which deserves to be mentioned here, ast.literal_eval: This can be used for safely evaluating strings containing Python expressions from untrusted sources without the need to parse the values oneself. That is, a safe 'eval' >>> import ast >>> ast.literal_eval("545.2222") 545.2222 >>> ast.literal_eval("31") 31 
Peter Mortensen
11#
Peter Mortensen Reply to 2015-03-21 08:33:48Z
 The YAML parser can help you figure out what datatype your string is. Use yaml.load(), and then you can use type(result) to test for type: >>> import yaml >>> a = "545.2222" >>> result = yaml.load(a) >>> result 545.22220000000004 >>> type(result) >>> b = "31" >>> result = yaml.load(b) >>> result 31 >>> type(result) >>> c = "HI" >>> result = yaml.load(c) >>> result 'HI' >>> type(result) 
Mark Chackerian
12#
Mark Chackerian Reply to 2016-04-06 23:29:46Z

## Localization and commas

You should consider the possibility of commas in the string representation of a number, for cases like float("545,545.2222") which throws an exception. Instead, use methods in locale to convert the strings to numbers and interpret commas correctly. The locale.atof method converts to a float in one step once the locale has been set for the desired number convention.

Example 1 -- United States number conventions

In the United States and the UK, commas can be used as a thousands separator. In this example with American locale, the comma is handled properly as a separator:

>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>


Example 2 -- European number conventions

In the majority of countries of the world, commas are used for decimal marks instead of periods. In this example with French locale, the comma is correctly handled as a decimal mark:

>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222


The method locale.atoi is also available, but the argument should be an integer.

Eric Leschinski
13#
Eric Leschinski Reply to 2017-01-28 16:00:48Z

## Python method to check if a string is a float:

def isfloat(value):
try:
float(value)
return True
except:
return False


A longer and more accurate name for this function could be: isConvertibleToFloat(value)

## What is, and is not a float in Python may surprise you:

val                   isfloat(val) Note
--------------------  ----------   --------------------------------
""                    False        Blank string
"127"                 True         Passed string
True                  True         Pure sweet Truth
"True"                False        Vile contemptible lie
False                 True         So false it becomes true
"123.456"             True         Decimal
"      -127    "      True         Spaces trimmed
"\t\n12\r\n"          True         whitespace ignored
"NaN"                 True         Not a number
"NaNanananaBATMAN"    False        I am Batman
"-iNF"                True         Negative infinity
"123.E4"              True         Exponential notation
".1"                  True         mantissa only
"1,234"               False        Commas gtfo
u'\x30'               True         Unicode is fine.
"NULL"                False        Null is not special
"6e7777777777777"     True         Shrunk to infinity
"1.797693e+308"       True         This is max value
"infinity"            True         Same as inf
"infinityandBEYOND"   False        Extra characters wreck it
"12.34.56"            False        Only one dot allowed
u'四'                  False        Japanese '4' is not a float.
"#56"                 False        Pound sign
"56%"                 False        Percent of what?
"0E0"                 True         Exponential, move dot 0 places
0**0                  True         0___0  Exponentiation
"-5e-5"               True         Raise to a negative number
"+1e1"                True         Plus is OK with exponent
"+1e1^5"              False        Fancy exponent not interpreted
"+1e1.3"              False        No decimals in exponent
"-+1"                 False        Make up your mind
"(1)"                 False        Parenthesis is bad


You think you know what numbers are? You are not so good as you think! Not big surprise.

SethMMorton
14#
SethMMorton Reply to 2017-08-20 02:48:49Z
 If you aren't averse to third-party modules, you could check out the fastnumbers module. It provides a function called fast_real that does exactly what this question is asking for and does it faster than a pure-Python implementation: >>> from fastnumbers import fast_real >>> fast_real("545.2222") 545.2222 >>> type(fast_real("545.2222")) float >>> fast_real("31") 31 >>> type(fast_real("31")) int 
Aaron Hall
15#
Aaron Hall Reply to 2017-01-27 15:07:40Z

In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 542.2222? Or parse the string "31" to an integer, 31? I just want to know how to parse a float string to a float, and (separately) an int string to an int.

It's good that you ask to do these separately. If you're mixing them, you may be setting yourself up for problems later. The simple answer is:

"545.2222" to float:

>>> float("545.2222")
545.2222


"31" to an integer:

>>> int("31")
31


# Other conversions, ints to and from strings and literals:

Conversions from various bases, and you should know the base in advance (10 is the default). Note you can prefix them with what Python expects for its literals (see below) or remove the prefix:

>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31


If you don't know the base in advance, but you do know they will have the correct prefix, Python can infer this for you if you pass 0 as the base:

>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31


### Non-Decimal (i.e. Integer) Literals from other Bases

If your motivation is to have your own code clearly represent hard-coded specific values, however, you may not need to convert from the bases - you can let Python do it for you automatically with the correct syntax.

You can use the apropos prefixes to get automatic conversion to integers with the following literals. These are valid for Python 2 and 3:

Binary, prefix 0b

>>> 0b11111
31


Octal, prefix 0o

>>> 0o37
31


Hexadecimal, prefix 0x

>>> 0x1f
31


This can be useful when describing binary flags, file permissions in code, or hex values for colors - for example, note no quotes:

>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215


### Making ambiguous Python 2 octals compatible with Python 3

If you see an integer that starts with a 0, in Python 2, this is (deprecated) octal syntax.

>>> 037
31


It is bad because it looks like the value should be 37. So in Python 3, it now raises a SyntaxError:

>>> 037
File "<stdin>", line 1
037
^
SyntaxError: invalid token


Convert your Python 2 octals to octals that work in both 2 and 3 with the 0o prefix:

>>> 0o37
31

user1132603
16#
user1132603 Reply to 2017-05-30 11:39:49Z
 def get_int_or_float(v): number_as_float = float(v) number_as_int = int(number_as_float) return number_as_int if number_as_float == number_as_int else number_as_float 
Jerry T
17#
Jerry T Reply to 2015-10-09 18:04:52Z
 def num(s): """num(s) num(3),num(3.7)-->3 num('3')-->3, num('3.7')-->3.7 num('3,700')-->ValueError num('3a'),num('a3'),-->ValueError num('3e4') --> 30000.0 """ try: return int(s) except ValueError: try: return float(s) except ValueError: raise ValueError('argument is not a string of number') 
Peter Mortensen
18#
Peter Mortensen Reply to 2017-01-14 15:06:15Z
 Use: >>> str_float = "545.2222" >>> float(str_float) 545.2222 >>> type(_) # Check its type >>> str_int = "31" >>> int(str_int) 31 >>> type(_) # Check its type 
Peter Mortensen
19#
Peter Mortensen Reply to 2017-01-14 15:05:50Z
 Use: def num(s): try: for each in s: yield int(each) except ValueError: yield float(each) a = num(["123.55","345","44"]) print a.next() print a.next()  This is the most Pythonic way I could come up with.
Community
20#
Community Reply to 2017-05-23 11:55:19Z
 This is a corrected version of https://stackoverflow.com/a/33017514/5973334 This will try to parse a string and return either int or float depending on what the string represents. It might rise parsing exceptions or have some unexpected behaviour.  def get_int_or_float(v): number_as_float = float(v) number_as_int = int(number_as_float) return number_as_int if number_as_float == number_as_int else number_as_float 
Shameem
21#
Shameem Reply to 2017-07-09 08:27:39Z
 I use this function for that def parse_str(s): try: return eval(str(s)) except: return  It will convert the string to its type value = parse_str('1') # Returns Integer value = parse_str('1.5') # Returns Float 
Sławomir Lenart
22#
Sławomir Lenart Reply to 2017-07-12 17:01:41Z
 I am surprised nobody mentioned regex because sometimes string must be prepared and normalized before casting to number import re def parseNumber(value, as_int=False): try: number = float(re.sub('[^.\-\d]', '', value)) if as_int: return int(number + 0.5) else: return number except ValueError: return float('nan') # or None if you wish  usage: parseNumber('13,345') > 13345.0 parseNumber('- 123 000') > -123000.0 parseNumber('99999\n') > 99999.0  and by the way, something to verify you have a number: import numbers def is_number(value): return isinstance(value, numbers.Number) # will work with int, float, long, Decimal 
 You need to login account before you can post.
Processed in 0.383244 second(s) , Gzip On .