问题
Given a string, "Hello4.2this.is random 24 text42", I want to return all ints or floats, [4.2, 24, 42]. All the other questions have solutions that return just 24. I want to return a float even if non-digit characters are next to the number. Since I am new to Python, I am trying to avoid regex or other complicated imports. I have no idea how to start. Please help. Here are some research attempts: Python: Extract numbers from a string, this didn't work since it doesn't recognize 4.2 and 42. There are other questions like the one mentioned, none of which sadly recognize 4.2 and 42.
回答1:
A regex from perldoc perlretut:
import re
re_float = re.compile("""(?x)
^
[+-]?\ * # first, match an optional sign *and space*
( # then match integers or f.p. mantissas:
\d+ # start out with a ...
(
\.\d* # mantissa of the form a.b or a.
)? # ? takes care of integers of the form a
|\.\d+ # mantissa of the form .b
)
([eE][+-]?\d+)? # finally, optionally match an exponent
$""")
m = re_float.match("4.5")
print m.group(0)
# -> 4.5
To get all numbers from a string:
str = "4.5 foo 123 abc .123"
print re.findall(r"[+-]? *(?:\d+(?:\.\d*)?|\.\d+)(?:[eE][+-]?\d+)?", str)
# -> ['4.5', ' 123', ' .123']
回答2:
Using regular expressions is likely to give you the most concise code for this problem. It is hard to beat the conciseness of
re.findall(r"[+-]? *(?:\d+(?:\.\d*)?|\.\d+)(?:[eE][+-]?\d+)?", str)
from pythad's answer.
However, you say "I am trying to avoid regex", so here's a solution that does not use regular expressions. It is obviously a bit longer than a solution using a regular expression (and probably much slower), but it is not complicated.
The code loops through the input character by character.
As it pulls each character from the string, it appends it to current (a string that holds the number currently being parsed) if appending it still maintains a valid number. When it encounters a character that cannot be appended to current, current is saved to a list of numbers, but only if current itself isn't one of '', '.', '-' or '-.'; these are strings that could potentially begin a number but are not themselves valid numbers.
When current is saved, a trailing 'e', 'e-' or 'e+' is removed. That will happen with a string such as '1.23eA'. While parsing that string, current will eventually become '1.23e', but then 'A' is encountered, which means the string does not contain a valid exponential part, so the 'e' is discarded.
After saving current, it is reset. Usually current is reset to '', but when the character that triggered current to be saved was '.' or '-', current is set to that character, because those characters could be the beginning of a new number.
Here's the function extract_numbers(s). The line before return numbers converts the list of strings to a list of integers and floating point values. If you want just the strings, remove that line.
def extract_numbers(s):
"""
Extract numbers from a string.
Examples
--------
>>> extract_numbers("Hello4.2this.is random 24 text42")
[4.2, 24, 42]
>>> extract_numbers("2.3+45-99")
[2.3, 45, -99]
>>> extract_numbers("Avogadro's number, 6.022e23, is greater than 1 million.")
[6.022e+23, 1]
"""
numbers = []
current = ''
for c in s.lower() + '!':
if (c.isdigit() or
(c == 'e' and ('e' not in current) and (current not in ['', '.', '-', '-.'])) or
(c == '.' and ('e' not in current) and ('.' not in current)) or
(c == '+' and current.endswith('e')) or
(c == '-' and ((current == '') or current.endswith('e')))):
current += c
else:
if current not in ['', '.', '-', '-.']:
if current.endswith('e'):
current = current[:-1]
elif current.endswith('e-') or current.endswith('e+'):
current = current[:-2]
numbers.append(current)
if c == '.' or c == '-':
current = c
else:
current = ''
# Convert from strings to actual python numbers.
numbers = [float(t) if ('.' in t or 'e' in t) else int(t) for t in numbers]
return numbers
来源：https://stackoverflow.com/questions/45001775/find-all-floats-or-ints-in-a-given-string