Efficient way to count number of int and float in string using Python

Question

Is there any efficient way to find count of int and float in string, without using two different list comprehension

st = '2.00 00 DBEL 215 Frox Lyxo 2.000 2.00'

reg = r'\d+(?:\.\d*)?'

out = re.compile(reg).findall(st)

# output as  ['2.00', '00', '215', '2.000', '2.00']

int_in_string = len([i for i in out if isinstance(eval(str(i)), int)])
fl_in_string = len([i for i in out if isinstance(eval(str(i)), float)])

print('no of int : ', int_in_string, 'no of float : ', fl_in_string)

And the usage of `eval` is by no means recommended as it is not *safe*. — Ma0, Aug 09 '18 at 08:22
@Ev.Kounis I know but without eval it does not match string is int or float. — utks009, Aug 09 '18 at 08:23
Maybe [this question](https://stackoverflow.com/questions/354038/how-do-i-check-if-a-string-is-a-number-float) will help you, you just have to implement another twin function for int. It is based on [handling exceptions](https://docs.python.org/3/tutorial/errors.html). — Cezar Cobuz, Aug 09 '18 at 08:26

score 2 · Accepted Answer · answered Aug 09 '18 at 08:25

Take a look at the following:

st = '2.00 00 DBEL 215 Frox Lyxo 2.000 2.00'

floats, ints = 0, 0
for word in st.split():
    try:
        int(word)
    except ValueError:
        try:
            float(word)
        except ValueError:
            pass
        else:
            floats += 1
    else:
        ints += 1

print('no of int : ', ints, 'no of float : ', floats)  # no of int :  2 no of float :  3

My code takes a different approach. regex is not used. The string is split on whitespace and the individual words are cast to float and int inside try-blocks. If the tries are successful (else-part of the block), a counter is increased.

score 0 · Answer 2 · answered Aug 09 '18 at 08:27

You don't need to actually save the numbers again on int_in_string and fl_in_string, so you could use two simple counters

int_in_string = 0
fl_in_string   = 0
for i in out:
    if isinstance(eval(str(i)), int):
        int_in_string += 1
    elif isinstance(eval(str(i)), float):
        fl_in_string += 1

score 0 · Answer 3 · answered Aug 09 '18 at 08:35

If all your ints follow the pattern \d+ and all your floats follow the pattern \d+\.\d+, you can count them as follows

>>> num_ints = sum(1 for _ in re.finditer(r'(?<=\W)\d+(?=\W)', st))
>>> num_floats = sum(1 for _ in re.finditer(r'(?<=\W)\d+\.\d+(?=\W)', st))
>>> print (num_ints, num_floats)
6 1

score 0 · Answer 4 · answered Aug 09 '18 at 09:14

0

Assuming regex is NOT a must.

st = '2.00 00 DBEL 215 Frox Lyxo 2.000 2.00'
s = [i.isdigit() for i in st.split(' ') if not(i.isalpha())]
print('no of int : ', s.count(True), 'no of float : ', s.count(False))

answered Aug 09 '18 at 09:14

LChiu

1
1

Efficient way to count number of int and float in string using Python

4 Answers4