python integer between two values

Question

I found a thread on this but am not having any luck getting it working.

min = 9
max = 10
a = ['8','9','10','11']

for x in a:
    if max >= x > min:
        print 'one'
    else:
        d = (max >= x > min)
        print d, x
    if (x > min >= max):
   #if (min < x >= max):
        print x
    else:
        print x, ' is equal to or greater than', max

Outputs:

False 8
8  is equal to or greater than 10
False 9
9  is equal to or greater than 10
False 10
10  is equal to or greater than 10
False 11
11  is equal to or greater than 10

this thread working code? indicates syntax needs to be:

if 10000 <= number <= 30000:
pass

I've tried every combination of signs I can think of, and the returns are always True or False for all, which is wrong.

I've also tried this (much longer) code:

min = 9
max = 10
a = ['8','9','10','11']

for x in a:
    print 'X is:', x
    if int(x) == max:
        print 'max found:', x
    elif int(x) < max:
        if int(x) > min:
            print 'min:', x
    elif int(x) < min:
        print 'under range', x
    else:
        print 'out of range', x

with output also unexpected, since I expect to catch all cases:

X is: 8
X is: 9
X is: 10
max found: 10
X is: 11
out of range 11

Ugh! How can I check all items 'properly' and return under, at, over my min, max?

There aren't any integers that are > min and < max. Perhaps you meant `>=` and `<=`? — Bryan Oakley, Jan 20 '16 at 17:12
wait, you want a integer between 9 and 10?? you know that there is none, right? — Copperfield, Jan 20 '16 at 17:13
Doh. I had min set to 8 before I started pulling hair out. Thx! — shawn, Jan 20 '16 at 17:14

mementum · Accepted Answer · 2016-01-20T17:29:15.460

You need to compare int, int or str, str. Past that I would recommend bisect which gives you insertion points.

For example:

import bisect

minint, maxint = 9, 10
a = ['8','9','10','11']

b = [int(x) for x in a]

minidx = bisect.bisect_left(b, minint)
maxidx = bisect.bisect_left(b, maxint)

print('below:', a[0:minidx])
print('between:', a[minidx:maxidx])
print('above:', a[maxidx:])

which outputs:

below: ['8']
between: ['9']
above: ['10', '11']

Depending on how you define the comparison <= or < at each of the ends you need to apply either bisect_left or bisect_right

If we change the first bisect to right:

import bisect

minint, maxint = 9, 10
a = ['8','9','10','11']

b = [int(x) for x in a]

minidx = bisect.bisect_right(b, minint)
maxidx = bisect.bisect_left(b, maxint)

print('below:', a[0:minidx])
print('between:', a[minidx:maxidx])
print('above:', a[maxidx:])

The output changes and shows that nothing is in between 9 and 10

below: ['8', '9']
between: []
above: ['10', '11']

The docs (for 2.7): https://docs.python.org/2/library/bisect.html

Apart from the type error in my first code chunk - it turns out I don't actually need to check within, only case A or B. This code does what it's supposed to do: if len(rec) == max_rec_len: authored.append(rec) elif len(rec) == min_rec_len: records.append(rec) — shawn, Jan 20 '16 at 17:28

python integer between two values

1 Answers1