Pandas how to filter dataframe with certain range of numbers in a column of dataframe

pd.cut works fine here, especially as your intervals are not overlapping:

intervals = pd.IntervalIndex.from_tuples([(200, 230), (1000, 1150), (350, 370), (100, 130)])

# if the values do not fall within the intervals, it is a null
# hence the isna check to keep only the null matches
# thanks to @corralien for the include_lowest=True suggestion

data_df.loc[pd.cut(data_df.sbin, intervals, include_lowest=True).isna()]

   partname  sbin
0     part1   234
7     part8  1457
8     part9   789
9    part10  1367
10   part11   467

New version

Use np.logical_and and any to select values in ranges and invert the mask to keep other ones.

intervals = [(100, 130), (200, 230), (350, 370), (1000, 1150)]
m = np.any([np.logical_and(data_df['sbin'] >= l, data_df['sbin'] <= u)
                                    for l, u in intervals], axis=0)
out = data_df.loc[~m]

Note any can be replaced by np.logical_or.reduce:

intervals = [(100, 130), (200, 230), (350, 370), (1000, 1150)]
m = np.logical_or.reduce([np.logical_and(data_df['sbin'] >= l, data_df['sbin'] <= u)
                                    for l, u in intervals])
out = data_df.loc[~m]

Output result:

>>> out
   partname  sbin
0     part1   234
7     part8  1457
8     part9   789
9    part10  1367
10   part11   467

Old version

Not work with float numbers as is

Use np.where and in1d:

intervals = [(100, 130), (200, 230), (350, 370), (1000, 1150)]
m = np.hstack([np.arange(l, u+1) for l, u in intervals])
out = data_df.loc[~np.in1d(data_df['sbin'], m)]

Performance: for 100k records:

data_df = pd.DataFrame({'sbin': np.random.randint(0, 2000, 100000)})

def exclude_range_danimesejo():
    intervals = sorted([(200, 230), (1000, 1150), (350, 370), (100, 130)])
    intervals = np.array(intervals).flatten()
    mask = (np.searchsorted(intervals, data_df['sbin']) % 2 == 0) & ~np.in1d(data_df['sbin'], intervals[::2])
    return data_df.loc[mask]

def exclude_range_sammywemmy():
    intervals = pd.IntervalIndex.from_tuples([(200, 230), (1000, 1150), (350, 370), (100, 130)])
    return data_df.loc[pd.cut(data_df.sbin, intervals, include_lowest=True).isna()]

def exclude_range_corralien():
    intervals = [(100, 130), (200, 230), (350, 370), (1000, 1150)]
    m = np.hstack([np.arange(l, u+1) for l, u in intervals])
    return data_df.loc[~np.in1d(data_df['sbin'], m)]

def exclude_range_corralien2():
    intervals = [(100, 130), (200, 230), (350, 370), (1000, 1150)]
    m = np.any([np.logical_and(data_df['sbin'] >= l, data_df['sbin'] <= u)
                                        for l, u in intervals], axis=0)
    return data_df.loc[~m]

>>> %timeit exclude_range_danimesejo()
2.66 ms ± 18.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

>>> %timeit exclude_range_sammywemmy()
63.6 ms ± 549 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

>>> %timeit exclude_range_corralien()
6.87 ms ± 58.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

>>> %timeit exclude_range_corralien2()
2.26 ms ± 8.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

For non-overlapping intervals, you could use np.searchsorted:

# sort the non overlapping intervals
intervals = sorted([(200, 230), (1000, 1150), (350, 370), (100, 130)])

# flatten
intervals = np.array(intervals).flatten()

# search in the intervals, if the index is even is not in the intervals
mask = (np.searchsorted(intervals, data_df['sbin']) % 2 == 0) & ~np.in1d(data_df['sbin'], intervals[::2])

print(data_df.loc[mask])

Output

   partname  sbin
0     part1   234
7     part8  1457
8     part9   789
9    part10  1367
10   part11   467

Timings (on a 3,1 GHz Intel Core i7)

%timeit exclude_range_danimesejo()
3.45 ms ± 13.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Setup (in addition to the one by @Corralien)

def exclude_range_danimesejo():
    intervals = sorted([(200, 230), (1000, 1150), (350, 370), (100, 130)])
    intervals = np.array(intervals).flatten()
    mask = (np.searchsorted(intervals, data_df['sbin']) % 2 == 0) & ~np.in1d(data_df['sbin'], intervals[::2])
    return data_df.loc[mask]

The overall complexity of this approach is O((n + m) log n) where n is the size of the list of intervals and m the length of the sbin column.

What about using a loop to apply your interval conditions?

# this will be broadcast to a boolean array where it is `True` if sbin
# is in any of the intervals in the list and `False` otherwise
is_in_intervals = False
intervals = [(200, 230), (1000, 1150), (350, 370), (100, 130)]

for interval in intervals:
    is_in_intervals |= (data_df.sbin >= interval[0]) & (data_df.sbin <= interval[1])

data_df[~is_in_intervals]