I have a DataFrame which contains first and last number of some intervals. how can I get how many times each number was in each interval?

try:

df1 = pd.DataFrame(data=[i for i in range(1,98)])

df:

    0   1       2
0   0   1.0     25.0
1   1   1.0     31.0
2   2   2.0     97.0
3   3   1.0     25.0
4   4   1.0     26.0

df1.head():

    0
0   1
1   2
2   3
3   4
4   5

res = df1[0].apply(lambda x: sum((df[1]<=x) & (df[2]>=x)))

res:

0     4
1     5
2     5
3     5
4     5
     ..
92    1
93    1
94    1
95    1
96    1
Name: 0, Length: 97, dtype: int64

You can form ranges per row and then explode them. Counting the values gives the final result:

result = (pd.Series(np.arange(first, second+1) for first, second in df.to_numpy())
                   .explode()
                   .value_counts(sort=False))

to get

>>> result

1.0     4
2.0     5
3.0     5
4.0     5
5.0     5
       ..
93.0    1
94.0    1
95.0    1
96.0    1
97.0    1

This won't necessarily include all the values in 1..97 because if a number falls out of every interval, it won't be counted. To guarantee an index of 1..97, we can reindex with the min and max values (i.e., 1 and 97 here) and put 0 to those that didn't appear:

values = df.to_numpy()
min_, max_ = values.min(), values.max()

result = result.reindex(np.arange(min_, max_+1), fill_value=0)

and final note is the np.aranges can be replaced with range if the values in the frame are integers and therefore df = df.astype(int) loses no information. If not, np.arange is needed. Also, np.arange encapsulates range so it can be used in either case.

try:

df1 = df.groupby(['ColumnName']).count()

This will return the column you choose as the index and get a count of the matching intervals

First we need to know how many intervals open and close at each value:

>>> df
    col_1 col_2
0     1.0  25.0
1.0   1.0  31.0
2.0   2.0  97.0
3.0   1.0  25.0
4.0   1.0  26.0
>>> idx = pd.RangeIndex(1, 98)
>>> opencount = df['col_1'].value_counts().reindex(idx, fill_value=0)
>>> closecount = df['col_2'].value_counts().reindex(idx, fill_value=0)
>>> opencount
1     3
2     1
3     0
4     0
5     0
 ..
93    0
94    0
95    0
96    0
97    0
Name: col_1, Length: 97, dtype: int64
>>> closecount
1     0
2     0
3     0
4     0
5     0
 ..
93    0
94    0
95    0
96    0
97    1
Name: col_2, Length: 97, dtype: int64

Note that we used reindex to add zeros at all the values not available in col_1 and col_2.

If in fact the end of the interval is contained in the interval (as per comments) you can simply shift the closecount down by 1:

>>> closecount = closecount.shift(fill_value=0)
>>> closecount
1     0
2     0
3     0
4     0
5     0
     ..
93    0
94    0
95    0
96    0
97    0
Name: col_2, Length: 97, dtype: int64

Then we can compute the number of intervals at each point as the sum of intervals having opened before, minus the sum of intervals having closed before. This can be done with cumsum

>>> opencount.cumsum() - closecount.cumsum()
1     4
2     5
3     5
4     5
5     5
     ..
93    1
94    1
95    1
96    1
97    1
Length: 97, dtype: int64

You can do as following. First create a new dataframe with a column from 1 to 97

>>> df2 = pd.DataFrame(list(range(1,98,1)), dtype=float, columns=["range"])
>>> df2
    range
0     1.0
1     2.0
2     3.0
3     4.0
4     5.0
..    ...
92   93.0
93   94.0
94   95.0
95   96.0
96   97.0

To get counts of values from your original dataframe

>>> dfg = df.groupby('val').agg('count')
>>> dfg
     range
val
1.0      4
2.0      1

Now map from your df, and at the same time fill missing values with zeros. Then cast to integer to match OP's expected format.

>>> df2["count"] = df2["range"].map(dfg['range']).fillna(0)
>>> df2 = df2.astype(int)
>>> df2
    range  count
0       1      4
1       2      1
2       3      0
3       4      0
4       5      0
..    ...    ...
92     93      0
93     94      0
94     95      0
95     96      0
96     97      0

And if needed, cast to integer