user2958395 - 5 months ago 15

Python Question

If there is a file with ranges sorted by the first column (no overlap of ranges):

`1 10`

12 15

18 19

And another, sorted by the first column (can have overlaps):

`1 5`

2 10

12 13

13 20

I would like to determine for each line (range)in the second file, if this line(range) intersects with any of the ranges in the first file. I did the following so far

`df_1 = pd.read_csv('range1.txt',sep=' ')`

df_2 = pd.read_csv('range2.txt',sep=' ')

for i in xrange(len(df_1)):

start_1 = df_1.iloc[i,0]

stop_1 = df_1.iloc[i, 1]

for j in xrange(len(df_2)):

start_2 = df_2.iloc[j,0]

stop_2 = df_2.iloc[j, 1]

if start_2 > stop_1:

break

elif stop_2 < start_1:

continue

else:

# add ranges from second file to list

This I know can be terribly inefficient, so I was wondering if there is a more computationally efficient/faster way to solve this.

Answer

@Olivier Pellier-Cuit has provided a link to fast overlap test. If you need membership check instead of overlap test, use this algorithm.

So using this algorithm we can do the following:

```
df1['m'] = (df1.a + df1.b)
df1['d'] = (df1.b - df1.a)
df2['m'] = (df2.a + df2.b)
df2['d'] = (df2.b - df2.a)
df2[['m','d']].apply(lambda x: (np.abs(df1.m - x.m) < df1.d +x.d).any(), axis=1)
```

PS i've slightly simplified the calculations of `m`

and `d`

by getting rid of `division by 2`

, because it can be done eliminating common terms.

Output:

```
In [105]: df2[['m','d']].apply(lambda x: (np.abs(df1.m - x.m) < df1.d +x.d).any(), axis=1)
Out[105]:
0 True
1 True
2 True
3 True
4 False
dtype: bool
```

setup:

```
df1 = pd.read_csv(io.StringIO("""
a b
1 10
12 15
18 19
"""), delim_whitespace=True)
df2 = pd.read_csv(io.StringIO("""
a b
1 5
2 10
12 13
13 20
50 60
"""), delim_whitespace=True)
```

NOTE: i've intentionally added a pair (50, 60) to the DF2, which doesn't overlap with any interval from DF1

Data frames with calculated `m`

and `d`

columns:

```
In [106]: df1
Out[106]:
a b m d
0 1 10 11 9
1 12 15 27 3
2 18 19 37 1
In [107]: df2
Out[107]:
a b m d
0 1 5 6 4
1 2 10 12 8
2 12 13 25 1
3 13 20 33 7
4 50 60 110 10
```