'Filter dataframe with multiple conditions including OR
I wrote a little script that loops through constraints to filter a dataframe. Example and follow up explaining the issue are below.
constraints = [['stand','==','L'],['zone','<','20']]
for x in constraints:
vari = x[2]
df = df.query("{0} {1} @vari".format(x[0],x[1]))
| zone | stand | speed | type | |
|---|---|---|---|---|
| 0 | 2 | L | 83.7 | CH |
| 1 | 7 | L | 95.9 | SI |
| 2 | 14 | L | 94.9 | FS |
| 3 | 11 | L | 93.3 | FS |
| 4 | 13 | L | 86.9 | CH |
| 5 | 7 | L | 96.4 | SI |
| 6 | 13 | L | 82.6 | SL |
I can't figure out a way to filter when there is an OR condition. For example, in the table above I'd like to return a dataframe using the constraints in the code example along with any rows that contain SI or CH in the type column. Does anyone have ideas on how to accomplish this? Any help would be greatly appreciated.
Solution 1:[1]
This seems to have gotten the job done but there is probably a much better way of going about it.
for x in constraints:
vari = x[2]
if isinstance(vari,list):
frame = frame[frame[x[0]].isin(vari)]
else:
frame = frame.query("{0} {1} @vari".format(x[0],x[1]))
Solution 2:[2]
IIUC (see my question in the comment) you can do it like this:
Made a little different df to show you the result (I guess the table you show is already filtered)
df = pd.DataFrame(
{'zone': {0: 2, 1: 11, 2: 25, 3: 11, 4: 23, 5: 7, 6: 13},
'stand': {0: 'L', 1: 'L', 2: 'L', 3: 'C', 4: 'L', 5: 'K', 6: 'L'},
'speed': {0: 83.7, 1: 95.9, 2: 94.9, 3: 93.3, 4: 86.9, 5: 96.4, 6: 82.6},
'type': {0: 'CH', 1: 'SI', 2: 'FS', 3: 'FS', 4: 'CH', 5: 'SI', 6: 'SL'}})
print(df)
zone stand speed type
0 2 L 83.7 CH
1 11 L 95.9 SI
2 25 L 94.9 FS
3 11 C 93.3 FS
4 23 L 86.9 CH
5 7 K 96.4 SI
6 13 L 82.6 SL
res = df.loc[ ( (df['type']=='SI') | (df['type']=='CH') ) & ( (df['zone']<20) & (df['stand']=='L') ) ]
print(res)
zone stand speed type
0 2 L 83.7 CH
1 11 L 95.9 SI
Let me know if that is what you are searching for.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Nick |
| Solution 2 | Rabinzel |
