问题
I have a pandas dataframe containing the following data:
matchID server court speed
1 1 A 100
1 2 D 200
1 3 D 300
1 4 A 100
1 1 A 120
1 2 A 250
1 3 D 110
1 4 D 100
2 1 A 100
2 2 D 200
2 3 D 300
2 4 A 100
2 1 A 120
2 2 A 250
2 3 D 110
2 4 D 100
I would like to add two new columns containing the mean based on two conditions. The column meanSpeedCourtA13 shall contain the mean speed of servers 1 and 3 where court = A. This would bee (100 + 120) / 2 = 110. The second column named meanSpeedCourtD13 shall contain the mean speed of servers 1 and 3 where court = D. This would be (300 + 110) / 2 = 205.
Please note that this should be done for each matchID, hence, a groupby is also required. this means that solutions containing iloc() cannot be used.
The resulting dataframe should look as follows:
matchID server court speed meanSpeedCourtA13 meanSpeedCourtD13
1 1 A 100 110 205
1 2 D 200 110 205
1 3 D 300 110 205
1 4 A 100 110 205
1 1 A 120 110 205
1 2 A 250 110 205
1 3 D 110 110 205
1 4 D 100 110 205
2 1 A 100 110 205
2 2 D 200 110 205
2 3 D 300 110 205
2 4 A 100 110 205
2 1 A 120 110 205
2 2 A 250 110 205
2 3 D 110 110 205
2 4 D 100 110 205
回答1:
Ok this got a bit more complicated. Normally I'd try something with transform but I'd be glad if someone had something better than the following:
Use groupby and send df to func where df.loc is used, lastly use pd.concat to glue the dataframe together again:
import pandas as pd
data = {'matchID': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 2, 9: 2, 10: 2,
11: 2, 12: 2, 13: 2, 14: 2, 15: 2},
'court': {0: 'A', 1: 'D', 2: 'D', 3: 'A', 4: 'A', 5: 'A', 6: 'D', 7: 'D', 8: 'A',
9: 'D', 10: 'D', 11: 'A', 12: 'A', 13: 'A', 14: 'D', 15: 'D'},
'speed': {0: 100, 1: 200, 2: 300, 3: 100, 4: 120, 5: 250, 6: 110, 7: 100, 8: 100,
9: 200, 10: 300, 11: 100, 12: 120, 13: 250, 14: 110, 15: 100},
'server': {0: 1, 1: 2, 2: 3, 3: 4, 4: 1, 5: 2, 6: 3, 7: 4, 8: 1, 9: 2, 10: 3,
11: 4, 12: 1, 13: 2, 14: 3, 15: 4}}
df = pd.DataFrame(data)
def func(dfx):
dfx['meanSpeedCourtA13'],dfx['meanSpeedCourtD13'] = \
(dfx.loc[(dfx.server.isin((1,3))) & (dfx.court == 'A'),'speed'].mean(),
dfx.loc[(dfx.server.isin((1,3))) & (dfx.court == 'D'),'speed'].mean())
return dfx
newdf = pd.concat(func(dfx) for _, dfx in df.groupby('matchID'))
print(newdf)
Returns
court matchID server speed meanSpeedCourtA13 meanSpeedCourtD13
0 A 1 1 100 110.00 205.00
1 D 1 2 200 110.00 205.00
2 D 1 3 300 110.00 205.00
3 A 1 4 100 110.00 205.00
4 A 1 1 120 110.00 205.00
5 A 1 2 250 110.00 205.00
6 D 1 3 110 110.00 205.00
7 D 1 4 100 110.00 205.00
8 A 2 1 100 110.00 205.00
9 D 2 2 200 110.00 205.00
10 D 2 3 300 110.00 205.00
11 A 2 4 100 110.00 205.00
12 A 2 1 120 110.00 205.00
13 A 2 2 250 110.00 205.00
14 D 2 3 110 110.00 205.00
15 D 2 4 100 110.00 205.00
回答2:
You can get the mean by groupby and assign the values by getting the item() i.e
vals = df[df['server'].isin([1,3])].groupby(['court'])['speed'].mean().to_frame()
df['A13'],df['D13'] = vals.query("court=='A'")['speed'].item(), vals.query("court=='D'")['speed'].item()
matchID server court speed A13 D13
0 1 1 A 100 110.0 205.0
1 1 2 D 200 110.0 205.0
2 1 3 D 300 110.0 205.0
3 1 4 A 100 110.0 205.0
4 1 1 A 120 110.0 205.0
5 1 2 A 250 110.0 205.0
6 1 3 D 110 110.0 205.0
7 1 4 D 100 110.0 205.0
8 2 1 A 100 110.0 205.0
9 2 2 D 200 110.0 205.0
10 2 3 D 300 110.0 205.0
11 2 4 A 100 110.0 205.0
12 2 1 A 120 110.0 205.0
13 2 2 A 250 110.0 205.0
14 2 3 D 110 110.0 205.0
15 2 4 D 100 110.0 205.0
回答3:
With groupby, we can still use loc to select the intended parts that we want to replace but put the whole computation within a for loop from df.groupby("matchID").
for id, subg in df.groupby("matchID"):
df.loc[df.matchID==id, "meanSpeedCourtA13"] = (subg
.where(subg.server.isin([1,3])).where(subg.court == "A").speed.mean())
df.loc[df.matchID==id, "meanSpeedCourtD13"] = (subg
.where(subg.server.isin([1,3])).where(subg.court == "D").speed.mean())
Specail thanks to @Dark to point it out that I was hard coding groupby.
For loc, it can be used to select values based on information from 2 axes: rows and columns. By convention on the documentation, the sequence to put information is rows first and columns second. For example, in df.loc[df.matchID==id, "meanSpeedCourtD13"], df.matchID==id is about selecting rows that have matchID being id and that "meanSpeedCourtD13" specifies a column we want to look into.
Side notes about calculating mean:
- for each group
subg where(subg.server.isin([1,3]))then filter out server not in [1 ,3].where(subg.court == "A")further to do filtering on court.- finally call
meanto compute mean from speed.
As an alternative, you can use np.where to assign values to each matchID in [1, 2]. This works only for binary matchID. It is roughly the same speed with the groupby method above tested on my computer. To save space, we only demonstrate with "meanSpeedCourtA13" column.
# First we calculate the means
# Calculate mean for Group with mathcID being 1
meanSpeedCourtA13_ID1 = (df[df.matchID==1].
where(df.server.isin([1,3])).where(df.court == "A").speed.mean())
# Calculate mean for Group with matchID being 2
meanSpeedCourtA13_ID2 = (df[df.matchID==2].
where(df.server.isin([1,3])).where(df.court == "A").speed.mean())
# Use np.where to allocate values to each matchID in [1, 2]
df["meanSpeedCourtA13"] = np.where(df.matchID == 1,
meanSpeedCourtA13_ID1, meanSpeedCourtA13_ID2)
For np.where(condition, x, y), it will return x if condition is met, y otherwise. See np.where for documentation.
来源：https://stackoverflow.com/questions/48038958/python-pandas-average-based-on-condition-into-new-column