I am using the split-apply-combine pattern in pandas to group my df by a custom aggregation function.
But this returns an undesired DataFrame with the grouped column existing twice: In an MultiIndex and the columns.
The following is a simplified example of my problem.
Say, I have this df
df = pd.DataFrame([[1,2],[3,4],[1,5]], columns=['A','B']))
A B
0 1 2
1 3 4
2 1 5
I want to group by column A and keep only those rows where B has an even value. Thus the desired df is this:
B
A
1 2
3 4
The custom function my_combine_func should do the filtering. But applying it after a groupby, leads to an MultiIndex with the former Index in the second level. And thus column A existing two times.
my_combine_func = group[group['B'] % 2 == 0]
df.groupby(['A']).apply(my_combine_func)
A B
A
1 0 1 2
3 1 3 4
How to apply a custom group function and have the desired df?
It's easier to use apply here so you get a boolean array back:
df[df.groupby('A')['B'].apply(lambda x: x % 2 == 0)]
A B
0 1 2
1 3 4
Related
I have the case where I want to sanity check labeled data. I have hundreds of features and want to find points which have the same features but different label. These found cluster of disagreeing labels should then be numbered and put into a new dataframe.
This isn't hard but I am wondering what the most elegant solution for this is.
Here an example:
import pandas as pd
df = pd.DataFrame({
"feature_1" : [0,0,0,4,4,2],
"feature_2" : [0,5,5,1,1,3],
"label" : ["A","A","B","B","D","A"]
})
result_df = pd.DataFrame({
"cluster_index" : [0,0,1,1],
"feature_1" : [0,0,4,4],
"feature_2" : [5,5,1,1],
"label" : ["A","B","B","D"]
})
In order to get the output you want (both de-duplication and cluster_index), you can use a groupby approach:
g = df.groupby(['feature_1', 'feature_2'])['label']
(df.assign(cluster_index=g.ngroup()) # get group name
.loc[g.transform('size').gt(1)] # filter the non-duplicates
# line below only to have a nice cluster_index range (0,1…)
.assign(cluster_index= lambda d: d['cluster_index'].factorize()[0])
)
output:
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
First get all duplicated values per feature columns and then if necessary remove duplciated by all columns (here in sample data not necessary), last add GroupBy.ngroup for groups indices:
df = df[df.duplicated(['feature_1','feature_2'],keep=False)].drop_duplicates()
df['cluster_index'] = df.groupby(['feature_1', 'feature_2'])['label'].ngroup()
print (df)
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
My problem is similar to split a dataframe into chunks of N rows problem, expect that the number of rows in each chunk will be different. I have a datafame as such:
A
B
C
1
2
0
1
2
1
1
2
2
1
2
0
1
2
1
1
2
2
1
2
3
1
2
4
1
2
0
A and B are just whatever don't pay attention. Column C though starts at 0 and increments with each row until it suddenly resets to 0. So in the dataframe included the first 3 rows are a new dataframe, then the next 5 are a second new dataframe, and this continues as my dataframe adds more and more rows.
To finish off the question,
df = [x for _, x in df.groupby(df['C'].eq(0).cumsum())]
allows me to group all the subgroups and then with this groupby I can select each subgroups as a separate dataframe.
I am removing consecutive duplicates in groups in a dataframe. I am looking for a faster way than this:
def remove_consecutive_dupes(subdf):
dupe_ids = [ "A", "B" ]
is_duped = (subdf[dupe_ids].shift(-1) == subdf[dupe_ids]).all(axis=1)
subdf = subdf[~is_duped]
return subdf
# dataframe with columns key, A, B
df.groupby("key").apply(remove_consecutive_dupes).reset_index()
Is it possible to remove these without grouping first? Applying the above function to each group individually takes a lot of time, especially if the group count is like half the row count. Is there a way to do this operation on the entire dataframe at once?
A simple example for the algorithm if the above was not clear:
input:
key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5
5 x 1 2
output:
key A B
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
5 x 1 2
Row 2 was dropped because A=1 B=2 was also the previous row in group x.
Row 5 will not be dropped because it is not a consecutive duplicate in group x.
According to your code, you drop only lines if they appear below each other if
they are grouped by the key. So rows with another key inbetween do not influence this logic. But doing this, you want to preserve the original order of the records.
I guess the biggest influence in the runtime is the call of your function and
possibly not the grouping itself.
If you want to avoid this, you can try the following approach:
# create a column to restore the original order of the dataframe
df.reset_index(drop=True, inplace=True)
df.reset_index(drop=False, inplace=True)
df.columns= ['original_order'] + list(df.columns[1:])
# add a group column, that contains consecutive numbers if
# two consecutive rows differ in at least one of the columns
# key, A, B
compare_columns= ['key', 'A', 'B']
df.sort_values(['key', 'original_order'], inplace=True)
df['group']= (df[compare_columns] != df[compare_columns].shift(1)).any(axis=1).cumsum()
df.drop_duplicates(['group'], keep='first', inplace=True)
df.drop(columns=['group'], inplace=True)
# now just restore the original index and it's order
df.set_index('original_order', inplace=True)
df.sort_index(inplace=True)
df
Testing this, results in:
key A B
original_order
0 x 1 2
1 y 1 4
3 x 1 4
4 y 2 5
If you don't like the index name above (original_order), you just need to add the following line to remove it:
df.index.name= None
Testdata:
from io import StringIO
infile= StringIO(
""" key A B
0 x 1 2
1 y 1 4
2 x 1 2
3 x 1 4
4 y 2 5"""
)
df= pd.read_csv(infile, sep='\s+') #.set_index('Date')
df
I have this dataframe and I need to drop all duplicates but I need to keep first AND last values
For example:
1 0
2 0
3 0
4 0
output:
1 0
4 0
I tried df.column.drop_duplicates(keep=("first","last")) but it doesn't word, it returns
ValueError: keep must be either "first", "last" or False
Does anyone know any turn around for this?
Thanks
You could use the panda's concat function to create a dataframe with both the first and last values.
pd.concat([
df['X'].drop_duplicates(keep='first'),
df['X'].drop_duplicates(keep='last'),
])
you can't drop both first and last... so trick is too concat data frames of first and last.
When you concat one has to handle creating duplicate of non-duplicates. So only concat unique indexes in 2nd Dataframe. (not sure if Merge/Join would work better?)
import pandas as pd
d = {1:0,2:0,10:1, 3:0,4:0}
df = pd.DataFrame.from_dict(d, orient='index', columns=['cnt'])
print(df)
cnt
1 0
2 0
10 1
3 0
4 0
Then do this:
d1 = df.drop_duplicates(keep=("first"))
d2 = df.drop_duplicates(keep=("last"))
d3 = pd.concat([d1,d2.loc[set(d2.index) - set(d1.index)]])
d3
Out[60]:
cnt
1 0
10 1
4 0
Use a groupby on your column named column, then reindex. If you ever want to check for duplicate values in more than one column, you can extend the columns you include in your groupby.
df = pd.DataFrame({'column':[0,0,0,0]})
Input:
column
0 0
1 0
2 0
3 0
df.groupby('column', as_index=False).apply(lambda x: x if len(x)==1 else x.iloc[[0, -1]]).reset_index(level=0, drop=True)
Output:
column
0 0
3 0
I have a dataframe
A B C
1 2 3
2 3 4
3 8 7
I want to take only rows where there is a sequence of 3,4 in columns C (in this scenario - first two rows)
What will be the best way to do so?
You can use rolling for general solution working with any pattern:
pat = np.asarray([3,4])
N = len(pat)
mask= (df['C'].rolling(window=N , min_periods=N)
.apply(lambda x: (x==pat).all(), raw=True)
.mask(lambda x: x == 0)
.bfill(limit=N-1)
.fillna(0)
.astype(bool))
df = df[mask]
print (df)
A B C
0 1 2 3
1 2 3 4
Explanation:
use rolling.apply and test pattern
replace 0s to NaNs by mask
use bfill with limit for filling first NANs values by last previous one
fillna NaNs to 0
last cast to bool by astype
Use shift
In [1085]: s = df.eq(3).any(1) & df.shift(-1).eq(4).any(1)
In [1086]: df[s | s.shift()]
Out[1086]:
A B C
0 1 2 3
1 2 3 4