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.
Related
I have a csv with around 8 million of rows, something like that:
a b c
0 2 3
and I wanted to generate from it new rows based on the second and the third value so I will get:
a b c
0 2 3
0 3 3
0 4 3
0 5 3
which is basically just itereating through every row(in this example one row), and then creating a new row with a value of b+i, where i is between 0 to the value of c including c itself.
c column is irelevant after the rows have been generated, problem is that it has million of rows, and doing that might generate many rows, so how can I do it efficenly? (loops are too slow for that amount of data).
thanks
You can reindex on the repeated index:
out = df.loc[df.index.repeat(df['c']+1)]
out['b'] += out.groupby(level=0).cumcount()
print(out)
Output (reset index if you want):
a b c
0 0 2 3
0 0 3 3
0 0 4 3
0 0 5 3
Note since you blow your data up by the c column and you already have 8 million rows, your new dataframe can be too big on its own.
I have the following dataframe:
d = {'value': [1,1,1,1,1,1,1,1,1,1], 'flag_1': [0,1,0,1,1,1,0,1,1,1],'flag_2':[1,0,1,1,1,1,1,0,1,1],'index':[1,2,3,4,5,6,7,8,9,10]}
df = pd.DataFrame(data=d)
I need to perform the following filter on it:
If flag 1 and flag 2 are equal keep the row with the maximum index from the consecutive indices. Below for rows 4,5,6 and rows 9,10 flag 1 and flag 2 are equal. From the group of consecutive indices 4,5,6 therefore I wish to keep only row 6 and drop rows 4 and 5. For the next group of rows 9 and 10 I wish to keep only row 10. The rows where flag 1 and 2 are not equal should all be retained. I want my final output to look as shown below:
I am really not sure how to achieve what is required so I would be grateful for any advice on how to do it.
IIUC, you can compare consecutive rows with shift. This solution requires a sorted index.
In [5]: df[~df[['flag_1', 'flag_2']].eq(df[['flag_1', 'flag_2']].shift(-1)).all(axis=1)]
Out[5]:
value flag_1 flag_2 index
0 1 0 1 1
1 1 1 0 2
2 1 0 1 3
5 1 1 1 6
6 1 0 1 7
7 1 1 0 8
9 1 1 1 10
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
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 a super strange problem which I spent the last hour trying to solve, but with no success. It is even more strange since I can't replicate it on a small scale.
I have a large DataFrame (150,000 entries). I took out a subset of it and did some manipulation. the subset was saved as a different variable, x.
x is smaller than the df, but its index is in the same range as the df. I'm now trying to assign x back to the DataFrame replacing values in the same column:
rep_Callers['true_vpID'] = x.true_vpID
This inserts all the different values in x to the right place in df, but instead of keeping the df.true_vpID values that are not in x, it is filling them with NaNs. So I tried a different approach:
df.ix[x.index,'true_vpID'] = x.true_vpID
But instead of filling x values in the right place in df, the df.true_vpID gets filled with the first value of x and only it! I changed the first value of x several times to make sure this is indeed what is happening, and it is. I tried to replicate it on a small scale but it didn't work:
df = DataFrame({'a':ones(5),'b':range(5)})
a b
0 1 0
1 1 1
2 1 2
3 1 3
4 1 4
z =Series([random() for i in range(5)],index = range(5))
0 0.812561
1 0.862109
2 0.031268
3 0.575634
4 0.760752
df.ix[z.index[[1,3]],'b'] = z[[1,3]]
a b
0 1 0.000000
1 1 0.812561
2 1 2.000000
3 1 0.575634
4 1 4.000000
5 1 5.000000
I really tried it all, need some new suggestions...
Try using df.update(updated_df_or_series)
Also using a simple example, you can modify a DataFrame by doing an index query and modifying the resulting object.
df_1
a b
0 1 0
1 1 1
2 1 2
3 1 3
4 1 4
df_2 = df_1.ix[3:5]
df_2.b = df_2.b + 2
df_2
a b
3 1 5
4 1 6
df_1
a b
0 1 0
1 1 1
2 1 2
3 1 5
4 1 6