I have a dataframe with the following columns:
DATE ALFA BETA
2016-04-26 1 3
2016-04-27 3 0
2016-04-28 0 8
2016-04-29 4 2
2016-04-30 3 1
2016-05-01 -2 -5
2016-05-02 3 0
2016-05-03 3 3
2016-05-08 1 7
2016-05-11 3 1
2016-05-12 10 1
2016-05-13 4 2
I would like to group the data in a weekly range but treat the alpha and beta columns differently. I would like to calculate the sum of the numbers in the ALFA column for each week while for the BETA column I would like to calculate the difference between the beginning and the end of the week. I show you an example of the expected result.
DATE sum_ALFA diff_BETA
2016-04-26 12 3
2016-05-03 4 4
2016-05-11 17 1
I have tried this code but it calculates the sum for each column
df = df.resample('W', on='DATE').sum().reset_index().sort_values(by='DATE')
this is my dataset https://drive.google.com/uc?export=download&id=1fEqjINx9R5io7t_YxA9qShvNDxWRCUke
I'd guess I'm having a different locale here (hence my week is different), you can do:
df.resample("W", on="DATE",closed="left", label="left"
).agg({"ALFA":"sum", "BETA": lambda g: g.iloc[0] - g.iloc[-1]})
ALFA BETA
DATE
2016-04-24 11 2
2016-05-01 4 -8
2016-05-08 18 5
I think there is a solution for your data with my approach. Define
def get_series_first_minus_last(s):
try:
return s.iloc[0] - s.iloc[-1]
except IndexError:
return 0
and replace the lambda call just by the function call, i.e.
df.resample("W", on="DATE",closed="left", label="left"
).agg({"ALFA":"sum", "BETA": get_series_first_minus_last})
Note that in the newly defined function, you could also return nan if you'd prefer that.
Related
I have a pandas dataframe with several columns and I would like to know the number of columns above the date 2016-12-31 . Here is an example:
ID
Bill
Date 1
Date 2
Date 3
Date 4
Bill 2
4
6
2000-10-04
2000-11-05
1999-12-05
2001-05-04
8
6
8
2016-05-03
2017-08-09
2018-07-14
2015-09-12
17
12
14
2016-11-16
2017-05-04
2017-07-04
2018-07-04
35
And I would like to get this column
Count
0
2
3
Just create the mask and call sum on axis=1
date = pd.to_datetime('2016-12-31')
(df[['Date 1','Date 2','Date 3','Date 4']]>date).sum(1)
OUTPUT:
0 0
1 2
2 3
dtype: int64
If needed, call .to_frame('count') to create datarame with column as count
(df[['Date 1','Date 2','Date 3','Date 4']]>date).sum(1).to_frame('Count')
Count
0 0
1 2
2 3
Use df.filter to filter the Date* columns + .sum(axis=1)
(df.filter(like='Date') > '2016-12-31').sum(axis=1).to_frame(name='Count')
Result:
Count
0 0
1 2
2 3
You can do:
df['Count'] = (df.loc[:, [x for x in df.columns if 'Date' in x]] > '2016-12-31').sum(axis=1)
Output:
ID Bill Date 1 Date 2 Date 3 Date 4 Bill 2 Count
0 4 6 2000-10-04 2000-11-05 1999-12-05 2001-05-04 8 0
1 6 8 2016-05-03 2017-08-09 2018-07-14 2015-09-12 17 2
2 12 14 2016-11-16 2017-05-04 2017-07-04 2018-07-04 35 3
We select columns with 'Date' in the name. It's better when we have lots of columns like these and don't want to put them one by one. Then we compare it with lookup date and sum 'True' values.
Here is data
id
date
population
1
2021-5
21
2
2021-5
22
3
2021-5
23
4
2021-5
24
1
2021-4
17
2
2021-4
24
3
2021-4
18
4
2021-4
29
1
2021-3
20
2
2021-3
29
3
2021-3
17
4
2021-3
22
I want to calculate the monthly change regarding population in each id. so result will be:
id
date
delta
1
5
.2353
1
4
-.15
2
5
-.1519
2
4
-.2083
3
5
.2174
3
4
.0556
4
5
-.2083
4
4
.3182
delta := (this month - last month) / last month
How to approach this in pandas? I'm thinking of groupby but don't know what to do next
remember there might be more dates. but results is always
Use GroupBy.pct_change with sorting columns first before, last remove misisng rows by column delta:
df['date'] = pd.to_datetime(df['date'])
df = df.sort_values(['id','date'], ascending=[True, False])
df['delta'] = df.groupby('id')['population'].pct_change(-1)
df = df.dropna(subset=['delta'])
print (df)
id date population delta
0 1 2021-05-01 21 0.235294
4 1 2021-04-01 17 -0.150000
1 2 2021-05-01 22 -0.083333
5 2 2021-04-01 24 -0.172414
2 3 2021-05-01 23 0.277778
6 3 2021-04-01 18 0.058824
3 4 2021-05-01 24 -0.172414
7 4 2021-04-01 29 0.318182
Try this:
df.groupby('id')['population'].rolling(2).apply(lambda x: (x.iloc[0] - x.iloc[1]) / x.iloc[0]).dropna()
maybe you could try something like:
data['delta'] = data['population'].diff()
data['delta'] /= data['population']
with this approach the first line would be NaNs, but for the rest, this should work.
I have a dataframe that looks like this
ID | START | END
1 |2016-12-31|2017-02-30
2 |2017-01-30|2017-10-30
3 |2016-12-21|2018-12-30
I want to know the number of active IDs in each possible day. So basically count the number of overlapping time periods.
What I did to calculate this was creating a new data frame c_df with the columns date and count. The first column was populated using a range:
all_dates = pd.date_range(start=min(df['START']), end=max(df['END']))
Then for every line in my original data frame I calculated a different range for the start and end dates:
id_dates = pd.date_range(start=min(user['START']), end=max(user['END']))
I then used this range of dates to increment by one the corresponding count cell in c_df.
All these loops though are not very efficient for big data sets and look ugly. Is there a more efficient way of doing this?
If your dataframe is small enough so that performance is not a concern, create a date range for each row, then explode them and count how many times each date exists in the exploded series.
Requires pandas >= 0.25:
df.apply(lambda row: pd.date_range(row['START'], row['END']), axis=1) \
.explode() \
.value_counts() \
.sort_index()
If your dataframe is large, take advantage of numpy broadcasting to improve performance.
Work with any version of pandas:
dates = pd.date_range(df['START'].min(), df['END'].max()).values
start = df['START'].values[:, None]
end = df['END'].values[:, None]
mask = (start <= dates) & (dates <= end)
result = pd.DataFrame({
'Date': dates,
'Count': mask.sum(axis=0)
})
Create IntervalIndex and use genex or list comprehension with contains to check each date again each interval (Note: I made a smaller sample to test on this solution)
Sample `df`
Out[56]:
ID START END
0 1 2016-12-31 2017-01-20
1 2 2017-01-20 2017-01-30
2 3 2016-12-28 2017-02-03
3 4 2017-01-20 2017-01-25
iix = pd.IntervalIndex.from_arrays(df.START, df.END, closed='both')
all_dates = pd.date_range(start=min(df['START']), end=max(df['END']))
df_final = pd.DataFrame({'dates': all_dates,
'date_counts': (iix.contains(dt).sum() for dt in all_dates)})
In [58]: df_final
Out[58]:
dates date_counts
0 2016-12-28 1
1 2016-12-29 1
2 2016-12-30 1
3 2016-12-31 2
4 2017-01-01 2
5 2017-01-02 2
6 2017-01-03 2
7 2017-01-04 2
8 2017-01-05 2
9 2017-01-06 2
10 2017-01-07 2
11 2017-01-08 2
12 2017-01-09 2
13 2017-01-10 2
14 2017-01-11 2
15 2017-01-12 2
16 2017-01-13 2
17 2017-01-14 2
18 2017-01-15 2
19 2017-01-16 2
20 2017-01-17 2
21 2017-01-18 2
22 2017-01-19 2
23 2017-01-20 4
24 2017-01-21 3
25 2017-01-22 3
26 2017-01-23 3
27 2017-01-24 3
28 2017-01-25 3
29 2017-01-26 2
30 2017-01-27 2
31 2017-01-28 2
32 2017-01-29 2
33 2017-01-30 2
34 2017-01-31 1
35 2017-02-01 1
36 2017-02-02 1
37 2017-02-03 1
I have a dataframe which looks like this:
UserId Date_watched Days_not_watch
1 2010-09-11 5
1 2010-10-01 8
1 2010-10-28 1
2 2010-05-06 12
2 2010-05-18 5
3 2010-08-09 10
3 2010-09-25 5
I want to find out the no. of days the user gave as a gap, so I want a column for each row for each user and my dataframe should look something like this:
UserId Date_watched Days_not_watch Gap(2nd watch_date - 1st watch_date - days_not_watch)
1 2010-09-11 5 0 (First gap will be 0 for all users)
1 2010-10-01 8 15 (11th Sept+5=16th Sept; 1st Oct - 16th Sept=15days)
1 2010-10-28 1 9
2 2010-05-06 12 0
2 2010-05-18 5 0 (because 6th May+12 days=18th May)
3 2010-08-09 10 0
3 2010-09-25 4 36
3 2010-10-01 2 2
I have mentioned the formula for calculating the Gap beside the column name of the dataframe.
Here is one approach using groupby + shift:
# sort by date first
df['Date_watched'] = pd.to_datetime(df['Date_watched'])
df = df.sort_values(['UserId', 'Date_watched'])
# calculate groupwise start dates, shifted
grp = df.groupby('UserId')
starts = grp['Date_watched'].shift() + \
pd.to_timedelta(grp['Days_not_watch'].shift(), unit='d')
# calculate timedelta gaps
df['Gap'] = (df['Date_watched'] - starts).fillna(pd.Timedelta(0))
# convert to days and then integers
df['Gap'] = (df['Gap'] / pd.Timedelta('1 day')).astype(int)
print(df)
UserId Date_watched Days_not_watch Gap
0 1 2010-09-11 5 0
1 1 2010-10-01 8 15
2 1 2010-10-28 1 19
3 2 2010-05-06 12 0
4 2 2010-05-18 5 0
5 3 2010-08-09 10 0
6 3 2010-09-25 5 37
I have this pandas dataframe with daily asset prices:
Picture of head of Dataframe
I would like to create a pandas series (It could also be an additional column in the dataframe or some other datastructure) with the weakly average asset prices. This means I need to calculate the average on every 7 consecutive instances in the column and save it into a series.
Picture of how result should look like
As I am a complete newbie to python (and programming in general, for that matter), I really have no idea how to start.
I am very grateful for every tipp!
I believe need GroupBy.transform by modulo of numpy array create by numpy.arange for general solution also working with all indexes (e.g. with DatetimeIndex):
np.random.seed(2018)
rng = pd.date_range('2018-04-19', periods=20)
df = pd.DataFrame({'Date': rng[::-1],
'ClosingPrice': np.random.randint(4, size=20)})
#print (df)
df['weekly'] = df['ClosingPrice'].groupby(np.arange(len(df)) // 7).transform('mean')
print (df)
ClosingPrice Date weekly
0 2 2018-05-08 1.142857
1 2 2018-05-07 1.142857
2 2 2018-05-06 1.142857
3 1 2018-05-05 1.142857
4 1 2018-05-04 1.142857
5 0 2018-05-03 1.142857
6 0 2018-05-02 1.142857
7 2 2018-05-01 2.285714
8 1 2018-04-30 2.285714
9 1 2018-04-29 2.285714
10 3 2018-04-28 2.285714
11 3 2018-04-27 2.285714
12 3 2018-04-26 2.285714
13 3 2018-04-25 2.285714
14 1 2018-04-24 1.666667
15 0 2018-04-23 1.666667
16 3 2018-04-22 1.666667
17 2 2018-04-21 1.666667
18 2 2018-04-20 1.666667
19 2 2018-04-19 1.666667
Detail:
print (np.arange(len(df)) // 7)
[0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2]