I need to add a new feature that aggregates the last 5 data. When it adds 6th data, then it should forget the first data and consider only the last 5 data sets as shown below. Here is the dummy data frame, new_feature is the expected output.
id feature new_feature
1 a a
2 b a+b
3 c a+b+c
4 d a+b+c+d
5 e a+b+c+d+e
6 f b+c+d+e+f
7 g c+d+e+f+g
Use Series.rolling with min_periods=1 parameter and sum:
df = pd.DataFrame({'feature':[1,2,4,5,6,2,3,4,5]})
df['new_feature'] = df['feature'].rolling(5, min_periods=1).sum()
print (df)
feature new_feature
0 1 1.0
1 2 3.0
2 4 7.0
3 5 12.0
4 6 18.0
5 2 19.0
6 3 20.0
7 4 20.0
8 5 20.0
Related
I have this df
d={}
d['id']=['1','1','1','1','1','1','1','1','2','2','2','2','2','2','2','2']
d['qty']=[5,5,5,5,5,6,5,5,1,1,2,2,2,3,5,8]
I would like to create a column that is going to have the following non-equal value of column qty. Meaning that if qty is equal to 5 and its next row is 5 I am going to skip it and look until I find next value not equal to 5, In my case it is 6. And all this should be grouped by id
Here is the desired dataframe.
d['id']=['1','1','1','1','1','1','1','1','2','2','2','2','2','2','2','2']
d['qty']=[5,5,5,5,5,6,5,5,1,1,2,2,2,3,5,8]
d['qty2']=[6,6,6,6,6,5,'NAN','NAN',2,2,3,3,3,5,8,'NAN']
Any help is very much appreciated
You can groupby.shift, mask the identical values, and groupby.bfill:
# shift up per group
s = df.groupby('id')['qty'].shift(-1)
# keep only the different values and bfill per group
df['qty2'] = s.where(df['qty'].ne(s)).groupby(df['id']).bfill()
output:
id qty qty2
0 1 5 6.0
1 1 5 6.0
2 1 5 6.0
3 1 5 6.0
4 1 5 6.0
5 1 6 5.0
6 1 5 NaN
7 1 5 NaN
8 2 1 2.0
9 2 1 2.0
10 2 2 3.0
11 2 2 3.0
12 2 2 3.0
13 2 3 5.0
14 2 5 8.0
15 2 8 NaN
Lets say we want to compute the variable D in the dataframe below based on time values in variable B and C.
Here, second row of D is C2 - B1, the difference is 4 minutes and
third row = C3 - B2= 4 minutes,.. and so on.
There is no reference value for first row of D so its NA.
Issue:
We also want a NA value for the first row when the category value in variable A changes from 1 to 2. In other words, the value -183 must be replaced by NA.
A B C D
1 5:43:00 5:24:00 NA
1 6:19:00 5:47:00 4
1 6:53:00 6:23:00 4
1 7:29:00 6:55:00 2
1 8:03:00 7:31:00 2
1 8:43:00 8:05:00 2
2 6:07:00 5:40:00 -183
2 6:42:00 6:11:00 4
2 7:15:00 6:45:00 3
2 7:53:00 7:17:00 2
2 8:30:00 7:55:00 2
2 9:07:00 8:32:00 2
2 9:41:00 9:09:00 2
2 10:17:00 9:46:00 5
2 10:52:00 10:20:00 3
You can use:
# Compute delta
df['D'] = (pd.to_timedelta(df['C']).sub(pd.to_timedelta(df['B'].shift()))
.dt.total_seconds().div(60))
# Fill nan
df.loc[df['A'].ne(df['A'].shift()), 'D'] = np.nan
Output:
>>> df
A B C D
0 1 5:43:00 5:24:00 NaN
1 1 6:19:00 5:47:00 4.0
2 1 6:53:00 6:23:00 4.0
3 1 7:29:00 6:55:00 2.0
4 1 8:03:00 7:31:00 2.0
5 1 8:43:00 8:05:00 2.0
6 2 6:07:00 5:40:00 NaN
7 2 6:42:00 6:11:00 4.0
8 2 7:15:00 6:45:00 3.0
9 2 7:53:00 7:17:00 2.0
10 2 8:30:00 7:55:00 2.0
11 2 9:07:00 8:32:00 2.0
12 2 9:41:00 9:09:00 2.0
13 2 10:17:00 9:46:00 5.0
14 2 10:52:00 10:20:00 3.0
You can use the difference between datetime columns in pandas.
Having
df['B_dt'] = pd.to_datetime(df['B'])
df['C_dt'] = pd.to_datetime(df['C'])
Makes the following possible
>>> df['D'] = (df.groupby('A')
.apply(lambda s: (s['C_dt'] - s['B_dt'].shift()).dt.seconds / 60)
.reset_index(drop=True))
You can always drop these new columns later.
i have this dataframe:
a b c d
4 7 5 12
3 8 2 8
1 9 3 5
9 2 6 4
i want the column 'd' to become the difference between n-value of column a and n+1 value of column 'a'.
I tried this but it doesn't run:
for i in data.index-1:
data.iloc[i]['d']=data.iloc[i]['a']-data.iloc[i+1]['a']
can anyone help me?
Basically what you want is diff.
df = pd.DataFrame.from_dict({"a":[4,3,1,9]})
df["d"] = df["a"].diff(periods=-1)
print(df)
Output
a d
0 4 1.0
1 3 2.0
2 1 -8.0
3 9 NaN
lets try simple way:
df=pd.DataFrame.from_dict({'a':[2,4,8,15]})
diff=[]
for i in range(len(df)-1):
diff.append(df['a'][i+1]-df['a'][i])
diff.append(np.nan)
df['d']=diff
print(df)
a d
0 2 2.0
1 4 4.0
2 8 7.0
3 15 NaN
I have this dataframe.
from pandas import DataFrame
import pandas as pd
df = pd.DataFrame({'name': ['A','D','M','T','B','C','D','E','A','L'],
'id': [1,1,1,2,2,3,3,3,3,5],
'rate': [3.5,4.5,2.0,5.0,4.0,1.5,2.0,2.0,1.0,5.0]})
>> df
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 C 3 1.5
6 D 3 2.0
7 E 3 2.0
8 A 3 1.0
9 L 5 5.0
df = df.groupby('id')['rate'].mean()
what i want is this:
1) find mean of every 'id'.
2) give the number of ids (length) which has mean >= 3.
3) give back all rows of dataframe (where mean of any id >= 3.
Expected output:
Number of ids (length) where mean >= 3: 3
>> dataframe where (mean(id) >=3)
>>df
name id rate
0 A 1 3.0
1 D 1 4.0
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 L 5 5.0
Use GroupBy.transform for means by all groups with same size like original DataFrame, so possible filter by boolean indexing:
df = df[df.groupby('id')['rate'].transform('mean') >=3]
print (df)
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
9 L 5 5.0
Detail:
print (df.groupby('id')['rate'].transform('mean'))
0 3.333333
1 3.333333
2 3.333333
3 4.500000
4 4.500000
5 1.625000
6 1.625000
7 1.625000
8 1.625000
9 5.000000
Name: rate, dtype: float64
Alternative solution with DataFrameGroupBy.filter:
df = df.groupby('id').filter(lambda x: x['rate'].mean() >=3)
I am looking to create a new column in panda based on the value in the row. My sample data:
df=pd.DataFrame({"A":['a','a','a','a','a','a','b','b','b'],
"Sales":[2,3,7,1,4,3,5,6,9,10,11,8,7,13,14],
"Week":[1,2,3,4,5,11,1,2,3,4])
I want a new column "Last3WeekSales" corresponding to each week, having the sum of sales for the previous 3 weeks.
NOTE: Shift() won't work here as data for some weeks is missing.
Logic which I thought:
Checking the week no. in each row, then summing up the data from w-1, w-2, w-3.
Output required:
A Week Last3WeekSales
0 a 1 0
1 a 2 2
2 a 3 5
3 a 4 12
4 a 5 11
5 a 11 0
6 b 1 0
7 b 2 5
8 b 3 11
9 b 4 20
Use groupby, shift and rolling:
df['Last3WeekSales'] = df.groupby('A')['Sales']\
.apply(lambda x: x.shift(1)
.rolling(3, min_periods=1)
.sum())\
.fillna(0)
Output:
A Sales Week Last3WeekSales
0 a 2 1 0.0
1 a 3 2 2.0
2 a 7 3 5.0
3 a 1 4 12.0
4 a 4 5 11.0
5 a 3 6 12.0
6 b 5 1 0.0
7 b 6 2 5.0
8 b 9 3 11.0
you can use pandas.rolling_sum to sum over 3 last values, and shift(n) to shift your column by n times (1 in your case).
if we suppose you a column 'sales' with the sales of each week, the code would be :
df["Last3WeekSales"] = df.groupby("A")["sales"].apply(lambda x: pd.rolling_sum(x.shoft(1),3))