Starting from this data frame:
id
date
value
1
01.01.
2
2
01.01.
3
1
01.03.
5
2
01.03.
3
1
01.09.
5
2
01.09.
2
1
01.10.
5
2
01.10.
2
I would like to get a weekly sum of value:
id
date
value
1
01.01.
2
2
01.01.
3
1
01.03.
7
2
01.03.
6
1
01.09.
10
2
01.09.
5
1
01.10.
15
2
01.10.
8
I use this command, but it is not working:
df['value'] = df.groupby('id')['value'].rolling(7).sum()
Any ideas?
You can do groupby and apply.
df['date'] = pd.to_datetime(df['date'], format='%m.%d.')
df['value'] = (df.groupby('id', as_index=False, group_keys=False)
.apply(lambda g: g.rolling('7D', on='date')['value'].sum()))
Note that for 1900-01-10, the rolling window is 1900-01-04, 1900-01-05...1900-01-10
print(df)
id date value
0 1 1900-01-01 2.0
1 2 1900-01-01 3.0
2 1 1900-01-03 7.0
3 2 1900-01-03 6.0
4 1 1900-01-09 10.0
5 2 1900-01-09 5.0
6 1 1900-01-10 10.0
7 2 1900-01-10 4.0
Related
Assume I have the following data frame in pandas, with accumulated values over time for all ids:
id
date
value
1
01.01.1999
2
2
01.01.1999
3
3
01.01.1999
5
1
03.01.1999
5
2
03.01.1999
8
3
03.01.1999
7
And I want to have the following, the difference for each id to the previous date:
id
date
value
1
01.01.1999
2
2
01.01.1999
3
3
01.01.1999
5
1
03.01.1999
3
2
03.01.1999
5
3
03.01.1999
2
This is basically the difference. I can only apply something like this:
df["values"].diff().fillna(0)
But this would not include the date column. Any help?
IIUC, you want to groupby and diff
df['value'] = df.groupby('id')['value'].diff().fillna(df['value'])
print(df)
id date value
0 1 01.01.1999 2.0
1 2 01.01.1999 3.0
2 3 01.01.1999 5.0
3 1 03.01.1999 3.0
4 2 03.01.1999 5.0
5 3 03.01.1999 2.0
I have a dataframe which looks like this:
Name T1 T2 alpha
A 10 3 30
A 11 5 Nan
A 13 5 Nan
B 5 2 7
B 3 1 Nan
However need to divide the alpha column in equal parts to replace Nan values for each unique names in Name column like this: eg 30 for A becomes 10 each for corresponding row where A is present
Name T1 T2 alpha
A 10 3 10
A 11 5 10
A 13 5 10
B 5 2 3.5
B 3 1 3.5
I tried using explode but it is not working as I want it to look like, any idea here would help
With groupby and transform
df['alpha'] = pd.to_numeric(df['alpha'],errors = 'coerce').fillna(0).groupby(df['Name']).transform('mean')
df
Out[50]:
Name T1 T2 alpha
0 A 10 3 10.0
1 A 11 5 10.0
2 A 13 5 10.0
3 B 5 2 3.5
4 B 3 1 3.5
This'll get the job done:
df['alpha'] = (df['alpha'] / df['T2']).ffill()
Output:
Name T1 T2 alpha
0 A 10 3 10.0
1 A 11 5 10.0
2 A 13 5 10.0
3 B 5 2 3.5
4 B 3 1 3.5
I have a data frame as shown below
ID Class Score1 Score2 Name
1 A 9 7 Xavi
2 B 7 8 Alba
3 A 10 8 Messi
4 A 8 10 Neymar
5 A 7 8 Mbappe
6 C 4 6 Silva
7 C 3 2 Pique
8 B 5 7 Ramos
9 B 6 7 Serge
10 C 8 5 Ayala
11 A NaN 4 Casilas
12 A NaN 4 De_Gea
13 B NaN 2 Seaman
14 C NaN 7 Chilavert
15 B NaN 3 Courtous
From the above, I would like to calculate the number of players with scoer1 less than or equal to 6 in each Class along with count of non NaN rows (Class wise)
Expected output:
Class Total_Number Count_Non_NaN Score1_less_than_6_# Avg_score1
A 6 4 0 8.5
B 5 3 2 6
C 4 3 2 5
tried below code
df2 = df.groupby('Class').agg(Total_Number = ('Score1','size'),
Score1_less_than_6 = ('Score1',lambda x: x.between(0,6).sum()),
Avg_score1 = ('Score1','mean'))
df2 = df2.reset_index()
df2
Groupby and aggregate using a dictionary
df['s'] = df['Score1'].le(6)
df.groupby('Class').agg(**{'total_number': ('Score1', 'size'),
'count_non_nan': ('Score1', 'count'),
'score1_less_than_six': ('s', 'sum'),
'avg_score1': ('Score1', 'mean')})
total_number count_non_nan score1_less_than_six avg_score1
Class
A 6 4 0 8.5
B 5 3 2 6.0
C 4 3 2 5.0
Try:
x = df.groupby("Class", as_index=False).agg(
Total_Number=("Class", "count"),
Count_Non_NaN=("Score1", lambda x: x.notna().sum()),
Score1_less_than_6=("Score1", lambda x: (x <= 6).sum()),
Avg_score1=("Score1", "mean"),
)
print(x)
Prints:
Class Total_Number Count_Non_NaN Score1_less_than_6 Avg_score1
0 A 6 4.0 0.0 8.5
1 B 5 3.0 2.0 6.0
2 C 4 3.0 2.0 5.0
I have a dataframe
Intialise data of lists.
data = {'Id':['1', '2', '3', '4','5','6','7','8','9','10'], 'reply_id':[2, 2,2, 5,5,6,8,8,1,1]}
Create DataFrame
df = pd.DataFrame(data)
Id reply_id
0 1 2
1 2 2
2 3 2
3 4 5
4 5 5
5 6 6
6 7 8
7 8 8
8 9 1
9 10 1
I want to get total of reply_id in new for every Id.
Id=1 have 2 time occurrence in reply_id which i want in new column new
Desired output
Id reply_id new
0 1 2 2
1 2 2 3
2 3 2 0
3 4 5 0
4 5 5 2
5 6 6 1
6 7 8 0
7 8 8 2
8 9 1 0
9 10 1 0
I have done this line of code.
df['new'] = df.reply_id.eq(df.Id).astype(int).groupby(df.Id).transform('sum')
In this answer, I used Series.value_counts to count values in reply_id, and converted the result to a dict. Then, I used Series.map on the Id column to associate counts to Id. fillna(0) is used to fill values not present in reply_id
df['new'] = (df['Id']
.astype(int)
.map(df['reply_id'].value_counts().to_dict())
.fillna(0)
.astype(int))
Use, Series.groupby on the column reply_id, then use the aggregation function GroupBy.count to create a mapping series counts, finally use Series.map to map the values in Id column with their respective counts:
counts = df['reply_id'].groupby(df['reply_id']).count()
df['new'] = df['Id'].map(counts).fillna(0).astype(int)
Result:
# print(df)
Id reply_id new
0 1 2 2
1 2 2 3
2 3 2 0
3 4 5 0
4 5 5 2
5 6 6 1
6 7 8 0
7 8 8 2
8 9 1 0
9 10 1 0
I have an dataframe which contain the observed data as:
import pandas as pd
d = {'humanID': [1, 1, 2,2,2,2 ,2,2,2,2], 'dogID':
[1,2,1,5,4,6,7,20,9,7],'month': [1,1,2,3,1,2,3,1,2,2]}
df = pd.DataFrame(data=d)
The df is follow
humanID dogID month
0 1 1 1
1 1 2 1
2 2 1 2
3 2 5 3
4 2 4 1
5 2 6 2
6 2 7 3
7 2 20 1
8 2 9 2
9 2 7 2
We total have two human and twenty dog, and above df contains the observed data. For example:
The first row means: human1 adopt dog1 at January
The second row means: human1 adopt dog2 at January
The third row means: human2 adopt dog1 at Febuary
========================================================================
My goal is randomly generating two unobserved data for each (human, month) that are not appear in the original observed data.
like for human1 at January, he does't adopt the dog [3,4,5,6,7,..20] And I want to randomly create two unobserved sample (human, month) in triple form
humanID dogID month
1 20 1
1 10 1
However, the follow sample is not allowed since it appear in original df
humanID dogID month
1 2 1
For human1, he doesn't have any activity at Feb, so we don't need to sample the unobserved data.
For human2, he have activity for Jan, Feb and March. Therefore, for each month, we want to randomly create the unobserved data. For example, In Jan, human2 adopt dog1, dog4 and god 20. The two random unobserved samples can be
humanID dogID month
2 2 1
2 6 1
same process can be used for Feb and March.
I want to put all of the unobserved in one dataframe such as follow unobserved
humanID dogID month
0 1 20 1
1 1 10 1
2 2 2 1
3 2 6 1
4 2 13 2
5 2 16 2
6 2 1 3
7 2 20 3
Any fast way to do this?
PS: this is a code interview for a start-up company.
Using groupby and random.choices:
import random
dogs = list(range(1,21))
dfs = []
n_sample = 2
for i,d in df.groupby(['humanID', 'month']):
h_id, month = i
sample = pd.DataFrame([(h_id, dogID, month) for dogID in random.choices(list(set(dogs)-set(d['dogID'])), k=n_sample)])
dfs.append(sample)
new_df = pd.concat(dfs).reset_index(drop=True)
new_df.columns = ['humanID', 'dogID', 'month']
print(new_df)
humanID dogID month
0 1 11 1
1 1 5 1
2 2 19 1
3 2 18 1
4 2 15 2
5 2 14 2
6 2 16 3
7 2 18 3
If I understand you correctly, you can use np.random.permutation() for the dogID column to generate random permutations of the column,
df_new=df.copy()
df_new['dogID']=np.random.permutation(df.dogID)
print(df_new.sort_values('month'))
humanID dogID month
0 1 1 1
1 1 20 1
4 2 9 1
7 2 1 1
2 2 4 2
5 2 5 2
8 2 2 2
9 2 7 2
3 2 7 3
6 2 6 3
Or to create random sampling of missing values within the range of dogID:
df_new=df.copy()
a=np.random.permutation(range(df_new.dogID.min(),df_new.dogID.max()))
df_new['dogID']=np.random.choice(a,df_new.shape[0])
print(df_new.sort_values('month'))
humanID dogID month
0 1 18 1
1 1 16 1
4 2 1 1
7 2 8 1
2 2 4 2
5 2 2 2
8 2 16 2
9 2 14 2
3 2 4 3
6 2 12 3