Here is the code to merge two dataframes and get mean of a particular column like below,
pd.merge(df_old.loc[~df_old[‘COL1’].isnull()],
df_new,
on=’ID’)[‘COL1_y’].isnull().mean()
How do I repeat this for multiple columns (COL2, COL3…) without loop? I want to do it column wise. So that mean is there for each column
Here is my solution, is there a better way of doing this?
src = pd.Series()
for col in ['COL1', 'COL2', 'COL3']:
src[col]=pd.merge(df_old.loc[~df_old[col].isnull()],
df_new,
on=’ID’)[col+'_y'].isnull().mean()
IICU, you want to merge on multiple column at once.
import pandas as pd
df_old=pd.DataFrame(
[
[0,'1','2','3',10],
[0,'1','2','3',20],
[1,'1','2','3',30],
[1,'1','2','3',40],
[3,'1','2','3',50],
],
columns=['ID','COL1','COL2','COL3','x'])
df_new=pd.DataFrame(
[
[0,'1','2','3',50],
[0,'1','2','3',40],
[1,'1','2','3',30],
[1,'1','2','3',20],
[3,'1','2','3',10],
],columns=['ID','COL1','COL2','COL3','y'])
df_merge=pd.merge(df_old,df_new,on=['ID','COL1','COL2','COL3'], how='left')
df_merge
And you will get this
ID
COL1
COL2
COL3
x
y
0
0
1
2
3
10
50
1
0
1
2
3
10
40
2
0
1
2
3
20
50
3
0
1
2
3
20
40
4
1
1
2
3
30
30
5
1
1
2
3
30
20
6
1
1
2
3
40
30
7
1
1
2
3
40
20
8
3
1
2
3
50
10
And if you want the mean of y after merge,
df_merge['y'].mean()
Then you will get
32.22222222222222
Please state the result you expect, it is more easy to help you.
Hope this help.
Related
I'm using groupby on a pandas dataframe to drop all rows that don't have the minimum of a specific column. Something like this:
df1 = df.groupby("item", as_index=False)["diff"].min()
However, if I have more than those two columns, the other columns (e.g. otherstuff in my example) get dropped. Can I keep those columns using groupby, or am I going to have to find a different way to drop the rows?
My data looks like:
item diff otherstuff
0 1 2 1
1 1 1 2
2 1 3 7
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
and should end up like:
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
but what I'm getting is:
item diff
0 1 1
1 2 -6
2 3 0
I've been looking through the documentation and can't find anything. I tried:
df1 = df.groupby(["item", "otherstuff"], as_index=false)["diff"].min()
df1 = df.groupby("item", as_index=false)["diff"].min()["otherstuff"]
df1 = df.groupby("item", as_index=false)["otherstuff", "diff"].min()
But none of those work (I realized with the last one that the syntax is meant for aggregating after a group is created).
Method #1: use idxmin() to get the indices of the elements of minimum diff, and then select those:
>>> df.loc[df.groupby("item")["diff"].idxmin()]
item diff otherstuff
1 1 1 2
6 2 -6 2
7 3 0 0
[3 rows x 3 columns]
Method #2: sort by diff, and then take the first element in each item group:
>>> df.sort_values("diff").groupby("item", as_index=False).first()
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
[3 rows x 3 columns]
Note that the resulting indices are different even though the row content is the same.
You can use DataFrame.sort_values with DataFrame.drop_duplicates:
df = df.sort_values(by='diff').drop_duplicates(subset='item')
print (df)
item diff otherstuff
6 2 -6 2
7 3 0 0
1 1 1 2
If possible multiple minimal values per groups and want all min rows use boolean indexing with transform for minimal values per groups:
print (df)
item diff otherstuff
0 1 2 1
1 1 1 2 <-multiple min
2 1 1 7 <-multiple min
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
print (df.groupby("item")["diff"].transform('min'))
0 1
1 1
2 1
3 -6
4 -6
5 -6
6 -6
7 0
8 0
Name: diff, dtype: int64
df = df[df.groupby("item")["diff"].transform('min') == df['diff']]
print (df)
item diff otherstuff
1 1 1 2
2 1 1 7
6 2 -6 2
7 3 0 0
The above answer worked great if there is / you want one min. In my case there could be multiple mins and I wanted all rows equal to min which .idxmin() doesn't give you. This worked
def filter_group(dfg, col):
return dfg[dfg[col] == dfg[col].min()]
df = pd.DataFrame({'g': ['a'] * 6 + ['b'] * 6, 'v1': (list(range(3)) + list(range(3))) * 2, 'v2': range(12)})
df.groupby('g',group_keys=False).apply(lambda x: filter_group(x,'v1'))
As an aside, .filter() is also relevant to this question but didn't work for me.
I tried everyone's method and I couldn't get it to work properly. Instead I did the process step-by-step and ended up with the correct result.
df.sort_values(by='item', inplace=True, ignore_index=True)
df.drop_duplicates(subset='diff', inplace=True, ignore_index=True)
df.sort_values(by=['diff'], inplace=True, ignore_index=True)
For a little more explanation:
Sort items by the minimum value you want
Drop the duplicates of the column you want to sort with
Resort the data because the data is still sorted by the minimum values
If you know that all of your "items" have more than one record you can sort, then use duplicated:
df.sort_values(by='diff').duplicated(subset='item', keep='first')
I'm using groupby on a pandas dataframe to drop all rows that don't have the minimum of a specific column. Something like this:
df1 = df.groupby("item", as_index=False)["diff"].min()
However, if I have more than those two columns, the other columns (e.g. otherstuff in my example) get dropped. Can I keep those columns using groupby, or am I going to have to find a different way to drop the rows?
My data looks like:
item diff otherstuff
0 1 2 1
1 1 1 2
2 1 3 7
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
and should end up like:
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
but what I'm getting is:
item diff
0 1 1
1 2 -6
2 3 0
I've been looking through the documentation and can't find anything. I tried:
df1 = df.groupby(["item", "otherstuff"], as_index=false)["diff"].min()
df1 = df.groupby("item", as_index=false)["diff"].min()["otherstuff"]
df1 = df.groupby("item", as_index=false)["otherstuff", "diff"].min()
But none of those work (I realized with the last one that the syntax is meant for aggregating after a group is created).
Method #1: use idxmin() to get the indices of the elements of minimum diff, and then select those:
>>> df.loc[df.groupby("item")["diff"].idxmin()]
item diff otherstuff
1 1 1 2
6 2 -6 2
7 3 0 0
[3 rows x 3 columns]
Method #2: sort by diff, and then take the first element in each item group:
>>> df.sort_values("diff").groupby("item", as_index=False).first()
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
[3 rows x 3 columns]
Note that the resulting indices are different even though the row content is the same.
You can use DataFrame.sort_values with DataFrame.drop_duplicates:
df = df.sort_values(by='diff').drop_duplicates(subset='item')
print (df)
item diff otherstuff
6 2 -6 2
7 3 0 0
1 1 1 2
If possible multiple minimal values per groups and want all min rows use boolean indexing with transform for minimal values per groups:
print (df)
item diff otherstuff
0 1 2 1
1 1 1 2 <-multiple min
2 1 1 7 <-multiple min
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
print (df.groupby("item")["diff"].transform('min'))
0 1
1 1
2 1
3 -6
4 -6
5 -6
6 -6
7 0
8 0
Name: diff, dtype: int64
df = df[df.groupby("item")["diff"].transform('min') == df['diff']]
print (df)
item diff otherstuff
1 1 1 2
2 1 1 7
6 2 -6 2
7 3 0 0
The above answer worked great if there is / you want one min. In my case there could be multiple mins and I wanted all rows equal to min which .idxmin() doesn't give you. This worked
def filter_group(dfg, col):
return dfg[dfg[col] == dfg[col].min()]
df = pd.DataFrame({'g': ['a'] * 6 + ['b'] * 6, 'v1': (list(range(3)) + list(range(3))) * 2, 'v2': range(12)})
df.groupby('g',group_keys=False).apply(lambda x: filter_group(x,'v1'))
As an aside, .filter() is also relevant to this question but didn't work for me.
I tried everyone's method and I couldn't get it to work properly. Instead I did the process step-by-step and ended up with the correct result.
df.sort_values(by='item', inplace=True, ignore_index=True)
df.drop_duplicates(subset='diff', inplace=True, ignore_index=True)
df.sort_values(by=['diff'], inplace=True, ignore_index=True)
For a little more explanation:
Sort items by the minimum value you want
Drop the duplicates of the column you want to sort with
Resort the data because the data is still sorted by the minimum values
If you know that all of your "items" have more than one record you can sort, then use duplicated:
df.sort_values(by='diff').duplicated(subset='item', keep='first')
I have the following data:
dict={'A':[1,2,3,4,5],'B':[10,20,233,29,2],'C':[10,20,3040,230,238]...................}
and
df= pd.Dataframe(dict)
In this manner I have 20 columns with 5 numerical entry in each column
I want to have a new column where the value should come as the following logic:
0 A[0]*B[0]+A[0]*C[0] + A[0]*D[0].......
1 A[1]*B[1]+A[1]*C[1] + A[1]*D[1].......
2 A[2]*B[2]+A[2]*B[2] + A[2]*D[2].......
I tried in the following manner but manually I can not put 20 columns, so I wanted to know the way to apply a loop to get the desired output
:
lst=[]
for i in range(0,5):
j=df.A[i]*df.B[i]+ df.A[i]*df.C[i]+.......
lst.append(j)
i=i+1
A potential solution is the following. I am only taking the example you posted but is works fine for more. Your data is df
A B C
0 1 10 10
1 2 20 20
2 3 233 3040
3 4 29 230
4 5 2 238
You can create a new column, D by first subsetting your dataframe
add = df.loc[:, df.columns != 'A']
and then take the sum over all multiplications of the columns in D with column A in the following way:
df['D'] = df['A']*add.sum(axis=1)
which returns
A B C D
0 1 10 10 20
1 2 20 20 80
2 3 233 3040 9819
3 4 29 230 1036
4 5 2 238 1200
Suppose I have dataframe like this
>>> df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4],'value':[1,2,3,1,2,3,4,1,1]})
>>> df
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
Now I want top all records from each group using group id except last 3. That means I want to drop last 3 records from all groups. How can I do it using pandas group_by. This is dummy data.
Use GroupBy.cumcount for counter from back by ascending=False and then compare by Series.gt for greater values like 2, because python count from 0:
df = df[df.groupby('id').cumcount(ascending=False).gt(2)]
print (df)
id value
3 2 1
Details:
print (df.groupby('id').cumcount(ascending=False))
0 2
1 1
2 0
3 3
4 2
5 1
6 0
7 0
8 0
dtype: int64
With the DataFrame below as an example,
In [83]:
df = pd.DataFrame({'A':[1,1,2,2],'B':[1,2,1,2],'values':np.arange(10,30,5)})
df
Out[83]:
A B values
0 1 1 10
1 1 2 15
2 2 1 20
3 2 2 25
What would be a simple way to generate a new column containing some aggregation of the data over one of the columns?
For example, if I sum values over items in A
In [84]:
df.groupby('A').sum()['values']
Out[84]:
A
1 25
2 45
Name: values
How can I get
A B values sum_values_A
0 1 1 10 25
1 1 2 15 25
2 2 1 20 45
3 2 2 25 45
In [20]: df = pd.DataFrame({'A':[1,1,2,2],'B':[1,2,1,2],'values':np.arange(10,30,5)})
In [21]: df
Out[21]:
A B values
0 1 1 10
1 1 2 15
2 2 1 20
3 2 2 25
In [22]: df['sum_values_A'] = df.groupby('A')['values'].transform(np.sum)
In [23]: df
Out[23]:
A B values sum_values_A
0 1 1 10 25
1 1 2 15 25
2 2 1 20 45
3 2 2 25 45
I found a way using join:
In [101]:
aggregated = df.groupby('A').sum()['values']
aggregated.name = 'sum_values_A'
df.join(aggregated,on='A')
Out[101]:
A B values sum_values_A
0 1 1 10 25
1 1 2 15 25
2 2 1 20 45
3 2 2 25 45
Anyone has a simpler way to do it?
This is not so direct but I found it very intuitive (the use of map to create new columns from another column) and can be applied to many other cases:
gb = df.groupby('A').sum()['values']
def getvalue(x):
return gb[x]
df['sum'] = df['A'].map(getvalue)
df
In [15]: def sum_col(df, col, new_col):
....: df[new_col] = df[col].sum()
....: return df
In [16]: df.groupby("A").apply(sum_col, 'values', 'sum_values_A')
Out[16]:
A B values sum_values_A
0 1 1 10 25
1 1 2 15 25
2 2 1 20 45
3 2 2 25 45