I have a dataframe as below:
import pandas as pd
import numpy as np
import datetime
# intialise data of lists.
data = {'group' :["A","A","B","B","B"],
'A1_val' :[4,5,7,6,5],
'A1M_val' :[10,100,100,10,1],
'AB_val' :[4,5,7,6,5],
'ABM_val' :[10,100,100,10,1],
'AM_VAL' : [4,5,7,6,5]
}
# Create DataFrame
df1 = pd.DataFrame(data)
df1
group A1_val A1M_val AB_val ABM_val AM_VAL
0 A 4 10 4 10 4
1 A 5 100 5 100 5
2 B 7 100 7 100 7
3 B 6 10 6 10 6
4 B 5 1 5 1 5
Step 1: I want to create columns as below:
A1_agg_val = sum of A1_val + A1M_val (stripping M out of the column and if the name matches then sum it)
Similarly, AB_agg_val = AB_val + ABM_val
Since there is no matching columns for 'AM_VAL', AM_agg_val = AM_val
My expected output:
group A1_val A1M_val AB_val ABM_val AM_VAL A1_AGG_val AB_AGG_val A_AGG_val
0 A 4 10 4 10 4 14 14 4
1 A 5 100 5 100 5 105 105 5
2 B 7 100 7 100 7 107 107 7
3 B 6 10 6 10 6 16 16 6
4 B 5 1 5 1 5 6 6 5
you can use groupby on axis=1
out = (df1.assign(**df1.loc[:,df1.columns.str.lower().str.endswith('_val')]
.groupby(lambda x: x[:2],axis=1).sum().add_suffix('_agg_value')))
print(out)
group A1_val A1M_val AB_val ABM_val AM_VAL A1_agg_value AB_agg_value \
0 A 4 10 4 10 4 14 14
1 A 5 100 5 100 5 105 105
2 B 7 100 7 100 7 107 107
3 B 6 10 6 10 6 16 16
4 B 5 1 5 1 5 6 6
AM_agg_value
0 4
1 5
2 7
3 6
4 5
Related
I've been breaking my head over this simple thing. Ik we can assign a single value to multiple columns using .loc. But how to assign multiple columns with the same array.
Ik I can do this. Let's say we have a dataframe df in which I wish to replace some columns with the array arr:
df=pd.DataFrame({'a':[random.randint(1,25) for i in range(5)],'b':[random.randint(1,25) for i in range(5)],'c':[random.randint(1,25) for i in range(5)]})
>>df
a b c
0 14 8 5
1 10 25 9
2 14 14 8
3 10 6 7
4 4 18 2
arr = [i for i in range(5)]
#Suppose if I wish to replace columns `a` and `b` with array `arr`
df['a'],df['b']=[arr for j in range(2)]
Desired output:
a b c
0 0 0 16
1 1 1 10
2 2 2 1
3 3 3 20
4 4 4 11
Or I can also do this in a loopwise assignment. But is there a more efficient way without repetition or loops?
Let's try with assign:
cols = ['a', 'b']
df.assign(**dict.fromkeys(cols, arr))
a b c
0 0 0 5
1 1 1 9
2 2 2 8
3 3 3 7
4 4 4 2
I did an assign statement df.a = df.b = arr
df=pd.DataFrame({'a':[random.randint(1,25) for i in range(5)],'b':[random.randint(1,25) for i in range(5)],'c':[random.randint(1,25) for i in range(5)]})
arr = [i for i in range(5)]
df
a b c
0 2 8 18
1 17 15 25
2 6 5 17
3 12 15 25
4 10 10 6
df.a = df.b = arr
df
a b c
0 0 0 18
1 1 1 25
2 2 2 17
3 3 3 25
4 4 4 6
I have some data like this
df = pd.DataFrame({'class':['a','a','b','b','a','a','b','c','c'],'score':[3,5,6,7,8,9,10,11,14]})
df
class score
0 a 3
1 a 5
2 b 6
3 b 7
4 a 8
5 a 9
6 b 10
7 c 11
8 c 14
I want to use groupby function extract top n% data(descending by score),i know the nlargest can make it,but the number of every group is different,so i don't know how to do it
I tried this function
top_n = 0.5
g = df.groupby(['class'])['score'].apply(lambda x:x.nlargest(int(round(top_n*len(x))),keep='all')).reset_index()
g
class level_1 score
0 a 5 9
1 a 4 8
2 b 6 10
3 b 3 7
4 c 8 14
but it can not deal with big data(more than 10 million),it is very slow,how do i speed it,thank you!
Id B
1 6
2 13
1 6
2 6
1 6
2 6
1 10
2 6
2 6
2 6
I want a new columns say C where I can get a grouped value of B=6 at Id level
Jan18.loc[Jan18['Enquiry Purpose']==6].groupby(Jan18['Member Reference']).transform('count')
Id B No_of_6
1 6 3
2 13 5
1 6 3
2 6 5
1 6 3
2 6 5
1 10 3
2 6 5
2 6 5
2 6 5
Comapre values by Series.eq for ==, convert to integers and use GroupBy.transform for new column filled by sum per groups:
df['No_of_6'] = df['B'].eq(6).astype(int).groupby(df['Id']).transform('sum')
#alternative
#df['No_of_6'] = df.assign(B= df['B'].eq(6).astype(int)).groupby('Id')['B'].transform('sum')
print (df)
Id B No_of_6
0 1 6 3
1 2 13 5
2 1 6 3
3 2 6 5
4 1 6 3
5 2 6 5
6 1 10 3
7 2 6 5
8 2 6 5
9 2 6 5
Generally create boolean mask by your condition(s) and pass below:
mask = df['B'].eq(6)
#alternative
#mask = (df['B'] == 6)
df['No_of_6'] = mask.astype(int).groupby(df['Id']).transform('sum')
A solution using map. This solution will return NaN on groups of Id have no number of 6
df['No_of_6'] = df.Id.map(df[df.B.eq(6)].groupby('Id').B.count())
Out[113]:
Id B No_of_6
0 1 6 3
1 2 13 5
2 1 6 3
3 2 6 5
4 1 6 3
5 2 6 5
6 1 10 3
7 2 6 5
8 2 6 5
9 2 6 5
Hi all I have the following dataframe:
A | B | C
1 2 3
2 3 4
3 4 5
4 5 6
And I am trying to only repeat the last two rows of the data so that it looks like this:
A | B | C
1 2 3
2 3 4
3 4 5
3 4 5
4 5 6
4 5 6
I have tried using append, concat and repeat to no avail.
repeated = lambda x:x.repeat(2)
df.append(df[-2:].apply(repeated),ignore_index=True)
This returns the following dataframe, which is incorrect:
A | B | C
1 2 3
2 3 4
3 4 5
4 5 6
3 4 5
3 4 5
4 5 6
4 5 6
You can use numpy.repeat for repeating index and then create df1 by loc, last append to original, but before filter out last 2 rows by iloc:
df1 = df.loc[np.repeat(df.index[-2:].values, 2)]
print (df1)
A B C
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
print (df.iloc[:-2])
A B C
0 1 2 3
1 2 3 4
df = df.iloc[:-2].append(df1,ignore_index=True)
print (df)
A B C
0 1 2 3
1 2 3 4
2 3 4 5
3 3 4 5
4 4 5 6
5 4 5 6
If want use your code add iloc for filtering only last 2 rows:
repeated = lambda x:x.repeat(2)
df = df.iloc[:-2].append(df.iloc[-2:].apply(repeated),ignore_index=True)
print (df)
A B C
0 1 2 3
1 2 3 4
2 3 4 5
3 3 4 5
4 4 5 6
5 4 5 6
Use pd.concat and index slicing with .iloc:
pd.concat([df,df.iloc[-2:]]).sort_values(by='A')
Output:
A B C
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
I'm partial to manipulating the index into the pattern we are aiming for then asking the dataframe to take the new form.
Option 1
Use pd.DataFrame.reindex
df.reindex(df.index[:-2].append(df.index[-2:].repeat(2)))
A B C
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
Same thing in multiple lines
i = df.index
idx = i[:-2].append(i[-2:].repeat(2))
df.reindex(idx)
Could also use loc
i = df.index
idx = i[:-2].append(i[-2:].repeat(2))
df.loc[idx]
Option 2
Reconstruct from values. Only do this is all dtypes are the same.
i = np.arange(len(df))
idx = np.append(i[:-2], i[-2:].repeat(2))
pd.DataFrame(df.values[idx], df.index[idx])
0 1 2
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
Option 3
Can also use np.array in iloc
i = np.arange(len(df))
idx = np.append(i[:-2], i[-2:].repeat(2))
df.iloc[idx]
A B C
0 1 2 3
1 2 3 4
2 3 4 5
2 3 4 5
3 4 5 6
3 4 5 6
from pandas import *
import StringIO
df = read_csv(StringIO.StringIO('''id months state
1 1 C
1 2 3
1 3 6
1 4 9
2 1 C
2 2 C
2 3 3
2 4 6
2 5 9
2 6 9
2 7 9
2 8 C
'''), delimiter= '\t')
I want to create a column show the cumulative state of column state, by id.
id months state result
1 1 C C
1 2 3 C3
1 3 6 C36
1 4 9 C369
2 1 C C
2 2 C CC
2 3 3 CC3
2 4 6 CC36
2 5 9 CC69
2 6 9 CC699
2 7 9 CC6999
2 8 C CC6999C
Basically the cum concatenation of string columns. What is the best way to do it?
So long as the dtype is str then you can do the following:
In [17]:
df['result']=df.groupby('id')['state'].apply(lambda x: x.cumsum())
df
Out[17]:
id months state result
0 1 1 C C
1 1 2 3 C3
2 1 3 6 C36
3 1 4 9 C369
4 2 1 C C
5 2 2 C CC
6 2 3 3 CC3
7 2 4 6 CC36
8 2 5 9 CC369
9 2 6 9 CC3699
10 2 7 9 CC36999
11 2 8 C CC36999C
Essentially we groupby on 'id' column and then apply a lambda with a transform to return the cumsum. This will perform a cumulative concatenation of the string values and return a Series with it's index aligned to the original df so you can add it as a column