I'm new to programming and Pandas, I'd like to have an example of how to apply a grouping function that also applies some counters to reduce the following DataFrame:
child
groupName
state
name1
A
ok
name2
A
ko
name3
B
ok
to a new DataFrame like:
groupName
noOfChildren
noOfOk
noOfKo
A
2
1
1
B
1
1
0
Given the allChildren DataFrame, I can create the Series counting the entries by groupName:
childrenByGroupName= allChildren.groupby(['groupName'])['name'].count();
And also the Series to filter them by 'ok' state:
okChildrenByGroupName= allChildren.where(['state']=='ok').groupby(['groupName'])['name'].count();
But I cannot build the merged DataFrame as per the above expectation.
Any help?
Try:
pd.crosstab(df['groupName'], df['state'], margins='sum')
Output:
state ko ok All
groupName
A 1 1 2
B 0 1 1
All 1 2 3
and to (almost) match the expected output:
(pd.crosstab(df['groupName'], df['state'], margins='sum', margins_name='Children')
.drop('Children')
.add_prefix('noOf')
.reset_index()
)
you can try like this:
df1 = df.groupby(['groupName']).agg({'child': 'count', 'state': lambda x: x.value_counts().to_dict()}).add_prefix('noOf').reset_index()
df2 = pd.concat([df1.drop('noOfstate', axis=1), pd.DataFrame(df1['noOfstate'].tolist()).add_prefix('noOf')], axis=1).fillna(0)
df2:
groupName noOfchild noOfok noOfko
0 A 2 1 1.0
1 B 1 1 0.0
Related
I have the case where I want to sanity check labeled data. I have hundreds of features and want to find points which have the same features but different label. These found cluster of disagreeing labels should then be numbered and put into a new dataframe.
This isn't hard but I am wondering what the most elegant solution for this is.
Here an example:
import pandas as pd
df = pd.DataFrame({
"feature_1" : [0,0,0,4,4,2],
"feature_2" : [0,5,5,1,1,3],
"label" : ["A","A","B","B","D","A"]
})
result_df = pd.DataFrame({
"cluster_index" : [0,0,1,1],
"feature_1" : [0,0,4,4],
"feature_2" : [5,5,1,1],
"label" : ["A","B","B","D"]
})
In order to get the output you want (both de-duplication and cluster_index), you can use a groupby approach:
g = df.groupby(['feature_1', 'feature_2'])['label']
(df.assign(cluster_index=g.ngroup()) # get group name
.loc[g.transform('size').gt(1)] # filter the non-duplicates
# line below only to have a nice cluster_index range (0,1…)
.assign(cluster_index= lambda d: d['cluster_index'].factorize()[0])
)
output:
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
First get all duplicated values per feature columns and then if necessary remove duplciated by all columns (here in sample data not necessary), last add GroupBy.ngroup for groups indices:
df = df[df.duplicated(['feature_1','feature_2'],keep=False)].drop_duplicates()
df['cluster_index'] = df.groupby(['feature_1', 'feature_2'])['label'].ngroup()
print (df)
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
I have a dataframe where I need to sort the contents of one column (comma separated) alphabetically:
ID Data
1 Mo,Ab,ZZz
2 Ab,Ma,Bt
3 Xe,Aa
4 Xe,Re,Fi,Ab
Output:
ID Data
1 Ab,Mo,ZZz
2 Ab,Bt,Ma
3 Aa,Xe
4 Ab,Fi,Re,Xe
I have tried:
df.sort_values(by='Data')
But this does not work
You can split, sorting and then join back:
df['Data'] = df['Data'].apply(lambda x: ','.join(sorted(x.split(','))))
Or use list comprehension alternative:
df['Data'] = [','.join(sorted(x.split(','))) for x in df['Data']]
print (df)
ID Data
0 1 Ab,Mo,ZZz
1 2 Ab,Bt,Ma
2 3 Aa,Xe
3 4 Ab,Fi,Re,Xe
IIUC get_dummies
s=df.Data.str.get_dummies(',')
df['n']=s.dot(s.columns+',').str[:-1]
df
Out[216]:
ID Data n
0 1 Mo,Ab,ZZz Ab,Mo,ZZz
1 2 Ab,Ma,Bt Ab,Bt,Ma
2 3 Xe,Aa Aa,Xe
3 4 Xe,Re,Fi,Ab Ab,Fi,Re,Xe
IIUC you can use a list comprehension:
[','.join(sorted(i.split(','))) for i in df['Data']]
#['Ab,Mo,ZZz', 'Ab,Bt,Ma', 'Aa,Xe', 'Ab,Fi,Re,Xe']
using explode and sort_values
df["Sorted_Data"] = (
df["Data"].str.split(",").explode().sort_values().groupby(level=0).agg(','.join)
)
print(df)
ID Data Sorted_Data
0 1 Mo,Ab,ZZz Ab,Mo,ZZz
1 2 Ab,Ma,Bt Ab,Bt,Ma
2 3 Xe,Aa Aa,Xe
3 4 Xe,Re,Fi,Ab Ab,Fi,Re,Xe
Using row iteration:
for index, row in df.iterrows():
row['Data'] = ','.join(sorted(row['Data'].split(',')))
In [29]: df
Out[29]:
Data
0 Ab,Mo,ZZz
1 Ab,Bt,Ma
2 Aa,Xe
3 Ab,Fi,Re,Xe
I have 3 data frame:
df1
id,k,a,b,c
1,2,1,5,1
2,3,0,1,0
3,6,1,1,0
4,1,0,5,0
5,1,1,5,0
df2
name,a,b,c
p,4,6,8
q,1,2,3
df3
type,w_ave,vac,yak
n,3,5,6
v,2,1,4
from the multiplication, using pandas and numpy, I want to the output in df1:
id,k,a,b,c,w_ave,vac,yak
1,2,1,5,1,16,15,18
2,3,0,1,0,0,3,6
3,6,1,1,0,5,4,7
4,1,0,5,0,0,11,14
5,1,1,5,0,13,12,15
the conditions are:
The value of the new column will be =
#its not a code
df1["w_ave"][1] = df3["w_ave"]["v"]+ df1["a"][1]*df2["a"]["q"]+df1["b"][1]*df2["b"]["q"]+df1["c"][1]*df2["c"]["q"]
for output["w_ave"][1]= 2 +(1*1)+(5*2)+(1*3)
df3["w_ave"]["v"]=2
df1["a"][1]=1, df2["a"]["q"]=1 ;
df1["b"][1]=5, df2["b"]["q"]=2 ;
df1["c"][1]=1, df2["c"]["q"]=3 ;
Which means:
- a new column will be added in df1, from the name of the column from df3.
- for each row of the df1, the value of a, b, c will be multiplied with the same-named q value from df2. and summed together with the corresponding value of df3.
-the column name of df1 , matched will column name of df2 will be multiplied. The other not matched column will not be multiplied, like df1[k].
- However, if there is any 0 in df1["a"], the corresponding output will be zero.
I am struggling with this. It was tough to explain also. My attempts are very silly. I know this attempt will not work. However, I have added this:
import pandas as pd, numpy as np
data1 = "Sample_data1.csv"
data2 = "Sample_data2.csv"
data3 = "Sample_data3.csv"
folder = '~Sample_data/'
df1 =pd.read_csv(folder + data1)
df2 =pd.read_csv(folder + data2)
df3 =pd.read_csv(folder + data3)
df1= df2 * df1
Ok, so this will in no way resemble your desired output, but vectorizing the formula you provided:
df2=df2.set_index("name")
df3=df3.set_index("type")
df1["w_ave"] = df3.loc["v", "w_ave"]+ df1["a"].mul(df2.loc["q", "a"])+df1["b"].mul(df2.loc["q", "b"])+df1["c"].mul(df2.loc["q", "c"])
Outputs:
id k a b c w_ave
0 1 2 1 5 1 16
1 2 3 0 1 0 4
2 3 6 1 1 0 5
3 4 1 0 5 0 12
4 5 1 1 5 0 13
I have two pd df and I want to merge df2 to each row of df1 based on the ID in df1. The final df should look like in df3.
How do I do it? I tried merge, join and concat and didn't get want I wanted.
df1
ID Division
1 10
2 2
3 4
... ...
df2
Product type Level
1 0
1 1
1 2
2 0
2 1
2 2
2 3
df3
ID Product type Level Division
1 1 0 10
1 1 1 10
1 1 2 10
1 2 0 10
1 2 1 10
1 2 2 10
1 2 3 10
and repeat for ID 2 and ......
Looks like you are looking for a Cartesian product of two dataframes. The following approach should achieve what you want,
(df1.assign(key=1)
.merge(df2.assign(key=1))
.drop('key', axis=1))
Consider such an option:
set index in both DataFrames to 0,
perform an outer join (on indices, so the result is just the Cartesian
product),
reset index.
The code to do it is:
df1.index = [0] * df1.index.size
df2.index = [0] * df2.index.size
result = df1.join(df2, how='outer').reset_index(drop=True)
I have a pandas dataframe like this:
df = pd.DataFrame({'a':['A','A','A','B','B'],
'b':['Alabama','Alabama','Antioch','Brisbane','Boolean']})
I want to add two new columns say 'n1' and 'n2' so that all rows belonging to first group of column 'a' will have value 1 and next group will have value 2 in new column 'n1'. Same logic for column 'b' and 'n2' but nested within grouping of 'a'
The resulting dataframe should look like this (for some reason markdown table is not working hence a link):
How do I do this?
You can use pandas.factorize to Encode input values as an enumerated type or categorical variable; To get n2 column, just group by a and factorize b:
import pandas as pd
df['n1'] = pd.factorize(df.a)[0] + 1
df['n2'] = df.groupby('a').b.transform(lambda x: pd.factorize(x)[0] + 1)
df
# a b n1 n2
#0 A Alabama 1 1
#1 A Alabama 1 1
#2 A Antioch 1 2
#3 B Brisbane 2 1
#4 B Boolean 2 2
Or using astype category
df['n1'] = df.a.astype('category').cat.codes.add(1)
df['n2'] = df.groupby('a').b.transform(lambda x: x.astype('category').cat.codes.add(1))
df
Out[1254]:
a b n1 n2
0 A Alabama 1 1
1 A Alabama 1 1
2 A Antioch 1 2
3 B Brisbane 2 2
4 B Boolean 2 1