The 0's and 1's need to be transposed to there appropriate headers in python.
How can I achieve this and get the column final_list?
If there is always only one 1 per rows use DataFrame.dot:
df = pd.DataFrame({'a':[0,1,0],
'b':[1,0,0],
'c':[0,0,1]})
df['Final'] = df.dot(df.columns)
print (df)
a b c Final
0 0 1 0 b
1 1 0 0 a
2 0 0 1 c
If possible multiple 1 also add separator and then remove it by Series.str.rstrip from output Series:
df = pd.DataFrame({'a':[0,1,0],
'b':[1,1,0],
'c':[1,1,1]})
df['Final'] = df.dot(df.columns + ',').str.rstrip(',')
print (df)
a b c Final
0 0 1 1 b,c
1 1 1 1 a,b,c
2 0 0 1 c
Related
I have a pandas.DataFrame that looks like this:
A B C D E F
0 0 1 0 0 0
1 1 0 0 0 0
2 0 1 0 0 0
3 0 0 0 1 0
4 0 0 1 0 0
There are several rows that share a 1 in their columns and in each row there is only one 1 present. I want to merge the rows with each other so the resulting dataFrame would onyl consist of one row, that combines all the 1s of the dataframe, like this:
A B C D E F
0 1 1 1 1 0
Is there a smart, easy way to do this with pandas?
Use DataFrame.sum, then compare for greater or equal by Series.ge and last convert to 0,1 by Series.view:
s = df.sum().ge(1).view('i1')
Another idea if 0,1 values only is use DataFrame.any with convert mask to 0,1:
s = df.any().view('i1')
print (s)
A 1
B 1
C 1
D 1
E 1
F 0
dtype: int8
We can do
df.sum().ge(1).astype(int)
Out[316]:
A 1
B 1
C 1
D 1
E 1
F 0
dtype: int32
I have a data frame something like this:
df = pd.DataFrame({"ID":[1,1,2,2,2,3,3,3,3,3],
"IF_car":[1,0,0,1,0,0,0,1,0,1],
"IF_car_history":[0,0,0,1,0,0,0,1,0,1],
"observation":[0,0,0,1,0,0,0,2,0,3]})
I want output where I can trim rows in groupby with ID and condition on "IF_car_history" == 1
tried_df = df.groupby(['ID']).apply(lambda x: x.loc[:(x['IF_car_history'] == '1').idxmax(),:]).reset_index(drop = True)
I want to drop rows in a groupby by after i get ['IF_car_history'] == '1'
expected output:
Thanks
First compare values for mask m by Series.eq and then use GroupBy.cumsum, and for values before 1 compare by 0, last filter by boolean indexing, but because id necesary remove after last 1 is used swapped values by slicing with [::-1].
m = df['IF_car_history'].eq(1).iloc[::-1]
df1 = df[m.groupby(df['ID']).cumsum().ne(0).iloc[::-1]]
print (df1)
ID IF_car IF_car_history observation
2 2 0 0 0
3 2 1 1 1
5 3 0 0 0
6 3 0 0 0
7 3 1 1 2
8 3 0 0 0
9 3 1 1 3
I need some help with comparing two pandas dataframe
I have two dataframes
The first dataframe is
df1 =
a b c d
0 1 1 1 1
1 0 1 0 1
2 0 0 0 1
3 1 1 1 1
4 1 0 1 0
5 1 1 1 0
6 0 0 1 0
7 0 1 0 1
and the second dataframe is
df2 =
a b c d
0 1 1 1 1
1 1 0 1 0
2 0 0 1 0
I want to find the row index of dataframe 1 (df1) which the entire row is the same as the rows in dataframe 2 (df2). My expect result would be
0
3
4
6
The order of the above index does not need to be in order, all I want is the index of dataframe 1 (df1)
Is there a way without using for loop?
Thanks
Tommy
You can using merge
df1.merge(df2,indicator=True,how='left').loc[lambda x : x['_merge']=='both'].index
Out[459]: Int64Index([0, 3, 4, 6], dtype='int64')
I have the following dataframe loaded up in Pandas.
print(pandaDf)
id col1 col2 col3
12a a b d
22b d a b
33c c a b
I am trying to convert the values across multiple rows into its columns so the output would be like this :
Desired output:
id a b c d
12a 1 1 0 1
22b 1 1 0 0
33c 1 1 1 0
I have tried adding in a value column where the value = 1 and using a pivot table
pandaDf['value'] = 1
column = ['col1', 'col2', 'col3']
pandaDf.pivot_table(index = 'id', value = 'value', columns = column)
However, the resulting data frame is a multilevel index and the pandaDf.pivot() method does not allow multiple column values.
Please advise about how I could do this with an output of a single level index.
Thanks for taking the time to read this and I apologize if I have made any formatting errors in posting the question. I am still learning the proper stackoverflow syntax.
You can use One-Hot Encoding to solve this problem.
Here is one way to do this pd.get_dummies and some multiindex flatten and sum:
df1 = df.set_index('id')
df_out = pd.get_dummies(df1)
df_out.columns = df_out.columns.str.split('_', expand=True)
df_out = df_out.sum(level=1, axis=1).reset_index()
print(df_out)
Output:
id a c d b
0 12a 1 0 1 1
1 22b 1 0 1 1
2 33c 1 1 0 1
Using get_dummies
pd.get_dummies(df.set_index('id'),prefix='', prefix_sep='').sum(level=0,axis=1)
Out[81]:
a c d b
id
12a 1 0 1 1
22b 1 0 1 1
33c 1 1 0 1
Given the following data frame:
import pandas as pd
df=pd.DataFrame({'A':[0,4,4,4],
'B':[0,4,4,0],
'C':[0,4,4,4],
'D':[4,0,0,4],
'E':[4,0,0,0],
'Name':['a','a','b','c']})
df
A B C D E Name
0 0 0 0 4 4 a
1 4 4 4 0 0 a
2 4 4 4 0 0 b
3 4 0 4 4 0 c
I'd like to add a new field called "Match_Flag" which labels unique combinations of rows if they have complementary zero patterns (as with rows 0, 1, and 2) AND have the same name (just for rows 0 and 1). It uses the name of the rows that match.
The desired result is as follows:
A B C D E Name Match_Flag
0 0 0 0 4 4 a a
1 4 4 4 0 0 a a
2 4 4 4 0 0 b NaN
3 4 0 4 4 0 c NaN
Caveat:
The patterns may vary, but should still be complementary.
Thanks in advance!
UPDATE
Sorry for the confusion.
Here is some clarification:
The reason why rows 0 and 1 are "complementary" is that they have opposite patterns of zeros in their columns; 0,0,0,4,4 vs, 4,4,4,0,0.
The number 4 is arbitrary; it could just as easily be 0,0,0,4,2 and 65,770,23,0,0. So if 2 such rows are indeed complementary and they have the same name, I'd like for them to be flagged with that same name under the "Match_Flag" column.
You can identify a compliment if it's dot product is zero and it's element wise sum is nowhere zero.
def complements(df):
v = df.drop('Name', axis=1).values
n = v.shape[0]
row, col = np.triu_indices(n, 1)
# ensure two rows are complete
# their sum contains no zeros
c = ((v[row] + v[col]) != 0).all(1)
complete = set(row[c]).union(col[c])
# ensure two rows do not overlap
# their product is zero everywhere
o = (v[row] * v[col] == 0).all(1)
non_overlap = set(row[o]).union(col[o])
# we are a compliment iff we do
# not overlap and we are complete
complement = list(non_overlap.intersection(complete))
# return slice
return df.Name.iloc[complement]
Then groupby('Name') and apply our function
df['Match_Flag'] = df.groupby('Name', group_keys=False).apply(complements)