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0

I'm trying to reduce rows in an excel since I have repetetive data for all the columns/variables, except for 1 column - "link_id"

I have this:

A kk 323 11 44 linkA
A kk 323 11 44 linkB
A pp 444 22 88 linkZ

I would like this:

A kk 323 11 44 linkA; linkB
A pp 444 22 88 linkZ 

data_file = pd.read_excel("entrada.xlsx")
red = data_file["RED"]
tipus = data_file["TIPUS"]
label = data_file["Label"]
idnode = data_file["IDNode"]
agrupacio = data_file["AGRUPACIO"]
sumofshape = data_file["SumOfShape_Area"]
sumofarea = data_file["SumOfAREA_NETA"]
minofnode = data_file["MinOfNOD_PDinamica"]
nodQ = data_file["NOD_Q"]
link_id = data_file["link_id"]

my_data =[red, tipus, label, idnode, agrupacio, sumofshape, sumofarea, minofnode, nodQ, link_id]
df = pd.concat(my_data, axis=1)

python pandas jupyter-notebook

share|improve this question

edited Mar 27 at 18:40

Patrícia Almeida

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

add a comment | 

0

I'm trying to reduce rows in an excel since I have repetetive data for all the columns/variables, except for 1 column - "link_id"

I have this:

A kk 323 11 44 linkA
A kk 323 11 44 linkB
A pp 444 22 88 linkZ

I would like this:

A kk 323 11 44 linkA; linkB
A pp 444 22 88 linkZ 

data_file = pd.read_excel("entrada.xlsx")
red = data_file["RED"]
tipus = data_file["TIPUS"]
label = data_file["Label"]
idnode = data_file["IDNode"]
agrupacio = data_file["AGRUPACIO"]
sumofshape = data_file["SumOfShape_Area"]
sumofarea = data_file["SumOfAREA_NETA"]
minofnode = data_file["MinOfNOD_PDinamica"]
nodQ = data_file["NOD_Q"]
link_id = data_file["link_id"]

my_data =[red, tipus, label, idnode, agrupacio, sumofshape, sumofarea, minofnode, nodQ, link_id]
df = pd.concat(my_data, axis=1)

python pandas jupyter-notebook

share|improve this question

edited Mar 27 at 18:40

Patrícia Almeida

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

add a comment | 

I'm trying to reduce rows in an excel since I have repetetive data for all the columns/variables, except for 1 column - "link_id"

I have this:

A kk 323 11 44 linkA
A kk 323 11 44 linkB
A pp 444 22 88 linkZ

I would like this:

A kk 323 11 44 linkA; linkB
A pp 444 22 88 linkZ 

data_file = pd.read_excel("entrada.xlsx")
red = data_file["RED"]
tipus = data_file["TIPUS"]
label = data_file["Label"]
idnode = data_file["IDNode"]
agrupacio = data_file["AGRUPACIO"]
sumofshape = data_file["SumOfShape_Area"]
sumofarea = data_file["SumOfAREA_NETA"]
minofnode = data_file["MinOfNOD_PDinamica"]
nodQ = data_file["NOD_Q"]
link_id = data_file["link_id"]

my_data =[red, tipus, label, idnode, agrupacio, sumofshape, sumofarea, minofnode, nodQ, link_id]
df = pd.concat(my_data, axis=1)

python pandas jupyter-notebook

share|improve this question

edited Mar 27 at 18:40

Patrícia Almeida

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

A kk 323 11 44 linkA
A kk 323 11 44 linkB
A pp 444 22 88 linkZ

A kk 323 11 44 linkA; linkB
A pp 444 22 88 linkZ 

data_file = pd.read_excel("entrada.xlsx")
red = data_file["RED"]
tipus = data_file["TIPUS"]
label = data_file["Label"]
idnode = data_file["IDNode"]
agrupacio = data_file["AGRUPACIO"]
sumofshape = data_file["SumOfShape_Area"]
sumofarea = data_file["SumOfAREA_NETA"]
minofnode = data_file["MinOfNOD_PDinamica"]
nodQ = data_file["NOD_Q"]
link_id = data_file["link_id"]

my_data =[red, tipus, label, idnode, agrupacio, sumofshape, sumofarea, minofnode, nodQ, link_id]
df = pd.concat(my_data, axis=1)

share|improve this question

edited Mar 27 at 18:40

Patrícia Almeida

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

share|improve this question

edited Mar 27 at 18:40

Patrícia Almeida

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

asked Mar 27 at 17:38

Patrícia AlmeidaPatrícia Almeida

522 bronze badges

1 Answer
1

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oldest

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0

Using dummy columns names, and letting df denote your dataframe:

df = pd.DataFrame([['A', 'kk', '323', '11', '44', 'linkA' ],
['A', 'kk', '323', '11', '44', 'linkB' ],
['A', 'pp', '444', '22', '88', 'linkZ' ]])
df.columns = ['A', 'B', 'C', 'D', 'E', 'link_id']

Assuming you want all the data to match before combining links:

def col_to_list(df): 
 df['link_id2'] = [df['link_id'].tolist()]*len(df)
 return df

df = df.groupby([i for i in df.columns if i not in 'link_id']).apply(lambda df: col_to_list(df))
df = df.drop(columns = ['link_id']).drop_duplicates(subset= [i for i in df.columns if i not in 'link_id2']).reset_index(drop = True)
df

Output:

 A B C D E link_id2
0 A kk 323 11 44 [linkA, linkB]
1 A pp 444 22 88 [linkZ]

share|improve this answer

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges

add a comment | 

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0

Using dummy columns names, and letting df denote your dataframe:

df = pd.DataFrame([['A', 'kk', '323', '11', '44', 'linkA' ],
['A', 'kk', '323', '11', '44', 'linkB' ],
['A', 'pp', '444', '22', '88', 'linkZ' ]])
df.columns = ['A', 'B', 'C', 'D', 'E', 'link_id']

Assuming you want all the data to match before combining links:

def col_to_list(df): 
 df['link_id2'] = [df['link_id'].tolist()]*len(df)
 return df

df = df.groupby([i for i in df.columns if i not in 'link_id']).apply(lambda df: col_to_list(df))
df = df.drop(columns = ['link_id']).drop_duplicates(subset= [i for i in df.columns if i not in 'link_id2']).reset_index(drop = True)
df

Output:

 A B C D E link_id2
0 A kk 323 11 44 [linkA, linkB]
1 A pp 444 22 88 [linkZ]

share|improve this answer

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges

add a comment | 

0

Using dummy columns names, and letting df denote your dataframe:

df = pd.DataFrame([['A', 'kk', '323', '11', '44', 'linkA' ],
['A', 'kk', '323', '11', '44', 'linkB' ],
['A', 'pp', '444', '22', '88', 'linkZ' ]])
df.columns = ['A', 'B', 'C', 'D', 'E', 'link_id']

Assuming you want all the data to match before combining links:

def col_to_list(df): 
 df['link_id2'] = [df['link_id'].tolist()]*len(df)
 return df

df = df.groupby([i for i in df.columns if i not in 'link_id']).apply(lambda df: col_to_list(df))
df = df.drop(columns = ['link_id']).drop_duplicates(subset= [i for i in df.columns if i not in 'link_id2']).reset_index(drop = True)
df

Output:

 A B C D E link_id2
0 A kk 323 11 44 [linkA, linkB]
1 A pp 444 22 88 [linkZ]

share|improve this answer

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges

add a comment | 

Using dummy columns names, and letting df denote your dataframe:

df = pd.DataFrame([['A', 'kk', '323', '11', '44', 'linkA' ],
['A', 'kk', '323', '11', '44', 'linkB' ],
['A', 'pp', '444', '22', '88', 'linkZ' ]])
df.columns = ['A', 'B', 'C', 'D', 'E', 'link_id']

Assuming you want all the data to match before combining links:

def col_to_list(df): 
 df['link_id2'] = [df['link_id'].tolist()]*len(df)
 return df

df = df.groupby([i for i in df.columns if i not in 'link_id']).apply(lambda df: col_to_list(df))
df = df.drop(columns = ['link_id']).drop_duplicates(subset= [i for i in df.columns if i not in 'link_id2']).reset_index(drop = True)
df

Output:

 A B C D E link_id2
0 A kk 323 11 44 [linkA, linkB]
1 A pp 444 22 88 [linkZ]

share|improve this answer

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges

df = pd.DataFrame([['A', 'kk', '323', '11', '44', 'linkA' ],
['A', 'kk', '323', '11', '44', 'linkB' ],
['A', 'pp', '444', '22', '88', 'linkZ' ]])
df.columns = ['A', 'B', 'C', 'D', 'E', 'link_id']

def col_to_list(df): 
 df['link_id2'] = [df['link_id'].tolist()]*len(df)
 return df

df = df.groupby([i for i in df.columns if i not in 'link_id']).apply(lambda df: col_to_list(df))
df = df.drop(columns = ['link_id']).drop_duplicates(subset= [i for i in df.columns if i not in 'link_id2']).reset_index(drop = True)
df

 A B C D E link_id2
0 A kk 323 11 44 [linkA, linkB]
1 A pp 444 22 88 [linkZ]

share|improve this answer

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges

answered Mar 27 at 20:29

Parmandeep ChaddhaParmandeep Chaddha

9633 bronze badges