Argthtjtr

I want to split dataframe by uneven number of rows using row index.

The below code:

groups = df.groupby((np.arange(len(df.index))/l[1]).astype(int))

works only for uniform number of rows.

df



a b c  

1 1 1  

2 2 2  

3 3 3  

4 4 4  

5 5 5  

6 6 6  

7 7 7  



l = [2, 5, 7]



df1  

1 1 1  

2 2 2  



df2  

3,3,3  

4,4,4  

5,5,5  



df3  

6,6,6  

7,7,7  



df4  

8,8,8

You could use list comprehension with a little modications your list, l, first.

print(df)



   a  b  c

0  1  1  1

1  2  2  2

2  3  3  3

3  4  4  4

4  5  5  5

5  6  6  6

6  7  7  7

7  8  8  8





l = [2,5,7]

l_mod = [0] + l + [max(l)+1]



list_of_dfs = [df.iloc[l_mod[n]:l_mod[n+1]] for n in range(len(l_mod)-1)]

Output:

list_of_dfs[0]



   a  b  c

0  1  1  1

1  2  2  2



list_of_dfs[1]



   a  b  c

2  3  3  3

3  4  4  4

4  5  5  5



list_of_dfs[2]



   a  b  c

5  6  6  6

6  7  7  7



list_of_dfs[3]



   a  b  c

7  8  8  8

I think this is you are looking for.,

l = [2, 5, 7]

dfs=

i=0

for val in l:

    if i==0:

        temp=df.iloc[:val]

        dfs.append(temp)

    elif i==len(l):

        temp=df.iloc[val]

        dfs.append(temp)        

    else:

        temp=df.iloc[l[i-1]:val]

        dfs.append(temp)

    i+=1

Output:

   a  b  c

0  1  1  1

1  2  2  2

   a  b  c

2  3  3  3

3  4  4  4

4  5  5  5

   a  b  c

5  6  6  6

6  7  7  7

Another Solution:

l = [2, 5, 7]

t= np.arange(l[-1])

l.reverse()

for val in l:

    t[:val]=val

temp=pd.DataFrame(t)

temp=pd.concat([df,temp],axis=1)

for u,v in temp.groupby(0):

    print v

Output:

   a  b  c  0

0  1  1  1  2

1  2  2  2  2

   a  b  c  0

2  3  3  3  5

3  4  4  4  5

4  5  5  5  5

   a  b  c  0

5  6  6  6  7

6  7  7  7  7

Do this:

l = [2,5,7]

c = 0

d = dict()  # A dictionary to hold multiple dataframes



In [477]: for i in l:

     ...:     if c == 0:

     ...:         index_list = df[df.a <= i].index

     ...:     else:

     ...:         index_list = df[(df.a > l[c-1]) & (df.a <= l[c])].index

     ...:     min_index = index_list[0]

     ...:     max_index = index_list[-1] + 1

     ...:     d[i] = df.iloc[min_index:max_index]

     ...:     c += 1

     ...:     





In [479]: for key in d.keys():

     ...:     print(d[key])

     ...:     

   a  b  c

0  1  1  1

1  2  2  2

   a  b  c

2  3  3  3

3  4  4  4

4  5  5  5

   a  b  c

5  6  6  6

6  7  7  7

You can create an array to use for indexing via NumPy:

import pandas as pd, numpy as np



df = pd.DataFrame(np.arange(24).reshape((8, 3)), columns=list('abc'))



L = [2, 5, 7]

idx = np.cumsum(np.in1d(np.arange(len(df.index)), L))



for _, chunk in df.groupby(idx):

    print(chunk, 'n')



   a  b  c

0  0  1  2

1  3  4  5 



    a   b   c

2   6   7   8

3   9  10  11

4  12  13  14 



    a   b   c

5  15  16  17

6  18  19  20 



    a   b   c

7  21  22  23 

Instead of defining a new variable for each dataframe, you can use a dictionary:

d = dict(tuple(df.groupby(idx)))



print(d[1])  # print second groupby value



    a   b   c

2   6   7   8

3   9  10  11

4  12  13  14

I think this is what you need:

df = pd.DataFrame({'a': np.arange(1, 8),

                  'b': np.arange(1, 8),

                  'c': np.arange(1, 8)})

df.head()

    a   b   c

0   1   1   1

1   2   2   2

2   3   3   3

3   4   4   4

4   5   5   5

5   6   6   6

6   7   7   7



last_check = 0

dfs = 

for ind in [2, 5, 7]:

    dfs.append(df.loc[last_check:ind-1])

    last_check = ind

Although list comprehension are much more efficient than a for loop, the last_check is necessary if you don't have a pattern in your list of indices.

dfs[0]



    a   b   c

0   1   1   1

1   2   2   2



dfs[2]



    a   b   c

5   6   6   6

6   7   7   7