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
l_mod = [0] + l + [len(df)]. Now, in this instance, max(l)+1 and len(df) coincide, but if generalised you might lose rows. And as a second note, it could be worth passing it on set to ensure that no duplicate indicies exist (like having [0] 2 times). Great solution btw, you got my upvote :)
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
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
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
groupby if splitting indices are known a priori
df.loc?