Cumsum entire table and reset at zero - python-2.7

Question

I have following data frame.
d = pd.DataFrame({'one' : [0,1,1,1,0,1],'two' : [0,0,1,0,1,1]})
d
one two
0 0 0
1 1 0
2 1 1
3 1 0
4 0 1
5 1 1
I want cumulative sum which resets at zero
desired output should be
pd.DataFrame({'one' : [0,1,2,3,0,1],'two' : [0,0,1,0,1,2]})
one two
0 0 0
1 1 0
2 2 1
3 3 0
4 0 1
5 1 2
i have tried using group by but it does not work for entire table.

Answer 1

df2 = df.apply(lambda x: x.groupby((~x.astype(bool)).cumsum()).cumsum())
print(df2)
Output:
one two
0 0 0
1 1 0
2 2 1
3 3 0
4 0 1
5 1 2

Answer 2

pandas
def cum_reset_pd(df):
csum = df.cumsum()
return (csum - csum.where(df == 0).ffill()).astype(d.dtypes)
cum_reset_pd(d)
one two
0 0 0
1 1 0
2 2 1
3 3 0
4 0 1
5 1 2
numpy
def cum_reset_np(df):
v = df.values
z = np.zeros_like(v)
j, i = np.where(v.T)
r = np.arange(1, i.size + 1)
p = np.where(
np.append(False, (np.diff(i) != 1) | (np.diff(j) != 0))
)[0]
b = np.append(0, np.append(p, r.size))
z[i, j] = r - b[:-1].repeat(np.diff(b))
return pd.DataFrame(z, df.index, df.columns)
cum_reset_np(d)
one two
0 0 0
1 1 0
2 2 1
3 3 0
4 0 1
5 1 2
Why go through this trouble?
because it's quicker!

Answer 3

This one is without using Pandas, but using NumPy and list comprehensions:
import numpy as np
d = {'one': [0,1,1,1,0,1], 'two': [0,0,1,0,1,1]}
out = {}
for key in d.keys():
l = d[key]
indices = np.argwhere(np.array(l)==0).flatten()
indices = np.append(indices, len(l))
out[key] = np.concatenate([np.cumsum(l[indices[n-1]:indices[n]]) \
for n in range(1, indices.shape[0])]).ravel()
print(out)
First, I find all occurences of 0 (positions to split the lists), then I calculate cumsum of the resulting sublists and insert them into a new dict.

Answer 4

This should do it:
d = {'one' : [0,1,1,1,0,1],'two' : [0,0,1,0,1,1]}
one = d['one']
two = d['two']
i = 0
new_one = []
for item in one:
if item == 0:
i = 0
else:
i += item
new_one.append(i)
j = 0
new_two = []
for item in two:
if item == 0:
j = 0
else:
j += item
new_two.append(j)
d['one'], d['two'] = new_one, new_two
df = pd.DataFrame(d)