counting NaNs in 'for' loop in python - python-3.x

Question

I am trying to iterate over the rows in df and count consecutive rows when a certain value is NaN or 0 and start the count over if the value will change from NaN or 0. I would like to get something like this:
Value Period
0 1
0 2
0 3
NaN 4
21 NaN
4 NaN
0 1
0 2
NaN 3
I wrote the function which takes a dataframe as an argument and returns it with an additional column which denotes the count:
def calc_period(df):
period_x = []
sum_x = 0
for i in range(1,df.shape[0]):
if df.iloc[i,0] == np.nan or df.iloc[i,0] == 0:
sum_x += 1
period_x.append(sum_x)
else:
period_x.append(None)
sum_x = 0
period_x.append(sum_x)
df['period_x'] = period_x
return df
The function works well when the value is 0. But when the value is NaN the count is also NaN and I get the following result:
Value Period
0 1
0 2
0 3
NaN NaN
NaN NaN

Answer 1

Here is a revised version of your code:
import pandas as pd
import numpy as np
import math
def is_nan_or_zero(val):
return math.isnan(val) or val == 0
def calc_period(df):
is_first_nan_or_zero = is_nan_or_zero(df.iloc[0, 0])
period_x = [1 if is_first_nan_or_zero else np.nan]
sum_x = 1 if is_first_nan_or_zero else 0
for i in range(1,df.shape[0]):
val = df.iloc[i,0]
if is_nan_or_zero(val):
sum_x += 1
period_x.append(sum_x)
else:
period_x.append(None)
sum_x = 0
df['period_x'] = period_x
return df
There were 2 fixes:
Replacing df.iloc[i,0] == np.nan with math.isnan(val)
Remove period_x.append(sum_x) at the end, and add the first period value instead (since we start iterating from the second value)