How to fill in discrepancies between two lists with an “NA”

Question

I have a bunch of c and e python lists that I have to compare, in which case the e list length is always greater than or equal to the c list. What I want to do is compare these two lists and, if their lengths are not equal, I want to fill in the c list "gaps" with an "NA".

For instance, if we look at these two lists:

e = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', '14']
c = ['2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13']

I'd want the c list to fill in an "NA" for the values it's missing (and preserve the order), like so:

c = ['NA', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', 'NA']

Answer 1

I'd use pandas and pd.Series.where for the mask. First convert them to series e = pd.Series(e) , then

s = pd.concat([e,c],sort=True).drop_duplicates() #remove sort=True for versions < 0.23
s.where(s.isin(c))

outputs

0     NaN
1       2
2       3
3       4
4       5
5       6
6       7
7       8
8       9
9      10
10     11
11     12
12     13
13    NaN

---

You can also follow a classic, pure-python approach. Use two pointers, iterate through them simultaneously and compare the values.

p1 = 0
p2 = 0

f = []
while (p1 < len(e)) and (p2 < len(c)):

    vale = e[p1]
    valc = c[p2]

    if vale < valc: 
        f.append('NA')
        p1 += 1
    elif vale == valc:
        f.append(valc)
        p1 += 1
        p2 += 1
    else:
        p2 += 1

    if p1 == len(e): f.extend(c[p2:])
    if p2 == len(c): f.extend(['NA']*(len(e)-p1))