Python: map four arrays

Question

What I want to do is somehow map two 2D arrays. I have two pairs of corresponding arrays:

x1=[1,2,3,4,5]
y1=[7,9,10,17,4]

x2=[3,4,5,6,7]
y2=[5,4,1,13,12]

What I would like my program to do is:

Step 1. compare x1 with x2 and find matching elements

[3,4,5]

Step 2. For those matching elements perform a substraction of matching y-elements:

[y1(x)-y2(x)]=[y1(3)-y2(3), y1(4)-y2(4), y1(5)-y2(5)]

which is the same as: [10-5, 17-4, 4-1]=[5,13,3]

Step 3. Return another two arrays, that has unused elements from x1, y1, x2, y2 and those after substraction. Expected result:

x = [1,2,3,4,5,6,7] y = [7,9,5,13,3,13,12]

Is there an easy way to do something like that, maybe using map()? Those are all separate arrays.

@azro I have no idea how to start, due to the fact there are four arrays. I'm trying to zip() them both into dictionaries and maybe map them together then? Not sure if it's gonna work, maybe there's a better idea to do that. — n_inspring
– n_inspring, Commented Apr 12, 2020 at 19:03
What the values of x and y for the 3rd point ? do the steps so we understand — azro
– azro, Commented Apr 12, 2020 at 19:05
@azro actually the steps are done above. Not sure how would I describe them better than this. — n_inspring
– n_inspring, Commented Apr 12, 2020 at 19:09
The step to get the array x and y of the end, there is no step explained, just a sentence, bu I think i get it — azro
– azro, Commented Apr 12, 2020 at 19:09

azro · Accepted Answer · 2020-04-12 19:06:56Z

1

To get common elements between 2 lists, use set and operator &

x_common_vals = set(x1) & set(x2)
print(x_common_vals) #{3, 4, 5}

For the substraction operation, for each value you need to get its index in x1 and x2 and get the value at that index in y1 and y2, like y1[x1.index(value)]
```
y_sub = list(map(lambda v: y1[x1.index(v)] - y2[x2.index(v)], x_common_vals))
print(y_sub)  # [5, 13, 3]
```
WorkInProgress

answered Apr 12, 2020 at 19:06

azro

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Samer Ayoub · Accepted Answer · 2020-04-12 22:09:39Z

1

print( [ y1[x1.index(elt)]-y2[x2.index(elt)] for elt in x1 if elt in x2 ] )

To get all keys, but not very efficient

[ y1[x1.index(elt)]-y2[x2.index(elt)] if elt in x2 else y1[x1.index(elt)] for elt in x1 ]+[ y2[x2.index(elt)] for elt in x2 if elt not in x1 ]

Another method for both parts, using dictionaries to be more efficient

d1 = {x:y for x,y in zip(x1,y1)}
d2 = {x:y for x,y in zip(x2,y2)}
d = d2.copy()
d = { k:(d1[k]-v if k in d1 else v) for k,v in d.items() }
res = d1.copy()
res.update(d)
X, Y = list(res.keys()), list(res.values())
print(X, Y, sep="\n")

If you have huge date, then use pandas dataframes

import pandas as pd

df1 = pd.DataFrame({"k":x1, "v1":y1}).set_index("k")
df2 = pd.DataFrame({"k":x2, "v2":y2}).set_index("k")

def f(row):
    a, b = row[0], row[1]
    if a and b:
        return a - b
    elif a:
        return a
    else:
        return b

df = df1.join(other=df2, how="outer").fillna('')
df = pd.DataFrame({"v":df.apply(f, axis=1)} )
print(df)

edited Apr 12, 2020 at 22:09

answered Apr 12, 2020 at 19:31

Samer Ayoub

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1 Comment

azro Over a year ago

Please when you want to show the result of you code, 1.an image is the not the best, and also your captures show also the code (so twice), and the result can be pasted as code-text below the code, you won't be more trusted if you show picture ;)

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