CODEWITHSUNDEEP
pythonlistfilter
pirateofebay
pirateofebay

Reputation: 1090

Filter Multiple Lists for Element

I have 2 lists containing JSON values, allBuilds and rawRequirements. Each list has a length of 1201 items.

The contents of allBuilds looks like this:

{'id': 19553, 'buildTypeId': 'AlpsWeb_AlpsWebDeployWebDev'}
{'id': 24456, 'buildTypeId': 'AlpsWeb_AlpsWebDeployWebDevTomcat8', }
None
None
{'id': 19356, 'buildTypeId': 'AlpsWeb_AlpsWebDeployWebQa',}
{'id': 19357, 'buildTypeId': 'AlpsWeb_AlpsWebDeployWebQa',}
None
{'id': 19357, 'buildTypeId': 'AlpsWeb_AlpsWebDeployWebQa',}

I need to filter the contents of allBuilds to remove None while also removing the corresponding element from rawRequirements. I tried

successfulBuilds = list(filter(None, allBuilds))

This removes None but now the length of the new list successfulBuilds is 972 while rawRequirements is still 1201.

How can I filter allBuilds for None and also remove the corresponding list element in rawRequirements?

---- EDIT ----

Here's an example of what I'm trying to do:

allBuilds |  rawRequirements
id        |  Requirement value1
none      |  Requirement value2 <-- 
none      |  Requirement value3 <-- 
id        |  Requirement value4
none      |  Requirement value5 <-- 
id        |  Requirement value6

I need to remove the element in rawRequirements at the same column where "none" exists in allBuilds

Upvotes: 0

Views: 134

Answers (3)

STreu
STreu

Reputation: 128

You can use zip:

zipped =  zip(allBuilds,rawRequirements)
filtered_zip =  ((el1, el2) for ab, rR in zipped if el1 != None))
filtered_allBuilds, filteredRequirements = zip(*filtered_zip)

Upvotes: 2

j-i-l
j-i-l

Reputation: 10957

I recommend using zip:

allBuilds, rawRequirements = zip(*((ab, rR) for ab, rR in zip(allBuilds, rawRequirements) if ab is not None))

zip takes some getting used to but it is very efficient for filtering or sorting multiple lists.

Upvotes: 3

sahasrara62
sahasrara62

Reputation: 11228

you can use this

a=[1,2,3,None,4,5,6,7,None,10]
b = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

new_a =[]
new_b =[]

for i, v in enumerate(a):
    if v is not None:
        new_a.append(v)
        new_b.append(b[i])

print(new_a,new_b,sep='\n')

output

[1, 2, 3, 4, 5, 6, 7, 10] #a
[0, 1, 2, 4, 5, 6, 7, 9] #b

Upvotes: 2

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