How to check if list element is present in array in python

Question

I have a list

bulk_order
Out[230]: 
[3    523
Name: order_id, dtype: object]

And I have an array. Its a series but I am accessing it with values

clusters_order_ids.values
Out[231]: 
array([['520', '521', '524', '527', '532'], ['528'], ['531'],
   ['525', '526', '533'], ['519', '523', '529', '534', '535'], ['530']],     
dtype=object)

Now I want to check whether list element 523 is present in above array and if it is there I want to delete it.

I am doing following in python

bulk_order in clusters_order_ids.values

But It gives me False output

clusters_order_ids.values is a list of lists, not numbers. — tglaria
– tglaria, Commented Jan 14, 2016 at 16:02
Use any() to just check if the element exists. any(bulk_order in id_list for id_list in clusters_order_ids) — Prajwal K R
– Prajwal K R, Commented Jan 14, 2016 at 16:20

stoffen · Accepted Answer · 2016-01-14 17:23:24Z

3

Your list is not a list, but a list of lists.

If you want to delete the whole list containing something in list ['523']:

orders = [['520', '521', '524', '527', '532'], ['528'], ['531'], ['525', '526', '533'], ['519', '523', '529', '534', '535'], ['530']]
remove_order_with_ids = ['523'] # or bulk_order 
orders = [order for order in orders if not set(remove_order_with_ids).intersection(set(order))]
print orders
# [['520', '521', '524', '527', '532'], ['528'], ['531'], ['525', '526', '533'], ['530']]

If you only want to delete items in ['523'] from the inner list(s):

orders = [['520', '521', '524', '527', '532'], ['528'], ['531'], ['525', '526', '533'], ['519', '523', '529', '534', '535'], ['530']]
remove_order_with_id = ['523'] # or bulk_order 
new_orders = []
for order in orders:
    new_orders.append([item for item in order if item not in remove_order_with_id])
print new_orders
# [['520', '521', '524', '527', '532'], ['528'], ['531'], ['525', '526', '533'], ['519', '529', '534', '535'], ['530']]

edited Jan 14, 2016 at 17:23

answered Jan 14, 2016 at 16:24

stoffen

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7 Comments

Neil Over a year ago

It doesn't delete either. neither entire list containing 523 nor 523

Neil Over a year ago

Works for remove_order_with_id = '523' but not for remove_order_with_id = bulk_order

stoffen Over a year ago

Could it be that you need to access the order_id from the bulk order: remove_order_with_id = bulk_order.order_id?

Neil Over a year ago

bulk_order is a list.

stoffen Over a year ago

Is bulk_order a list of order_ids, also containing the number 3? In that case, what should happen if the array contains 3?

|

Community · Accepted Answer · 2017-05-23 11:45:01Z

3

Try this (from Making a flat list out of list of lists in Python):

l = clusters_order_ids.values
out = [item for sublist in l for item in sublist]
print (bulk_order in out)

For the deletion, you'd have to enter on each list so:

for sublist in clusters_order_ids.values:
    if bulk_order in sublist:
        sublist.remove(bulk_order)
        if not sublist:
            #do something to remove the empty list
        break;

edited May 23, 2017 at 11:45

CommunityBot

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answered Jan 14, 2016 at 16:03

tglaria

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3 Comments

Prajwal K R Over a year ago

Use any() to just check if the element exists. any(bulk_order in id_list for id_list in clusters_order_ids)

Neil Over a year ago

@tglaria It gives me output as False

tglaria Over a year ago

@user2927983 What gives you output as False?

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