How to compare two list of dictionaries for the key exist in one and not in another

Question

I have 2 list of dictionary
parent list pa = [{'key1': '1A', 'key2': 2},{'key1': '1B', 'key3': 2},{'key1': '1C', 'key4': 2}]
child list ch = [{'key1': '1A', 'keyvalue2': 2},{'key1': '1B', 'keyvalue23': 2},{'key1': '1D', 'key4': 2}]
I need to get the key1 which is present in pa and not in ch

Code is below

for each in pa:
    for each_1 in ch:
        if each['key1'] not in each_1['key1']:
            print(each['key1'])

My out is

1A
1A
1B
1B
1C
1C
1C

My expected out is 1C which present in pa and not in ch

I have created two list and compare which is working. but it’s not the case like code below. I don’t want to create a extra space

pa_list = []
for each in pa:
  if each['key1'] not in pa_list:
    pa_list.append(each['key1'])

ch_list = []
for each in ch:
  if each['key1'] not in ch_list:
    ch_list.append(each['key1'])

[i for i in pa_list if i not in ch_list]

Asked By: abd

||

Source

Answer 1

Starting from your original code, I suggest you to use the for/else loop, not very well-known, but quite useful in your case:

pa = [{'key1': '1A', 'key2': 2},{'key1': '1B', 'key3': 2},{'key1': '1C', 'key4': 2}]
ch = [{'key1': '1A', 'keyvalue2': 2},{'key1': '1B', 'keyvalue23': 2},{'key1': '1D', 'key4': 2}]

for i in pa:
    for j in ch:
        if (i['key1'] == j['key1']):
            break
    else:
        print(i['key1'])

It outputs only ‘1C’ without intermediate variable, as expected.

Answered By: Laurent H.

How to compare two list of dictionaries for the key exist in one and not in another

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