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Accepted
Piero Pajares
Asked: 2020-05-29 14:33:25 +0800 CST 2020-05-29 14:33:25 +0800 CST

Loop through dynamic object list in python

  • 772

I come from javascript and I would like to know how I can go through that array that in python is known as list.

"dispatches_types": [
  {
    "status_id": 2
  },
  {
    "status_id": 1
  },
  {
    "status_id": 1
  },
  {
    "status_id": 1
  },
  {
    "status_id": 2
  }
 ]

The problem I have is that I want to add a counter that increases when status_idit is 1 but this is dynamic since sometimes it can bring that status and sometimes not. Sometimes it can even bring up the empty list.

So far I am trying this way:

    cont = 0
    if len(dispatch_types) > 0:
        for status1 in dispatch_types:
            if 1 in status1:
                cont +=1
    else:
        cont = 0
python

2 Answers

  1. Best Answer

    There are several forms that are more "phytonic", or at least that take better advantage of the language tools, also collaterally tend to be easier to read. First of all, your example is a dictionary whose key "dispatches_types"contains a list of other dictionaries. Something like this:

    d = {
  "dispatches_types": [
     {"status_id": 2},
     {"status_id": 1},
     {"status_id": 1},
     {"status_id": 1},
     {"status_id": 2}
 ]
}

    What you are looking for is to determine how many status_id == 1are in the mentioned list. One way is as follows:

    print(sum([1 for e in d["dispatches_types"] if e["status_id"]==1]))

3
    • First, with [1 for e in d["dispatches_types"] if e["status_id"]==1]we use a list comprehension to construct a new list with a value 1for every time we havestatus_id == 1
    • And with sum()we simply add the list.
    • This form is consistent with an empty list or with non- status_id == 1existent ones, in both cases it will return0

    Also. you may find it interesting, the use of the classCounter()

    from collections import Counter
print(Counter(e["status_id"] for e in d["dispatches_types"]))

Counter({1: 3, 2: 2})

    He Counterwill return you from the list e["status_id"] for e in d["dispatches_types"]. a dictionary-like object, with the occurrences for eachstatus_id

    • 1

  2. I solved it this way but I still have doubts:

            cont = 0
        if len(dispatch_types) > 0:
            for status1 in dispatch_types:
                if status1["status_id"] == 1:
                    cont +=1
                    print(cont)
        else:
            cont = 0
            print(cont)
    • 0