CODEWITHSUNDEEP
pythonpython-3.xdictionary
Mandie Driskill
Mandie Driskill

Reputation: 83

Replacing items in a list in a dictionary with items from another dictionary

I have the following dictionaries:

d1 = {"00f_5" :[1,2,3], "00f_6": [1,2,3]}

d2 = [{"marker":"00f_5",1: 'AAA'},{"marker":"00f_6", 1: 'CCC'},{"marker":"00f_5", 2:"AAC"}]

I would like the following output:

d1 = {"00f_5" :["AAA","AAC",3], "00f_6": ["CCC",2,3]}

I have tried multiple attempts, but I couldn't get it. Any help would be much appreciated.

Upvotes: 1

Views: 87

Answers (3)

pylang
pylang

Reputation: 44505

I would not suggest overwriting the first dict. Just make a new one (d3).

Given

d1 = {"00f_5" :[1,2,3], "00f_6": [1,2,3]}

lst = [
    {"marker":"00f_5",1: 'AAA'}, 
    {"marker":"00f_6", 1: 'CCC'},
    {"marker":"00f_5", 2:"AAC"}
]

Code

d3 = {}
for d in lst:
    key = d["marker"]
    value = d1[key]
    for k, v in d.items():     
    if not isinstance(k, int):                         # skip to numeric keys
            continue
        if key not in d3:                              # fill missing entries
            d3[key] = value
        idx = k - 1
        d3[key][idx] = v                               # overwrite list item

d3
{'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]}

Upvotes: 0

blhsing
blhsing

Reputation: 106553

You can build a dict that maps the marker to a sub-dict that maps numeric keys to the 3-letter codes, so that given a marker and a numeric key, you can use dict.get method to get the mapped code if it exists:

d = {}
for s in d2:
    d.setdefault(s['marker'], {}).update(s)
d1 = {k: [d[k].get(i, i) for i in l] for k, l in d1.items()}

so that given:

d1 = {"00f_5" :[1,2,3], "00f_6": [1,2,3]}
d2 = [{"marker":"00f_5",1: 'AAA'},{"marker":"00f_6", 1: 'CCC'},{"marker":"00f_5", 2:"AAC"}]

d1 will become:

{'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]}

Upvotes: 3

2xB
2xB

Reputation: 296

Although I will not give you the full program, you might try to understand and play with the following code:

d2 = [{"marker":"00f_5",1: 'AAA'},{"marker":"00f_6", 1: 'CCC'},{"marker":"00f_5", 2:"AAC"}]

for d in d2:
  for key, value in d.items():
    if key == "marker":
      marker = value
    else:
      reference = value
      content = key
  print("Marker: {}, Reference: {}, Content: {}".format(marker, reference, content))

Upvotes: 0

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