计算字典列表中特定字典值的出现次数并使用该计数创建一个新字典

Question

I have a list of dictionaries and what I want to do is to look for an specific value in all the dictionaries that my list have. Then I want to create a new dictionary with that value.

Input:

{ "Element": ["Box"]
  "Element2": ["Pen"]
}

{ "Element": ["Box"]
  "Element2": ["Pencil"]
}

]

Expected Output:

[
{ There are two box 
 There is one pen 
There is one pencil 
}
{ "Element": ["Box"]
  "Element2": ["Pen"]
}

{ "Element": ["Box"]
  "Element2": ["Pencil"]
}

] ```

How would you do that? Thanks in advance

Answer 1

To simplify the code, the values of the dictionaries are strings instead of list here, but maybe some snippet helps you.

from functools import reduce

dictlist=[{ "Element": "Box", "Element2": "Pen"},{ "Element": "Box","Element2": "Pencil"}]

vals=reduce(lambda x,y:x+y,[list(dic.values()) for dic in dictlist])
for val in set(vals):
    print(f"There is {vals.count(val)} {val}")

Outputs:

There is 1 Pen
There is 1 Pencil
There is 2 Box

Answer 2

You can simply get a list of all the values using dict.values() , and the use .count(text) to count the amount of occurrences of one type of element. Counting multiple ones might be what you want though.

Let's use your example.

elements = [
{ "Element": ["Box"], "Element2": ["Pen"]},
{ "Element": ["Box"], "Element2": ["Pencil"]}]

total = []
for i in elements:
    for j in i.values():
        total.extend(j)

counts = {}
for i in total:
    if i in counts.keys():
        counts[i]+=1
    else:
        counts[i] = 1

print(counts)

EDIT: After seeing Luka's request about an element3 NOT being added to the count, here is the adjusted code, with a list of blocked elements.

elements = [
{ "Element": ["Box"], "Element2": ["Pen"], "Element3": ["Scissor"]},
{ "Element": ["Box"], "Element2": ["Pencil"], "Element3": ["Highlighter"]}]

blocked = ["Element3"]
total = []
for i in elements:
    for key, value in i.items():
        if key not in blocked:
            total.extend(value)

counts = {}
for i in total:
    if i in counts.keys():
        counts[i]+=1
    else:
        counts[i] = 1

print(counts)