Question

在列表dict下方，如果我们有重复的名称，则需要获取具有指定名称的dict。如果名称是单个名称，则需要获取不具有指定名称的字典

<form class="form-style-9" 
       [formGroup]="productForm"
      (ngSubmit)="addProduct(productForm.value)"
      enctype="multipart/form-data"
      >
    <ul>
        <li><span>Product Name</span>
            <input type="text" name="productname" class="field-style field-full align-none" placeholder="Name" formControlName="productname" />
        </li>
        <span>Product Category</span><br>
        <select formControlName="productcategory">
          <option value="Clothing">Clothing</option>
          <option value="Electronics">Electronics</option>
          <option value="Books">Books</option>
          <option value="Toys">Toys</option>
        </select>
        <li><span>Product Description</span>
            <textarea name="productdescription" class="field-style" placeholder="Product Description" formControlName="productdescription"></textarea>
        </li>
        <li><span>Product Image</span>
            <input type="file" name="productimage" class="field-style field-full align-none" placeholder="Image" formControlName="productimage"/>
        </li>
        <li><span>Product Price</span>
            <input type="number" name="productprice" class="field-style field-full align-none" placeholder="Product Price" formControlName="productprice" />
        </li>
        <li>
        <input type="submit" value="Add Product" />
        </li>
    </ul>
</form>

我的输出应如下所示。

lst = [{'desig': '', 'name': 'William'}, {'desig': 'Chairman of the Board', 'name': 'William'}, {'desig': '', 'name': 'English'}, {'desig': 'Director', 'name': 'English'}, {'desig': '', 'name': 'Charles '}]

Answer 1

以下应解决您的问题。请注意，如果名称不完全相同，则将它们计为不同的名称。在您的原始帖子中，第二个项目的名称为“ William”，而第一个项目的名称为“ William”，在m后面有一个空格！

data = [{'desig': '', 'name': 'William'}, {'desig': 'Chairman of the Board', 'name': 'William'},
        {'desig': '', 'name': 'English'}, {'desig': 'Director', 'name': 'English'},
        {'desig': '', 'name': 'Charles '}]

cleaned_data = []
names_added = []
for entry in data:
    if entry['name'] in names_added:
        if entry['desig'] != '':  # the != '' is actually not necessary, but included for clarity
            i = names_added.index(entry['name'])
            cleaned_data[i]=entry
    else:
        cleaned_data.append(entry)
        names_added.append(entry['name'])

print(cleaned_data)

Answer 2

from collections import defaultdict

lst = [{'desig': '', 'name': 'William'}, {'desig': 'Chairman of the Board', 'name': 'William'}, {'desig': '', 'name': 'English'}, {'desig': 'Director', 'name': 'English'}, {'desig': '', 'name': 'Charles '}]
d = defaultdict(list)
for i in lst:
    d[i['name']].append(i['desig'])
l = [{'desig': sorted(v)[-1], 'name': k}  for k, v in d.items()]

print(l)

打印：

[{'desig': 'Chairman of the Board', 'name': 'William'}, {'desig': 'Director', 'name': 'English'}, {'desig': '', 'name': 'Charles '}]

Answer 3

>>> lst = [{'desig': '', 'name': 'William'}, {'desig': 'Chairman of the Board', 'name': 'William'}, {'desig': '', 'name': 'English'}, {'desig': 'Director', 'name': 'English'}, {'desig': '', 'name': 'Charles '}]
>>> names=set()
>>> list(dict(names.add(d['name']) or (d['name'],d) for d in lst if not d['name'] in names or d['desig']).values())
[{'desig': 'Chairman of the Board', 'name': 'William'}, {'desig': 'Director', 'name': 'English'}, {'desig': '', 'name': 'Charles '}]

根据名称删除重复列表字典

3 个答案: