how to get the value multiple array in a dictonary

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Given:

[{'chin': [(297, 322), (297, 339), (299, 356), (301, 373), (305, 390), (312, 405), (325, 415), (342, 418),
                 (362, 420), (383, 421), (402, 421), (418, 415), (427, 403), (433, 387), (437, 370), (440, 352),
                 (443, 335)],
        'left_eyebrow': [(306, 296), (313, 282), (326, 276), (342, 276), (357, 281)],
        'right_eyebrow': [(378, 283), (393, 278), (410, 281), (423, 289), (431, 303)],
        'nose_bridge': [(366, 297), (365, 305), (364, 314), (363, 323)],
        'nose_tip': [(347, 337), (355, 340), (364, 341), (373, 340), (382, 339)],
        'left_eye': [(321, 304), (328, 296), (339, 296), (349, 304), (339, 305), (328, 305)],
        'right_eye': [(386, 307), (397, 299), (407, 300), (415, 309), (407, 309), (396, 308)],
        'top_lip': [(332, 363), (343, 355), (355, 351), (363, 353), (372, 352), (385, 357), (399, 366), (394, 365),
                    (372, 359), (363, 359), (355, 359), (337, 362)],
        'bottom_lip': [(399, 366), (385, 368), (372, 369), (363, 369), (354, 368), (343, 366), (332, 363), (337, 362),
                       (354, 360), (363, 361), (372, 361), (394, 365)]}]

I want to get all the values of coordinates, removing the chin, etc. just co-ordinates like (297, 322)

You can loop through your lst element and then find the values of the dictionaries it contains like this:

coords = []
for d in lst:
    coord.append(d.values())

print(coords)

Swift Dictionary (With Examples), A dictionary is simply a container that can hold multiple data as key-value pair You may use dictionary instead of array when you need to look up value with Now, you get the capital city from the collection by searching with the key country. There are two possible approaches, depending on how you want to treat duplications in the set of values for a key. The following approach allows such duplications: d1 = {} d1.setdefault (key, []).append (value) while this approach automatically eliminates duplications: d2 = {} d2.setdefault (key, {}) [value] = 1.

This becomes a lot easier if the code is formatted correctly:

lst = [
    {
        'chin': [
            (297, 322), (297, 339), (299, 356), (301, 373), (305, 390), (312, 405), (325, 415), (342, 418), (362, 420), (383, 421), (402, 421), (418, 415), (427, 403), (433, 387), (437, 370), (440, 352), (443, 335)
        ],
        'left_eyebrow': [
            (306, 296), (313, 282), (326, 276), (342, 276), (357, 281)
        ],
        'right_eyebrow': [
            (378, 283), (393, 278), (410, 281), (423, 289), (431, 303)
        ],
        'nose_bridge': [
            (366, 297), (365, 305), (364, 314), (363, 323)
        ],
        'nose_tip': [
            (347, 337), (355, 340), (364, 341), (373, 340), (382, 339)
        ],
        'left_eye': [
            (321, 304), (328, 296), (339, 296), (349, 304), (339, 305), (328, 305)
        ],
        'right_eye': [
            (386, 307), (397, 299), (407, 300), (415, 309), (407, 309), (396, 308)
        ],
        'top_lip': [
            (332, 363), (343, 355), (355, 351), (363, 353), (372, 352), (385, 357), (399, 366), (394, 365), (372, 359), (363, 359), (355, 359), (337, 362)
        ],
        'bottom_lip': [
            (399, 366), (385, 368), (372, 369), (363, 369), (354, 368), (343, 366), (332, 363), (337, 362), (354, 360), (363, 361), (372, 361), (394, 365)
        ]
    }
]

It's a list containing a dictionary where the value of each key is a list of tuples.

for k in lst[0].keys():
    print(lst[0][k])

To skip chin:

for k in lst[0].keys():
    if k == "chin":
        continue

    print(lst[0][k])

20. Dictionaries — How to Think Like a Computer Scientist: Learning , If we wanted to find a value associated with a key, we would have to iterate over The values in a dictionary are accessed with keys, not with indices, so there is no need Dictionaries implement the associative array abstract data type. memo � The values() method returns a view object. The view object contains the values of the dictionary, as a list. The view object contains the values of the dictionary, as a list. The view object will reflect any changes done to the dictionary, see example below.

You could do this:

output = []
for k in lst[0].keys():
   for p in lst[0][k]:
       output.append(p)
print(output)

Python, Till now, we have seen the ways to creating dictionary in multiple ways and different operations on the key and values in dictionary. Now, let's see different ways� Method #1 : Using min() + list comprehension + values() The combination of above functions can be used to perform this particular task. In this, minimum value is extracted using the min function, while values of dictionary is extracted using values(). The list comprehension is used to iterate through the dictionary for matching keys with min value.

you can do this: first, separate all co-ordinates assigned to the chin, eye, ...etc, and append in a separate array.

then loop through all the arrays present in a single array put it in another array. now the final array is ready to use.

test = []
for a in landmark[0]:
    test.append(landmark[0][a])

final = []
for a in range(0, 17):
    final.append(test[0][a])

for b in range(1, 3):
    for a in range(0, 5):
        final.append(test[b][a])

for a in range(0, 4):
    final.append(test[3][a])

for a in range(0, 5):
    final.append(test[4][a])

for b in range(5, 7):
    for a in range(0, 6):
        final.append(test[b][a])

for b in range(7, 9):
    for a in range(0, 12):
        final.append(test[b][a])

length = len(final)
print(final)

Dictionaries in Python – Real Python, Once you have finished this tutorial, you should have a good sense of when a Defining a Dictionary; Accessing Dictionary Values; Dictionary Keys vs. of a data structure that is more generally known as an associative array. There is also no restriction against a particular value appearing in a dictionary multiple times:. The approach used here is to find two separate lists of keys and values. Then fetch the key using the position of the value in the val_list. As key at any position N in key_list will have corresponding value at position N in val_list. We can also fetch key from a value by matching all the values and then print the corresponding key to given value.

co_ordinates = []

for rec in you_data:
    for data in rec:
        co_ordinates.extend(rec[data])

Output:
[(297, 322), (297, 339), (299, 356), (301, 373), (305, 390), (312, 405), (325, 415), (342, 418), (362, 420), (383, 421), (402, 421), (418, 415), (427, 403), (433, 387), (437, 370), (440, 352), (443, 335), (306, 296), (313, 282), (326, 276), (342, 276), (357, 281), (378, 283), (393, 278), (410, 281), (423, 289), (431, 303), (366, 297), (365, 305), (364, 314), (363, 323), (347, 337), (355, 340), (364, 341), (373, 340), (382, 339), (321, 304), (328, 296), (339, 296), (349, 304), (339, 305), (328, 305), (386, 307), (397, 299), (407, 300), (415, 309), (407, 309), (396, 308), (332, 363), (343, 355), (355, 351), (363, 353), (372, 352), (385, 357), (399, 366), (394, 365), (372, 359), (363, 359), (355, 359), (337, 362), (399, 366), (385, 368), (372, 369), (363, 369), (354, 368), (343, 366), (332, 363), (337, 362), (354, 360), (363, 361), (372, 361), (394, 365)]

Associating Multiple Values with Each Key in a Dictionary, Problem. You need a dictionary that maps each key to multiple values. Solution. By nature, a dictionary is a one-to-one mapping, but it's not hard to make it� Description. Python dictionary method get() returns a value for the given key. If key is not available then returns default value None. Syntax. Following is the syntax for get() method −

Dictionaries as alternative to arrays, Tcl arrays are collections of variables, rather than values. Instead you have to use the array get and array set commands to convert them to a value In other words, we are using a one-dimensional array, with keys like "foo,2" and "bar,3". The third brackets ([]) are to read the value of any particular key. Another data type exists in Python to store multiple data which is called List. The list works like a numeric array and its index starts from 0 and maintains order. But the key values of the dictionary contain different types of values that don’t need to maintain any order.

Working with Array and Dictionary Collections in Swift 3, A guide to working with dictionaries and arrays in Swift 3. By joining the beta, you will get access to experimental features, at the risk of encountering bugs and � The above example finds the value of the key “one” of Dictionary in Python. This gives value in Dictionary for the single key given. However, to get more values with the keys, you have to use the get function again. Find the Value Using Index Operator in Python. In addition to the above method, you can also get the values by keys in

Collection Types — The Swift Programming Language (Swift 5.3), It also means you can be confident about the type of values you will retrieve from a Swift's array, set, and dictionary types are implemented as generic collections . The same value can appear in an array multiple times at different positions. Here we use more a more complex value type, a String array. Step 1: We create a Dictionary. It has Integer keys and String array values—so each int can point to an entire array. Step 2: We use ContainsKey. When the key type is Integer, we must pass an Integer to ContainsKey. Integer. Step 3: We get the String array value with Item, and then

Comments
  • I think the idea was to not include the string chin, not to ignore the coordinates assigned to chin.
  • @ScottHunter I believe that's one of many valid interpretations of the question.. ;)
  • Hence I think.