# ML | Binning or Discretization

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There are three data smoothing methods:

1. Binning. Binning methods smooth the sorted data values ‚Äã‚Äãby referring to their "neighborhood", that is, the values ‚Äã‚Äãaround them.
2. Regression: matches function data values. Linear regression involves finding the "best‚" line to match two attributes (or variables) so that one attribute can be used to predict the other.
3. Analyzing outliers : outliers can be detected by clustering, for example when similar values ‚Äã‚Äãare organized into groups or "clusters". Intuitively, values ‚Äã‚Äãthat fall outside the cluster set can be considered outliers.

Binning method for data smoothing —
Here we are dealing with Binning’s method for data smoothing. In this method, the data is first sorted and then the sorted values ‚Äã‚Äãare spread across multiple segments or cells . Since binning methods refer to a neighborhood of values, they perform local smoothing.

There are basically two types of binning —

1. Binning is the same width (or distance). The simplest approach is to divide the variable range into k intervals of equal width. Spacing width — it’s just the range [A, B] of the variable divided by k,
` w = (BA) / k `

Thus, the interval of the i- th interval will be ` [A + (i-1) w, A + iw] ` where i = 1, 2, 3‚Ä¶ ..k
Skewed data cannot be handled well with this method.

2. Binning of equal depth (or frequency): In binning of equal frequency, we divide the range [A, B] of a variable into intervals that contain (approximately) equal points; equal frequency may not be possible due to duplicate values.

#### How to smooth the data?

There are three approaches to performing smoothing:

1. Bin smoothing mean s: when bin smoothing, each value in the bin is replaced by the bin’s mean .
2. Bin-mean -median-mode-in-python-without-libraries/">median smoothing: in this method each bin value is replaced by its bin mean -median-mode-in-python-without-libraries/">median value.
3. Bin Smoothing: When bin boundary smoothing, the minimum and maximum values ‚Äã‚Äãin a given bin are defined as bin boundaries. Each bin value is then replaced with the closest cutoff value.

Sorted data by price (in dollars): 2, 6, 7, 9, 13, 20, 21, 25, 30

` Partition using equal frequency approach: Bin 1: 2, 6, 7 Bin 2: 9, 13, 20 Bin 3: 21, 24, 30 Smoothing by bin  mean : Bin 1: 5, 5, 5 Bin 2: 14 , 14, 14 Bin 3: 25, 25, 25 Smoothing by bin mean -median-mode-in-python-without-libraries/">median: Bin 1: 6, 6, 6 Bin 2: 13, 13, 13 Bin 3: 24, 24, 24 Smoothing by bin boundary: Bin 1: 2 , 7, 7 Bin 2: 9, 9, 20 Bin 3: 21, 21, 30 `

Binning can also be used as a sampling method ... Here discretization refers to the process of transforming or breaking down continuous attributes, features, or variables into discrete or nominal attributes / features / variables / intervals.
For example, attribute values ‚Äã‚Äãcan be sampled by applying equal width or equal frequency binning and then replacing each bin value with the mean or mean -median-mode-in-python-without-libraries/">median bin, as in antialiasing by mean bin value or smoothing by bin mean -median-mode-in-python-without-libraries/">medians, respectively. Continuous values ‚Äã‚Äãcan then be converted to a nominal or sampled value that matches the corresponding bin value.

Below is the Python implementation:

bin_ mean

 ` import ` ` numpy as np ` ` from ` ` sklearn.linear_model ` ` import ` ` LinearRegression ` ` from ` ` sklearn ` ` import ` ` linear_model ` ` # import statsmodels.api as sm ` ` import ` ` statistics ` ` import ` ` math ` ` from ` ` collections ` ` import ` ` OrderedDict ` ` x ` ` = ` ` [] ` ` print ` ` (` ` "enter the data" ` `) ` ` x ` ` = ` ` list ` ` (` ` map ` ` (` ` float ` `, ` ` input ` ` (). split ())) ` ` print ` ` (` ` "enter the number of bins" ` `) ` ` bi ` ` = ` ` int ` ` (` ` input ` ` ()) ` ` # X_dict will store data in sorted order ` ` X_dict ` ` = ` ` OrderedDict () ` ` # x_old will store the original data ` ` x_old ` ` = ` ` {} ` ` # x_new will store data after binning ` ` x_new ` ` = ` ` {} ` ` for ` ` i ` ` in ` ` range ` ` (` ` len ` ` (x)): ` ` X_dict [i] ` ` = ` ` x [i] ` ` x_old [i] ` ` = ` ` x [i] ` ` x_dict ` ` = ` ` sorted ` ` (X_dict.items (), key ` ` = ` ` lambda ` ` x: x [` ` 1 ` `]) ` ` # list to lists (bins) ` ` binn ` ` = ` ` [] ` ` # variable to find the average of each bin ` ` avrg ` ` = ` ` 0 ` ` i ` ` = ` ` 0 ` ` k ` ` = ` ` 0 ` ` num_of_data_in_each_bin ` ` = ` ` int ` ` (math.ceil (` ` len ` ` (x) ` ` / ` ` bi)) ` ` # executing binning ` ` for ` ` g, h ` ` in ` ` X_dict.items (): ` ` if ` ` (i & lt; num_of_data_in_each_bin): ` ` avrg ` ` = ` ` avrg ` ` + ` ` h ` ` i ` ` = ` ` i ` ` + ` ` 1 ` ` elif ` ` (i ` ` = ` ` = ` ` num_of_data_in_each_bin): ` ` k ` ` = ` ` k ` ` + ` ` 1 ` ` i ` ` = ` ` 0 ` ` binn.append (` ` round ` ` (avrg ` ` / ` ` num_of_data_in_each_bin, ` ` 3 ` `)) ` ` avrg ` ` = ` ` 0 ` ` avrg ` ` = ` ` avrg ` ` + ` ` h ` ` ` ` i ` ` = ` ` i ` ` + ` ` 1 ` ` rem ` ` = ` ` len ` ` (x) ` `% ` ` bi ` ` if ` ` (rem ` ` = ` ` = ` ` 0 ` `): ` ` ` ` binn.append (` ` round ` ` (avrg ` ` / ` ` num_of_data_in_each_bin, ` ` 3 ` `)) ` ` else ` `: ` ` binn.append (` ` round ` ` (avrg ` ` / ` ` rem, ` ` 3 ` `)) ` ` # save the new value of each data ` ` i ` ` = ` ` 0 ` ` j ` ` = ` ` 0 ` ` for ` ` g, h ` ` in ` ` X_dict.items (): ` ` ` ` if ` ` (i & lt; num_of_data_in_each_bin): ` ` x_new [g] ` ` = ` ` binn [j] ` ` i ` ` = ` ` i ` ` + ` ` 1 ` ` else ` `: ` ` i ` ` = ` ` 0 ` ` j ` ` = ` ` j ` ` + ` ` 1 ` ` x_new [g] ` ` = ` ` binn [j] ` ` i ` ` = ` ` i ` ` + ` ` 1 ` ` print ` ` (` `" number of data in each bin "` `) ` ` print ` ` (math.ceil (` ` le n ` ` (x) ` ` / ` ` bi)) ` ` for ` ` i ` ` in ` ` range ` ` (` ` 0 ` `, ` ` len ` ` (x)): ` ` print ` ` (` ` ’index {2} old value {0} new value {1}’ ` `. ` ` format ` ` (x_old [i], x_new [i], i)) `

bin_mean -median-mode-in-python-without-libraries/">median

` `

` import numpy as np from sklearn.linear_model import LinearRegression from sklearn import linear_model # import statsmodels.api as sm import statistics import math from collections import OrderedDict x = [] print ( "enter the data" ) x = list ( map ( float , input (). split ())) print ( " enter the number of bins " ) bi = int ( input ()) # X_dict will store data in sorted order X_dict = OrderedDict () # x_old will store the original data x_old = {} # x_new will store data after binning x_new = {} for i in range ( len (x)) : X_dict [i] = x [i] x_old [ i] = x [i] x_dict = sorted (X_dict.items (), key = lambda x: x [ 1 ]) # list of lists (bins) binn = [] # variable to find the average of each bin avrg = [] i = 0 k = 0 num_of_data_in_each_bin = int (math.ceil ( len (x) / bi)) # executing binning for g, h in X_dict.items (): if (i & lt; num_of_data_in_each_bin): avrg.append (h) i = i + 1 elif (i = = num_of_data_in_each_bin): k = k + 1 i = 0 binn. append (statistics.mean -median-mode-in-python-without-libraries/">median (avrg)) avrg = [] avrg.append (h) i = i + 1 binn.append (statistics.mean -median-mode-in-python-without-libraries/">median (avrg)) # save the new value of each of the data i = 0 j = 0 for g, h in X_dict.items (): if (i & lt; num_of_data_in_each_bin): x_new [g] = round (binn [j], 3 ) i = i + 1 else : i = 0 j = j + 1 x_new [g] = round (binn [j], 3 ) i = i + 1 print ( "number of data in each bin" ) print (math.ceil ( len (x) / bi)) for i in range ( 0 , len (x)): print ( ’index {2} old value {0} new value {1} ’ . format (x_old [i], x_new [i], i)) `

` ` bin_boundary

` `

``` import numpy as np from sklearn.linear_model import LinearRegression from sklearn import linear_model # import statsmodels.api as sm import statistics import math from collections import OrderedDict x = [] print ( "enter the data" ) x = list ( map ( float , input (). split ())) print ( " enter the number of bins " ) bi = int ( input ()) # X_dict will store data in sorted order X_dict = OrderedDict () # x_old will store the original data x_old = {} # x_new will store data after binning x_new = {} for i in range ( len ( x)): X_dict [i] = x [i] x_old [i] = x [i] x_dict = sorted (X_dict.items (), key = lambda x: x [ 1 ] ) # list of lists (bins) binn = [] # variable to find the average of each bin avrg = [] i = 0 k = 0 num_of_data_in_each_bin = int (math.ceil ( len (x) / bi)) for g, h in X_dict.items (): if (i & lt; num_of_data_in_each_bin) : avrg.append (h) i = i + 1 elif (i = = num_of_data_in_each_bin): k = k + 1 i = 0 code class = "undefined spaces"> x_old [i] = x [i] x_dict = sorted (X_dict.items (), key = lambda x: x [ 1 ]) best laptop for engineering students? ML | Binning or Discretization __del__: Questions __del__ How can I make a time delay in Python? 5 answers I would like to know how to put a time delay in a Python script. 2973 Answer #1 import time time.sleep(5) # Delays for 5 seconds. You can also use a float value. Here is another example where something is run approximately once a minute: import time while True: print("This prints once a minute.") time.sleep(60) # Delay for 1 minute (60 seconds). 2973 Answer #2 You can use the sleep() function in the time module. It can take a float argument for sub-second resolution. from time import sleep sleep(0.1) # Time in seconds ML | Binning or Discretization __del__: Questions __del__ How to delete a file or folder in Python? 5 answers How do I delete a file or folder in Python? 2639 Answer #1 os.remove() removes a file. os.rmdir() removes an empty directory. shutil.rmtree() deletes a directory and all its contents. Path objects from the Python 3.4+ pathlib module also expose these instance methods: pathlib.Path.unlink() removes a file or symbolic link. pathlib.Path.rmdir() removes an empty directory. __dict__ How do I merge two dictionaries in a single expression (taking union of dictionaries)? 5 answers By Carl Meyer I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place. >>> x = {"a": 1, "b": 2} >>> y = {"b": 10, "c": 11} >>> z = x.update(y) >>> print(z) None >>> x {"a": 1, "b": 10, "c": 11} How can I get that final merged dictionary in z, not x? (To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.) 5839 Answer #1 How can I merge two Python dictionaries in a single expression? For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x. In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method: z = x | y # NOTE: 3.9+ ONLY In Python 3.5 or greater: z = {**x, **y} In Python 2, (or 3.4 or lower) write a function: def merge_two_dicts(x, y): z = x.copy() # start with keys and values of x z.update(y) # modifies z with keys and values of y return z and now: z = merge_two_dicts(x, y) Explanation Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries: x = {"a": 1, "b": 2} y = {"b": 3, "c": 4} The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first. >>> z {"a": 1, "b": 3, "c": 4} A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is z = {**x, **y} And it is indeed a single expression. Note that we can merge in with literal notation as well: z = {**x, "foo": 1, "bar": 2, **y} and now: >>> z {"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4} It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document. However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process: z = x.copy() z.update(y) # which returns None since it mutates z In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result. Not yet on Python 3.5, but want a single expression If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function: def merge_two_dicts(x, y): """Given two dictionaries, merge them into a new dict as a shallow copy.""" z = x.copy() z.update(y) return z and then you have a single expression: z = merge_two_dicts(x, y) You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number: def merge_dicts(*dict_args): """ Given any number of dictionaries, shallow copy and merge into a new dict, precedence goes to key-value pairs in latter dictionaries. """ result = {} for dictionary in dict_args: result.update(dictionary) return result This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g: z = merge_dicts(a, b, c, d, e, f, g) and key-value pairs in g will take precedence over dictionaries a to f, and so on. Critiques of Other Answers Don"t use what you see in the formerly accepted answer: z = dict(x.items() + y.items()) In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists - >>> c = dict(a.items() + b.items()) Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items" and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power. Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this: >>> c = dict(a.items() | b.items()) This example demonstrates what happens when values are unhashable: >>> x = {"a": []} >>> y = {"b": []} >>> dict(x.items() | y.items()) Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: unhashable type: "list" Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets: >>> x = {"a": 2} >>> y = {"a": 1} >>> dict(x.items() | y.items()) {"a": 2} Another hack you should not use: z = dict(x, **y) This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic. Here"s an example of the usage being remediated in django. Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings. >>> c = dict(a, **b) Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: keyword arguments must be strings From the mailing list, Guido van Rossum, the creator of the language, wrote: I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism. and Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool. It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.: dict(a=1, b=10, c=11) instead of {"a": 1, "b": 10, "c": 11} Response to comments Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords. Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2: >>> foo(**{("a", "b"): None}) Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: foo() keywords must be strings >>> dict(**{("a", "b"): None}) {("a", "b"): None} This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change. I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints. More comments: dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts. My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated. {**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging. Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression. Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them: from copy import deepcopy def dict_of_dicts_merge(x, y): z = {} overlapping_keys = x.keys() & y.keys() for key in overlapping_keys: z[key] = dict_of_dicts_merge(x[key], y[key]) for key in x.keys() - overlapping_keys: z[key] = deepcopy(x[key]) for key in y.keys() - overlapping_keys: z[key] = deepcopy(y[key]) return z Usage: >>> x = {"a":{1:{}}, "b": {2:{}}} >>> y = {"b":{10:{}}, "c": {11:{}}} >>> dict_of_dicts_merge(x, y) {"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}} Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge". Less Performant But Correct Ad-hocs These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence) You can also chain the dictionaries manually inside a dict comprehension: {k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7 or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced): dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2 itertools.chain will chain the iterators over the key-value pairs in the correct order: from itertools import chain z = dict(chain(x.items(), y.items())) # iteritems in Python 2 Performance Analysis I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.) from timeit import repeat from itertools import chain x = dict.fromkeys("abcdefg") y = dict.fromkeys("efghijk") def merge_two_dicts(x, y): z = x.copy() z.update(y) return z min(repeat(lambda: {**x, **y})) min(repeat(lambda: merge_two_dicts(x, y))) min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()})) min(repeat(lambda: dict(chain(x.items(), y.items())))) min(repeat(lambda: dict(item for d in (x, y) for item in d.items()))) In Python 3.8.1, NixOS: >>> min(repeat(lambda: {**x, **y})) 1.0804965235292912 >>> min(repeat(lambda: merge_two_dicts(x, y))) 1.636518670246005 >>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()})) 3.1779992282390594 >>> min(repeat(lambda: dict(chain(x.items(), y.items())))) 2.740647904574871 >>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items()))) 4.266070580109954 \$ uname -a Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux Resources on Dictionaries My explanation of Python"s dictionary implementation, updated for 3.6. Answer on how to add new keys to a dictionary Mapping two lists into a dictionary The official Python docs on dictionaries The Dictionary Even Mightier - talk by Brandon Rhodes at Pycon 2017 Modern Python Dictionaries, A Confluence of Great Ideas - talk by Raymond Hettinger at Pycon 2017 5839 Answer #2 In your case, what you can do is: z = dict(list(x.items()) + list(y.items())) This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value: >>> x = {"a":1, "b": 2} >>> y = {"b":10, "c": 11} >>> z = dict(list(x.items()) + list(y.items())) >>> z {"a": 1, "c": 11, "b": 10} If you use Python 2, you can even remove the list() calls. To create z: >>> z = dict(x.items() + y.items()) >>> z {"a": 1, "c": 11, "b": 10} If you use Python version 3.9.0a4 or greater, then you can directly use: x = {"a":1, "b": 2} y = {"b":10, "c": 11} z = x | y print(z) {"a": 1, "c": 11, "b": 10} 5839 Answer #3 An alternative: z = x.copy() z.update(y) (adsbygoogle = window.adsbygoogle || []).push({}); Shop Learn programming in R: courses\$ Best Python online courses for 2022\$ Best laptop for Fortnite\$ Best laptop for Excel\$ Best laptop for Solidworks\$ Best laptop for Roblox\$ Best computer for crypto mining\$ Best laptop for Sims 4\$ (adsbygoogle = window.adsbygoogle || []).push({}); Latest questions NUMPYNUMPY psycopg2: insert multiple rows with one query 12 answers NUMPYNUMPY How to convert Nonetype to int or string? 12 answers NUMPYNUMPY How to specify multiple return types using type-hints 12 answers NUMPYNUMPY Javascript Error: IPython is not defined in JupyterLab 12 answers All questions (adsbygoogle = window.adsbygoogle || []).push({}); News 05/10/2022 How can a Data Scientist create a classifier if some of the data is mislabelled? 04/10/2022 Tips to master Git 03/10/2022 Using objects instead of True and False - truthy and falsy values in Python Wiki __del__ Python OpenCV | cv2.putText () method __del__ numpy.arctan2 () in Python __del__ Python | os.path.realpath () method around Python OpenCV | cv2.circle () method cvtcolor Python OpenCV cv2.cvtColor () method Python functions Python - Move item to the end of the list Counters time.perf_counter () function in Python __dict__ Check if one list is a subset of another in Python __del__ Python os.path.join () method © 2017—2022 Python Engineering Hub EN | ES | DE | FR | IT | RU | TR | PL | PT | JP | KR | CN | HI | NL Python.Engineering is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to amazon.com Python Loops Counters NumPy NLP Regular Expressions Wiki Tech news Python Wiki StackOverflow PHP JavaScript Books All books Computations Development Cryptography For dummies Big Data document.addEventListener("DOMContentLoaded", () => { let arrayCode = document.querySelectorAll('pre'); arrayCode.forEach(element => { element.classList.add("prettyprint"); }); }); window.dataLayer = window.dataLayer || []; function gtag(){dataLayer.push(arguments);} gtag('js', new Date()); gtag('config', 'G-Q022WLXW4X'); ```