eshangj/stackoverflow_q_and_a_sample huggingface dataset:
Question: Given an array of integers (with at least two elements), I need to choose at least one element of each adjacent pair of elements in the array in a way that costs me the least. Then, return the cost and elements chosen. For example, [50, 30, 40, 60, 10, 30, 10]  We choose 30 for the pair (50, 30), and for the pair (30, 40). Then we choose 40 for the pair (40, 60) and 10 for the pairs (60, 10), (10, 30). Lastly, 10 for the pair (30, 10). So we got 30 + 40 + 10 + 10 = 90. Another example, [60, 100, 70]  There are two possible selections: [60, 70] or [100]. But, the optimal solution would be [100] for a total of 100, which is less than 60 + 70. So, the algorithm should choose 100. My issue is that I only succeeded in making a code that returns the lowest cost without saving the elements used. My Code in Python arr = [50, 30, 40, 60, 10, 30, 10] min_sum = [0] * len(arr) min_sum[0] = arr[0] min_sum[1] = arr[1] for i in range(2, len(arr)): choice1 = min_sum[i-1] + arr[i] choice2 = min_sum[i-2] + arr[i] min_sum[i] = min(choice1, choice2) res = min(min_sum[-1], min_sum[-2]) print(res) 

Answer: Calling cost(i) the cost of the optimal solution when considering the array up to element i included only, there is a simple recurrence formula for this problem: cost(i) = min( arr[i] + cost(i-1), arr[i-1] + cost(i-2), )  For instance, the cost for array [50, 30, 40, 60, 10, 30, 10] is the minimum of 10 + cost for [50, 30, 40, 60, 10, 30] 30 + cost for [50, 30, 40, 60, 10]  The base cases for this recurrence are: cost(0) = 0 cost(1) = 0 cost(2) = min(arr[0],arr[1])  This recurrence relation leads to a simple iterative code: def cost(a): if len(a) <= 1: return 0 elif len(a) == 2: return min(a) cost_iminus2 = 0 cost_iminus1 = min(a[:2]) for i in range(2,len(a)): cost_i = min( arr[i] + cost_iminus1, arr[i-1] + cost_iminus2, ) cost_iminus2, cost_iminus1 = cost_iminus1, cost_i return cost_i  You can easily amend this code to also remember which elements were used in the sum: def selection(a): if len(a) < 2: return 0, [] elif len(a) == 2: return min(a), [min(a)] cost_iminus2, selection_iminus2 = 0, [] cost_iminus1, selection_iminus1 = min(a[:2]), [min(a[:2])] for i in range(2,len(a)): cost_i = min( a[i] + cost_iminus1, a[i-1] + cost_iminus2, ) if cost_i == a[i] + cost_iminus1: selection_i = [*selection_iminus1, a[i]] # need to copy elif cost_i == a[i-1] + cost_iminus2: selection_i = selection_iminus2 # no need to copy selection_i.append(a[i-1]) else: raise ValueError("unreachable branch") cost_iminus2, cost_iminus1 = cost_iminus1, cost_i selection_iminus2, selection_iminus1 = selection_iminus1, selection_i return cost_i, selection_i  Output: >>> selection([50, 30, 40, 60, 10, 30, 10]) (90, [30, 40, 10, 10]) >>> selection([60,100,70]) (100, [100])
Question: I have two columns in a dataframe: Pests and FieldComment. If the value of Pests is listed as 'None', then I'd like to search the FieldComment column for specific words and overwrite what's in the Pests column. If none of the words are found in the FieldComment column, the Pests column can remain as 'None'. Example: pests_list = ['Spiders', 'Rodents', 'Ants', 'Honey Bees']     Pests FieldComment     Spiders Performed service.   None Performed service for reported rodents.    The above would ideally turn into this:    Pests FieldComment     Spiders Performed service.   Rodents Performed service for reported rodents.    This is what I've tried so far but I can't quite get it: for w in df['FieldComment'].str.split(): for p in pests_list: if w.str.lower() == p.str.lower(): df['Pests'] = p  I've also tried: df.loc[df['Pests'] == 'None', "Pests"] = *[pest for pest in pest_list if pest in df['FieldComment']]  And finally: df.loc[df['Pests'] == 'None', "Pests"] = df.loc[df['Pests'] == 'None', "Pests"].apply(lambda x: pest for pest in pest_list if pest in df['FieldComment'] else 'None') 

Answer:  Convert the pests list into a set. Create a set with the words from FieldComment. Get the intersection of both sets and fill column Pests where it is null.  pests_set = set([p.lower() for p in pests_list]) df.loc[df["Pests"].isna(), "Pests"] = df["FieldComment"].apply( lambda x: ", ".join( set(x.strip(".").lower().split()).intersection(pests_set) ).capitalize() )   Pests FieldComment 0 Spiders Performed service. 1 Rodents Performed service for reported rodents.   This solution will join pest names with , if there is more than one in the FieldComment column. For this dataframe:  Pests FieldComment 0 Spiders Performed service. 1 None Performed service for rodents and spiders.  Result would be:  Pests FieldComment 0 Spiders Performed service. 1 Spiders, rodents Performed service for rodents and spiders.   Note that if the dataframe has a str 'None', and not the None keyword, you would have to modify the above code slightly, replacing df["Pests"].isna() with df["Pests"] == 'None'
Question: KubeVirt is an extension of Kubernetes, enabling it to manage virtual machines.I have deployed a KubeVirt cluster. I want to use Python Client to operate KubeVirt to manage virtual machines. Is this feasible? I use Kubernetes Python Client.But it does not seem to support the resource type of virtual machine.My cluster can create virtual machines through kubectl apply - f testvm.yaml Python code: from kubernetes import client, config, utils config.load_kube_config('config') k8s_client = client.ApiClient() file = 'yamlfile\\testvm.yaml' utils.create_from_yaml(k8s_client, file, verbose=True)  error: AttributeError: module 'kubernetes.client' has no attribute 'KubevirtIoV1Api' testvm.yaml: apiVersion: kubevirt.io/v1 kind: VirtualMachine metadata: name: testvm spec: running: false template: metadata: labels: kubevirt.io/size: small kubevirt.io/domain: testvm spec: domain: devices: disks: - name: containerdisk disk: bus: virtio - name: cloudinitdisk disk: bus: virtio interfaces: - name: default masquerade: {} resources: requests: memory: 64M networks: - name: default pod: {} volumes: - name: containerdisk containerDisk: image: quay.io/kubevirt/cirros-container-disk-demo - name: cloudinitdisk cloudInitNoCloud: userDataBase64: SGkuXG4= --- apiVersion: v1 kind: Service metadata: name: vm-ssh spec: type: NodePort ports: - protocol: TCP port: 24 targetPort: 22 name: test selector: kubevirt.io/domain: testvm kubevirt.io/size: small 

Answer: You're trying to create resources that use the kubevirt.io/v1 api. This isn't a core Kubernetes API; it's provided by the kubevirt custom resource definition, which means it's not supported by any of the static parts of the Python kubernetes client. But don't worry, that's what the DynamicClient is for! You can find some examples of working with the dynamic client here. Your code would end up looking something like: import yaml from kubernetes import config, dynamic from kubernetes.client import api_client # Creating a dynamic client client = dynamic.DynamicClient( api_client.ApiClient(configuration=config.load_kube_config()) ) # fetching the configmap api api = client.resources.get(api_version="kubevirt.io/v1", kind="VirtualMachine") path = 'testvm.yaml' with open(path) as fd: resource = yaml.safe_load(fd) api.create(body=resource)
Question: I have a minimal reproducible example. Input data: coordinates = [(50.0, 50.0, 100.0, 50.0), (70.0, 50.0, 75.0, 40.0, 80.0, 50.0)]  This is what these lines form in the drawing:  The task is to find a loop, i.e. a closed region, and display their boundaries in the terminal. In this case, the result should be as follows (it is represented as [(line1), (line2)]):  [(70.0, 50.0, 75.0, 40.0), (75.0, 40.0, 80.0, 50.0), (80.0, 50.0, 70.0, 50.0)]  I tried the cycle_basic method from the networkx library. But in this method it is necessary that the cycle vertices touch the vertices of other cycles. But here the vertices can touch any place of another cycle

Answer: IIUC, you want the induced / chordless_cycles. If so, here is a primal approach with networkx: points = MultiPoint(list(batched(chain.from_iterable(coordinates), 2))) lines = [ line for coo in coordinates for pop in pairwise(batched(coo, 2)) for gc in [split(LineString(pop), points)] for line in gc.geoms ] G = gdf_to_nx( gpd.GeoDataFrame(geometry=lines), multigraph=False, approach="primal" ) cycles = [ [ tuple(chain.from_iterable(pair)) for pair in pairwise(cyc) ] for cyc in nx.chordless_cycles(G) for cyc in [cyc + [cyc[0]]] ]   Output (cycles, clockwise): [ [ # top/green cycle (70.0, 50.0, 80.0, 50.0), (80.0, 50.0, 77.5, 45.0), (77.5, 45.0, 72.5, 45.0), (72.5, 45.0, 70.0, 50.0), ], [ # bottom/red cycle (72.5, 45.0, 77.5, 45.0), (77.5, 45.0, 75.0, 40.0), (75.0, 40.0, 72.5, 45.0), ], ]  Used input (coordinates) : from itertools import chain, batched, pairwise import geopandas as gpd from momepy import gdf_to_nx import networkx as nx from shapely import LineString, MultiPoint from shapely.ops import split coordinates = [ (50.0, 50.0, 100.0, 50.0), (70.0, 50.0, 75.0, 40.0, 80.0, 50.0), (72.5, 45.0, 77.5, 45.0) ]   Full code Old answer
Question: I didn't see the difference between freq in .shift() df = pd.DataFrame({"Col1": [10, 20, 15, 30, 45], "Col2": [13, 23, 18, 33, 48], "Col3": [17, 27, 22, 37, 52]}, index=pd.date_range("2020-01-01", "2020-01-05")) df.shift(periods=2, freq="infer") Col1 Col2 Col3 2020-01-03 10 13 17 2020-01-04 20 23 27 2020-01-05 15 18 22 2020-01-06 30 33 37 2020-01-07 45 48 52  return the same as df.shift(periods=2, freq="d") Col1 Col2 Col3 2020-01-03 10 13 17 2020-01-04 20 23 27 2020-01-05 15 18 22 2020-01-06 30 33 37 2020-01-07 45 48 52  Can someone explain what the freq='infer' parameter does?

Answer: freq='infer' means that the frequency is inferred from the index metadata.  freq DateOffset, tseries.offsets, timedelta, or str, optional Offset to use from the tseries module or time rule (e.g. ‘EOM’). If freq is specified then the index values are shifted but the data is not realigned. That is, use freq if you would like to extend the index when shifting and preserve the original data. If freq is specified as “infer” then it will be inferred from the freq or inferred_freq attributes of the index. If neither of those attributes exist, a ValueError is thrown.  Assuming this example with an Index that has a frequency of 2 weeks: df = pd.DataFrame({"Col1": [10, 20, 15, 30, 45], "Col2": [13, 23, 18, 33, 48], "Col3": [17, 27, 22, 37, 52]}, index=pd.date_range("2020-01-01", periods=5, freq='2W')) df.index # DatetimeIndex(['2020-01-05', ..., '2020-03-01'], dtype='datetime64[ns]', # freq='2W-SUN') # <- the important part  df.shift(periods=2, freq='infer') will shift 2 periods of 2W = 4 weeks: df.shift(periods=2, freq='infer') Col1 Col2 Col3 2020-02-02 10 13 17 # 4W after 2020-01-05 2020-02-16 20 23 27 2020-03-01 15 18 22 2020-03-15 30 33 37 2020-03-29 45 48 52 # 4W after 2020-03-01  In comparison a simple df.shift(periods=2) just shifts by 2 rows: df.shift(periods=2) Col1 Col2 Col3 2020-01-05 NaN NaN NaN 2020-01-19 NaN NaN NaN 2020-02-02 10.0 13.0 17.0 2020-02-16 20.0 23.0 27.0 2020-03-01 15.0 18.0 22.0  In your example, the default frequency of date_range is D, so this indeed gives the same output. Let's change it to freq='3D': df = pd.DataFrame({"Col1": [10, 20, 15, 30, 45], "Col2": [13, 23, 18, 33, 48], "Col3": [17, 27, 22, 37, 52]}, index=pd.date_range("2020-01-01", periods=5, freq='3D')) # shift by 2 * D df.shift(periods=2, freq='D') Col1 Col2 Col3 2020-01-03 10 13 17 # 2*1D after 2020-01-01 2020-01-05 20 23 27 2020-01-07 15 18 22 2020-01-09 30 33 37 2020-01-11 45 48 52 # shift by 2 * 3D = 6D df.shift(periods=2, freq='infer') Col1 Col2 Col3 2020-01-07 10 13 17 # 2*3D after 2020-01-01 2020-01-10 20 23 27 2020-01-13 15 18 22 2020-01-16 30 33 37 2020-01-19 45 48 52
Question: I'm having some problems when static type checking decorated functions. For instance, when I use an incorrect function argument name or type, I don't get any warning or error hints in the IDE, only at runtime. What steps will reproduce the problem?  Add a decorator to a function  Use a incorrect argument name or type   What is the expected result?  PyCharm should highlight the wrong argument name or wrong type.  What happens instead?  The attribute name is not highlighted.  Additional Info  This is with Python 3.10 and PyCharm 2023.3.4  Code sample follows: def myDecorator(func): def wrapper(*args, **kwargs): return func(*args, **kwargs) return wrapper @myDecorator def myFunc(x: int, y: int): print(x+y) myFunc(z=4, x="something") # Expecting "z=4" and x="something" to be highlighted  Edit It's now working with regular functions. However, I'm encountering the same problem if these functions are defined inside of a Class, as seen below: from typing import ParamSpec from typing import TypeVar from typing import Callable P = ParamSpec("P") T = TypeVar("T") def myDecorator(func: Callable[P, T]) -> Callable[P, T]: def wrapper(*args: P.args, **kwargs: P.kwargs) -> T: return func(*args, **kwargs) return wrapper class myClass: @myDecorator def myFunc2(self, x: int, y: int) -> None: print(x + y) @myDecorator def another_function(self): self.myFunc2(x=3) # showing type hinting: P and highlighting "x=3" as a wrong argument. 

Answer: Looks similar to this PyCharm issue. If you are willing to annotate the decorator with ParamSpec you can do ... from typing import ParamSpec from typing import TypeVar from typing import Callable P = ParamSpec("P") T = TypeVar("T") def myDecorator(func: Callable[P, T]) -> Callable[P, T]: def wrapper(*args: P.args, **kwargs: P.kwargs) -> T: return func(*args, **kwargs) return wrapper @myDecorator def myFunc(x: int, y: int) -> None: print(x + y) myFunc(z=4, x="something") # Expecting "z=4" and x="something" to be highlighted  This way PyCharm will report the mistake.
Question: Say, I have a function that should do some internal stuff and display the provided text: def display_text(text: str): ... print(text)  There's also a class with a convert() method: class String: def __init__(self, string: str): self.string = string def convert(self): return self.string  Now, can you type hint text argument in display_text with String, but if the provided parameter will be str, call convert and assign the returned value to text? Like that: def display_text(text: String): ... print(text)  It should be done without any additional code in a display_text function, just with type hinting. I've seen that in some libs but couldn't figure out how does it work. I tried searching through some libraries' (e.g. discord.py converters) code, searching similar questions on StackOverflow, only found out about typing.Protocol but still no idea how this conversion is done.

Answer:  Can you type hint text argument in display_text with String, but if the provided parameter will be str, call convert and assign the returned value to text without any additional code in a display_text function, just with type hinting  No, not just with type hinting, since type annotations are absolutely inert at runtime and do nothing (and with from __future__ import annotations, they're not even evaluated). If this is a trick question and that "in a display_text function" was the catch, then yes, you could @decorate your functions (or decorate a class holding them, or use a metaclass) to wrap functions using type annotations to cast arguments if needed. An example of such a decorator: import dataclasses import inspect from functools import wraps from typing import Any def convert_args(fn): @wraps(fn) def wrapper(*args, **kwargs): sig = inspect.signature(fn) bound_args = sig.bind(*args, **kwargs) bound_args.apply_defaults() for name, val in bound_args.arguments.items(): param = sig.parameters[name] if hasattr(param.annotation, "convert"): bound_args.arguments[name] = param.annotation.convert(val) return fn(*bound_args.args, **bound_args.kwargs) return wrapper @dataclasses.dataclass class Shouty: val: str @classmethod def convert(cls, val: Any): return Shouty(val=str(val).upper()) @dataclasses.dataclass class Shorten: val: str @classmethod def convert(cls, val: Any): return Shorten(val=str(val)[::2]) @convert_args def display_text(arg1: Shouty, arg2: Shorten): print(locals()) display_text("hello", "world, this is an example")  This prints out {'arg1': Shouty(val='HELLO'), 'arg2': Shorten(val='wrd hsi neape')}
Question: I'm trying to generate an optimal combination of p records in a vector of length n while simultaneously ensuring (constraining) that specific elements in the vector are included in the solution set p (based on a binary value for now) I have the equation "simu_total_volume" below that does work for ensuring the solution set never exceeds p records, but I am unable to figure out how to modify this equation to ensure that specific elements i in the vector are included in the solution set (even if non-optimal). x7 is the binary vector that selects which elements are included based on p. "Labor Day" is a vector of 0s except for one element that is equal to 1 (this corresponds to element I want to include in the solution set). I can enforce this vector to sum to 1 per the below, but am unsure how to integrate it into "simu_total_volume" so that the solution conforms to it. Apologies for not including all relevant information for reproducibility, but the full solution is very large. simu_total_volume = [m.Intermediate(( (m.max2(0,base_volume[i]*(m.exp(total_vol_fedi[i])-1)) * x3[i] + m.max2(0,base_volume[i]*(m.exp(total_vol_feao[i])-1)) * x4[i] + m.max2(0,base_volume[i]*(m.exp(total_vol_diso[i])-1)) * x5[i] + m.max2(0,base_volume[i]*(m.exp(total_vol_tpro[i])-1)) * x6[i]) + base_volume[i]) * x7[i]) for i in range(n)] labor_day = [m.Intermediate(x8[i] * el_cppg['holiday_labor_day_flag'].values[i]) for i in range(n)] #Require labor day to be in output m.Equation(sum(labor_day) == 1) #Limit max output m.Equation(sum(x7)<=p) m.Maximize(m.sum(simu_total_volume)) m.options.SOLVER=1 try: m.solve(disp = True) except: continue 

Answer: There are multiple ways to constrain particular elements within an array. Here is a complete example that optimizes the elements of X to minimize the total cost. Each element X can be [0,1] and two can be selected with sum(X)==2. from gekko import GEKKO m = GEKKO(remote=False) X = m.Array(m.Var,7,lb=0,ub=1,integer=True) c = [1.2,0.95,1.3,1.0,0.8,1.25,1.4] m.Equation(sum(X)==2) m.Minimize(sum([X[i]*c[i] for i in range(7)])) m.options.SOLVER=1 m.solve() print(f'X: {X}')  The solver picks the two lowest that correspond to c elements of 0.95 and 0.8: c = [1.2,0.95,1.3,1.0,0.8,1.25,1.4] X: [[0.0] [1.0] [0.0] [0.0] [1.0] [0.0] [0.0]]  Here are a few ways to constrain the solution such as enforcing that the last element is always selected: Add an equation m.Equation(X[-1]==1)  Set the upper and lower bounds to the specified solution X[-1].lower=1 X[-1].upper=1  Use the m.fix() function m.fix(X[-1],1)  Add an objective as a soft constraint Use this method if adding a hard constraint gives an infeasible solution. This will encourage the selection of a preferred option, but won't enforce it if the equations aren't satisfied. m.Minimize(100*(X[-1]-1)**2)  Results All of these methods return the correct solution that selects the last element (not optimal) and the least costly element. X: [[0.0] [0.0] [0.0] [0.0] [1.0] [0.0] [1.0]]
Question: Lets assume we have a pandas data frame df with the column 'A', and the following non-vectorized transformation function: def transform_a_to_b(a): ... return b  Then if we wanted to create the column 'B' using the transform on 'A', we could do the following: df['B'] = df['A'].apply(lambda x: transform_a_to_b(a))  What is a better way to do this if the transform takes a non-trivial amount of time, there are many duplicate values in the column 'A', and the transform always maps the duplicate a values to the same b value? Also assume there are more columns in the data frame so I do want to map the values back to every single row in the original data frame. I came up with the following solution, but I feel like something simpler should exist. transform_counts = 0 def transform_a_to_b(a): global transform_counts # Keep count of how many times this was called transform_counts += 1 return 2 * a # Test dataframe with several duplicates df = pd.DataFrame({ 'A': [1, 3, 2, 2, 3, 3, 2, 3, 1, 1, 1], }) # My solution: # Perform transformation only 3 times for the 3 unique A values and preserve order df = df.merge( df['A'].drop_duplicates().apply(lambda a: pd.Series( data=[a, transform_a_to_b(a)], index=['A', 'B'], )), on='A', how='left', )  After the function transform_counts is 3 and the df is as follows:  A B 0 1 2 1 3 6 2 2 4 3 2 4 4 3 6 5 3 6 6 2 4 7 3 6 8 1 2 9 1 2 10 1 2  I am not apposed to caching, if that is simplest, but I cannot change the original transformation definition.

Answer: Your approach is good, I would just use map+unique in place of merge+drop_duplicates. df['B'] = df['A'].map({k: transform_a_to_b(k) for k in df['A'].unique()})  A pythonic alternative would be to. cache your function: from functools import cache transform_counts = 0 @cache def transform_a_to_b(a): global transform_counts # Keep count of how many times this was called transform_counts += 1 return 2 * a df = pd.DataFrame({ 'A': [1, 3, 2, 2, 3, 3, 2, 3, 1, 1, 1], }) df['B'] = df['A'].map(transform_a_to_b) print(df)  Output:  A B 0 1 2 1 3 6 2 2 4 3 2 4 4 3 6 5 3 6 6 2 4 7 3 6 8 1 2 9 1 2 10 1 2
Question: I can't figure out how to link a CSS file to my HTML file. I'm am using Visual Studio Code, Python, and Flask to do this My project directory is like this: - templates - home.html - style.css - app.py  home.html <!DOCTYPE html> <html> <head> <link href='https://fonts.googleapis.com/css?family=Raleway:400, 600' rel='stylesheet' type='text/css'> <link href='assets/style.css' rel='stylesheet' type='text/css'/> </head> <body> ... </body>  style.css html, body { margin: 0; padding: 0; } header { background-color: #333333; position: fixed; width: 100%; z-index: 5; } h1 { font-family: 'Times New Roman', Times, serif; color:cadetblue }  app.py from flask import Flask, render_template app = Flask(__name__, template_folder="templates") @app.route("/home") @app.route("/") def index(): return render_template("home.html")  I tried using the templates folder, it didn't work I tried without using the templates folder, it didn't work. No error is raised, but the page just doesn't has the CSS style, for example the h1 tags keep being black with the default font, and they should be blue with the Times New Roman Font

Answer: In flask, your style.css file should be inside the 'static' folder. Then in your Html file you should link it like this: <link href="{{ url_for('static', filename='style.css') }}" rel="stylesheet" type="text/css"/>
Question: Ultimately I want to write a value into a database. I found a script to output desired data. However it does return it in a loop - but I only need one single value each time I trigger the script, not a loop. import tibber account = tibber.Account("tokenstring") home = account.homes[0] @home.event("live_measurement") async def process_data( data): print(data.power) home.start_live_feed(user_agent="Homey/10.0.0") print(home.event("live_measurement"))`  This will output values in a loop - but I only need one value each time the code runs. Help!

Answer: The start_live_feed function starts an infinite loop if called with default arguments. In other words, Python will not run any code after you call this function. The function takes an argument exit_condition. This is a function that is run after every time data is received. If it returns a truthy value, the loop will end and Python will continue running whatever comes after the line home.start_live_feed(...). So, to run it a single time, you just need to pass a function that always returns True. import tibber account = tibber.Account(TOKEN_STRING) home = account.homes[0] @home.event("live_measurement") async def process_data(data): # Database storing logic here home.start_live_feed(user_agent="Homey/10.0.0", exit_condition=lambda data: True)