eshangj/stackoverflow_q_and_a_sample huggingface dataset:
Question: I am trying to write some code that will generate all lists of a fixed length that have the property that the next integer in each list will either be same or an increment of the previous integer. All lists should start with 0. I can write code that does this for size 4 BUT it uses 3 nested loops. If I want to continue down this path I will have to have n - 1 nested loops. Just to be clear I will give some examples for length 5: [0, 0, 0, 0, 0] [0, 0, 1, 1, 1] [0, 1, 2, 2, 2] [0, 1, 2, 3 ,3]  The code will return all such lists of length 5. Instead of down voting, could you help me improve my question? Here is my (really bad but working) code: import math def index_list(x): IL = [] for i in [0,1]: for j in [0,1,2]: for k in [0,1,2,3]: if i <= j and j <= k and abs(i-j) < 2 and abs(j-k) < 2: IL.append([x, x + i, x + j, x + k]) return IL print(index_list(0))  this produces: [[0, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 1], [0, 0, 1, 2], [0, 1, 1, 1], [0, 1, 1, 2], [0, 1, 2, 2], [0, 1, 2, 3]]  I have included a argument x as i wish to have the code to sometimes do the same thing but with a different starting index.

Answer: I am providing you several solutions. I think you may like one. First: from itertools import product def generate_sequences(n, x=0): if n == 0: return [] deltas = product([0, 1], repeat=n-1) sequences = [] for delta_seq in deltas: sequence = [x] current = x for delta in delta_seq: current += delta sequence.append(current) sequences.append(sequence) return sequences def print_sequences(sequences, max_display=10): print(f"Total sequences: {len(sequences)}") print("Sample sequences:") for i, seq in enumerate(sequences[:max_display]): print(seq) if len(sequences) > max_display: print(f"... (showing first {max_display} of {len(sequences)})") if __name__ == "__main__": print("\nSequences starting with 0, length 4:") print_sequences(generate_sequences(4, 0))  Second: def generate_sequences(n, x=0): if n <= 0: return [] sequences = [] stack = [(x, [x], n - 1)] while stack: last_val, current_seq, remaining = stack.pop() if remaining == 0: sequences.append(current_seq) continue stack.append((last_val, current_seq + [last_val], remaining - 1)) stack.append((last_val + 1, current_seq + [last_val + 1], remaining - 1)) return sequences def print_sequences(sequences, max_display=10): print(f"Total sequences: {len(sequences)}") print("Sample sequences:") for i, seq in enumerate(sequences[:max_display]): print(seq) if len(sequences) > max_display: print(f"... (showing first {max_display} of {len(sequences)})") if __name__ == "__main__": for length in [4, 5]: print(f"\nSequences of length {length}:") sequences = generate_sequences(length) print_sequences(sequences) print("\nSequences starting with 5, length 3:") print_sequences(generate_sequences(3, 5))  Third: def generate_sequences(n, x=0): def helper(current, remaining): if remaining == 0: return [current] last = current[-1] next_values = [last] if remaining == 1 else [last, last + 1] sequences = [] for val in next_values: sequences.extend(helper(current + [val], remaining - 1)) return sequences if n == 0: return [] return helper([x], n - 1) print(generate_sequences(5))
Question: There are a lot of ways to check if string is a palindrome. Plenty of them listed here This question is not about "how" but rather about performance. I was assuing that is_palindrome should be twice faster than is_palindrome0 because it does len/2 iterations in the worst case. However, in reality, is_palindrome takes more than 30 seconds to check all the strings while is_palindrome0 gets the job done in less than half of a second! Test-case def is_palindrome(s): for i in range(len(s) // 2): if s[i] != s[len(s)-i-1]: return 0 return 1 def is_palindrome0(s): if s == s[::-1]: return 1 else: return 0 N = 500 L = 99999 sss = '101' * L import time start = time.time() print(sum([1 for i in range(N) if is_palindrome0(sss+sss[i:])])) end = time.time() print(f'{(end - start):.2f}') start = time.time() print(sum([1 for i in range(N) if is_palindrome(sss+sss[i:])])) end = time.time() print(f'{(end - start):.2f}')  Output 168 0.40 168 34.20  Any ideas why string slicing is so crazy fast? How to debug further? Apologies if I missed answered question with in-depth performance comparison. UPDATE. Taking into account Frank's comment. def is_palindrome(s): l = len(s) for i in range(l // 2): if s[i] != s[l-i-1]: return 0 return 1 def is_palindrome0(s): if s == s[::-1]: return 1 else: return 0 N = 500 L = 99999 sss = '101' * L import time start = time.time() print(sum([1 for i in range(N) if is_palindrome0(sss+sss[i:])])) end = time.time() print(f'{(end - start):.2f}') start = time.time() print(sum([1 for i in range(N) if is_palindrome(sss+sss[i:])])) end = time.time() print(f'{(end - start):.2f}')  A bit faster but still around 50x times slower. 168 0.41 168 25.11 

Answer: is_palindrome0 (the fast one) s[::-1] uses C-level optimizations under the hood (it's implemented in highly efficient C code in CPython). String comparisons (==) are also optimized for short-circuiting; they stop early if a mismatch is found. The entire operation is happening in compiled code with no explicit Python loops. is_palindrome (the slow one): This is pure Python, interpreted line by line. Each index access like s[i] and s[len(s) - i - 1] is an individual bytecode operation. Python for loops and indexing are orders of magnitude slower than equivalent operations in C. The main performance killer here is actually in the line is_palindrome(sss+sss[i:]). For each iteration, in ur code:  U create a new string via concatenation (sss+sss[i:])  This new string is ~200,000 characters long  Then ur function iterates through half of it
Question: I have a time series dataframe, I would like to take x random samples from column "temperature" from each day. I am able to do this with: daily_groups = df.groupby([pd.Grouper(key='time', freq='D')])['temperature'].apply(lambda x: x.sample(10))  This works if there are at least x samples for each day. If there are not I get the error "a must be greater than 0 unless no samples are taken". What I would like is, lets say I want 10 samples, get 10 if available, if not get as many as possible and if there are none skip this day. I don't want to up sample data. Also I don't know if its possible to return the original dataframe with the values filtered for items mention above, instead of returning the groupby series object. Thanks for any help.

Answer: If you want to randomly sample values from the "temperature" column for each day, up to 10 values per day, but also want to handle days with fewer than 10 entries, here's my suggestion on how to do it. This code checks how many rows there are per day — if there are fewer than 10, it just takes as many as possible. If there are zero, it skips that day entirely. The best part is that it gives you back the original DataFrame rows, not the Series from Groupby. import pandas as pd def sample_temperature(group, n=10): if len(group) == 0: return pd.DataFrame() # return empty if there is nothing in the group return group.sample(min(len(group), n)) # Make sure the 'time' column is in datetime format df['time'] = pd.to_datetime(df['time']) # Group by day and sample sampled_df = (df.groupby(pd.Grouper(key='time', freq='D')).apply(lambda g: sample_temperature(g, n=10)).reset_index(drop=True))
Question: Suppose I want to annotate this function: def add_one(value): match value: case int(): return value + 1 case str(): return value + " and one more" case _: raise TypeError()  I want to tell the type checker "This function can be called with an int (or subclass) or a str (or subclass). In the former case it returns an int, and in the latter a str." How can this be accomplished? Here are some of my failed attempts: Attempt 1 def add_one(value: int | str) -> int | str: ...  This is too loose. The type checker no longer knows that the returned type is similar to the argument. Passing an int might return a str. Attempt 2 def add_one[T: int | str](value: T) -> T: ...  This is incorrect. It doesn't return literally the same type as the argument. If passed an IntEnum it returns int, and for a StrEnum it returns str. Attempt 3 def add_one[T: (int, str)](value: T) -> T: ...  This is better, but now I can't call add_one with Union[int, str]. These ques-tions talk about the differences between bounds and constraints, but I lack the brainpower to use them to solve my problem. Attempt 4 @overload def add_one(value: int) -> int: ... @overload def add_one(value: str) -> str: ... def add_one(value: int | str) -> int | str: # Put real implementation here ...  This is the best I can do. It does the right thing, but requires me to type out the function signature for every possible type it handles. Doesn't seem like much for two types, but my real code already has 7 or 8, and I intend to add more. It also requires me to manually expand any Union. Is there some better way to tell the type checker "Here's a Union and a T. Make the T be whatever union arg matched."?

Answer: One solution I can think of is to mix attempts 2, 3 and 4: (playgrounds: Mypy, Pyright) @overload def add_one[T: (str, int, bytes)](value: T) -> T: ... @overload def add_one[T: str | int | bytes](value: T) -> T: ... def add_one(value: str | int | bytes) -> str | int | bytes: ...  class S(StrEnum): A = '' class I(IntEnum): B = 0 class B(bytes, Enum): C = b'' def f(si: str | int, sb: str | bytes, ib: int | bytes, sib: str | int | bytes) -> None: reveal_type(add_one('')) # str reveal_type(add_one(0)) # int reveal_type(add_one(b'')) # bytes reveal_type(add_one(S.A)) # str reveal_type(add_one(I.B)) # int reveal_type(add_one(B.C)) # bytes reveal_type(add_one(si)) # str | int reveal_type(add_one(sb)) # str | bytes reveal_type(add_one(ib)) # int | bytes reveal_type(add_one(sib)) # str | int | bytes  This has one minor problem in that the second overload would be marked as overlapping with the first. It should be fine to # type: ignore it. Additionally, unions of subtypes like S | I won't be handled correctly; an upcast (v: str | int = S.A if bool() else I.B) would be necessary to make it work.
Question: I am doing the below but getting memory issues. make frame data = {'link': [1,2,3,4,5,6,7], 'code': ['xx', 'xx', 'xy', '', 'aa', 'ab', 'aa'], 'Name': ['Tom', 'Tom', 'Tom', 'Tom', 'nick', 'nick', 'nick'], 'Age': [20,20,20,20, 21, 21, 21]} # Create DataFrame df = pd.DataFrame(data) print(df)  output  link code Name Age 0 1 xx Tom 20 1 2 xx Tom 20 2 3 xy Tom 20 3 4 Tom 20 4 5 aa nick 21 5 6 ab nick 21 6 7 aa nick 21  minimal code example that works on subset of data but not on full dataset. temp = df.groupby(['Name', 'Age'])['code'].apply(list).reset_index() pd.merge(df, temp, on=['Name', 'Age']).explode('code_y').replace(r'^\s*$', np.nan, regex=True).dropna(subset='code_y').drop_duplicates()  output  error when used on full dataset. ### MemoryError: Unable to allocate 5.34 TiB for an array with shape (733324768776,) and data type object  The apply list makes a big long list with duplicates. Is there a way to drop dups from the lists or is there maybe a better way to do this? update Going with this code as it seem to work best. # Select relevant columns and drop duplicates directly d = df[['code', 'Name', 'Age']].replace('', pd.NA).drop_duplicates() # Perform the merge and drop rows with missing 'code_y' in one step df.merge(d, how='outer', on=['Name', 'Age']).dropna(subset=['code_y']) 

Answer: Your method is inefficient as it explodes then drops the duplicates. Ensure to drop the duplicates first then merge: d = df.mask(df.eq(''))[['code', 'Name', 'Age']].drop_duplicates() df.merge(d, how = 'outer', on = ['Name', 'Age']).dropna(subset='code_y') link code_x Name Age code_y 0 1 xx Tom 20 xx 1 1 xx Tom 20 xy 3 2 xx Tom 20 xx 4 2 xx Tom 20 xy 6 3 xy Tom 20 xx 7 3 xy Tom 20 xy 9 4 Tom 20 xx 10 4 Tom 20 xy 12 5 aa nick 21 aa 13 5 aa nick 21 ab 14 6 ab nick 21 aa 15 6 ab nick 21 ab 16 7 aa nick 21 aa 17 7 aa nick 21 ab
Question: I'm using RandomizedSearchCV with RandomForestClassifier in scikit-learn. I want to make sure my results are reproducible across runs. Where should I set the random_state—in the classifier, in RandomizedSearchCV, or both? Example code: from sklearn.ensemble import RandomForestClassifier from sklearn.model_selection import RandomizedSearchCV clf = RandomForestClassifier() search = RandomizedSearchCV(clf, param_distributions=params, n_iter=10)  What's the best practice to ensure consistent results?

Answer: You can perform a simple test using as a starter code given in the RandomizedSearchCV examples. In the code, a random_state is set both, in the classifier, as well as in the RandomizedSearchCV. Writing a loop with let's say 50 iterations and printing outcomes, that is .best_params_ will show the following:  setting random_state both in RandomizedSearchCV and classifier/regressor will always give the same outcome  setting random_state in just one of those will provide different outcomes across iterations.   So the conclusion is, that if you need reproducibility, you need to set this parameter in both places as in both places separate random generators are used. Also it is worth to check some more information on the used numbers from this post, as well as official glossary concerning random_state. The code: from sklearn.datasets import load_iris from sklearn.linear_model import LogisticRegression from sklearn.model_selection import RandomizedSearchCV from scipy.stats import uniform iris = load_iris() for i in range(50): logistic = LogisticRegression(solver='saga', tol=1e-2, max_iter=200 ,random_state=0) distributions = dict(C=uniform(loc=0, scale=4), penalty=['l2', 'l1']) clf = RandomizedSearchCV(logistic, distributions, random_state=0) search = clf.fit(iris.data, iris.target) print(search.best_params_)
Question: I'm working on a classification problem using scikit-learn's RandomForestClassifier. I tried using RandomizedSearchCV for hyperparameter tuning, but the results were worse than when I manually set the parameters based on intuition and trial/error. Here's a simplified version of my code: from sklearn.ensemble import RandomForestClassifier from sklearn.model_selection import RandomizedSearchCV param_dist = { "n_estimators": [100, 200, 300], "max_depth": [None, 10, 20, 30], "min_samples_split": [2, 5, 10], "min_samples_leaf": [1, 2, 4], } clf = RandomForestClassifier(random_state=42) random_search = RandomizedSearchCV(clf, param_distributions=param_dist, n_iter=10, cv=5, scoring='accuracy') random_search.fit(X_train, y_train)  In multiple runs, this approach yields models with lower accuracy on my test set than my manually-tuned model. What are common pitfalls when using RandomizedSearchCV? How can I ensure reproducibility and robustness of the tuning process?

Answer: RandomizedSearchCV can give worse results than manual tuning due to a few common reasons:  Too few iterations – n_iter=10 may not explore enough parameter combinations.  Poor parameter grid – Your grid might miss optimal values or be too coarse.  Inconsistent random seeds – Different runs can yield different results if random_state isn’t set.  Improper CV splits – Use StratifiedKFold for balanced class sampling.  Wrong scoring metric – Make sure scoring aligns with your real objective (e.g., accuracy, f1).
Question: I'm working with a day of year column that ranges from 1 to 366 (to account for leap years). I need to convert this column into a date for a specific task and I would like to set it to a year that is very unlikely to appear in my time series. Is there a way to set it to the oldest leap year of pandas? import pandas as pd # here is an example where the data have already been converted to datetime object # I just missed the year to set dates = pd.Series(pd.to_datetime(['2023-05-01', '2021-12-15', '2019-07-20'])) first_leap_year = 2000 # this is where I don't know what to set new_dates = dates.apply(lambda d: d.replace(year=first_leap_year)) 

Answer: The documentation for the pandas.Timestamp type says:  Timestamp is the pandas equivalent of python’s Datetime and is interchangeable with it in most cases.  So we can look up the Python documentation for datetime objects, where we find:  Like a date object, datetime assumes the current Gregorian calendar extended in both directions; like a time object, datetime assumes there are exactly 3600*24 seconds in every day.  In other words, it assumes that the current rules for calculating leap years apply at any point in history, even though they were actually introduced in 1582, and adopted by different countries over the next few centuries. (The technical term for this is a "proleptic Gregorian calendar".) Standard Python has a datetime.MINYEAR constant:  The smallest year number allowed in a date or datetime object. MINYEAR is 1.  So the lowest year divisible by 4, (and not by 100, so meeting the Gregorian definition of leap year as well as the Julian one) would be 4. However, Pandas also has pandas.Timestamp.min:  Timestamp.min = Timestamp('1677-09-21 00:12:43.145224193')  (In case you're wondering, that's 263 nanoseconds before January 1, 1970, i.e. the limit of a 64-bit signed integer with nanosecond resolution.) So you probably want a year after 1677, meaning the earliest available year would be 1680.
Question: I'm new to LLMs and need to extract the file names of files that have been already ingested into PrivateGPT that the system uses to answer questions. I can list the doc_ids using :  from pgpt_python.client import PrivateGPTApi client = PrivateGPTApi(base_url="http://localhost:8001") # Health print(client.health.health()) # List ingested docs for doc in client.ingestion.list_ingested().data: print(doc.doc_id)  which gives an output of : e019a7be-3b0d-45b6-a1f6-195735a20725 8d59d127-3432-47e4-8beb-5465fdb1e72d 8df9068b-fa3c-42b5-b987-0bee211bce0a b56fb71a-fd3e-4cd9-9728-3b70ff045162 dbebc4a6-29af-4ac9-b311-0a53b74cbb4f e9f28c23-5d08-4660-aa4c-8f2389901583 68da2416-57ff-45c7-9af6-dc70175d1a15 f6880d64-434c-4527-8705-df67f19a6dfc 84e66ed4-aa0f-4058-a1b9-19d7a89d8d95 etc  However haven't been able to find the mechanism to list the current 5 files that the system has ingested. Any help appreciated.

Answer: According to PrivateGPT API reference here (https://docs.privategpt.dev/api-reference/api-reference/ingestion/ingest-file), in addition to the doc_id property, which seems to be what you're getting, you should also be able to get the file's metadata in doc_metadata property, where the file name should be. Try running this instead:  # List ingested docs for doc in client.ingestion.list_ingested().data: print(doc.doc_metadata)  Have a look if the file names are in the metadata property, then you can parse the property to output only the file name.
Question: Running Spyder 6.0.5 with Python version 3.11.11 and IPython 8.34.0. From within the Console window in Spyder I attempted to install Scipy with the following command: conda install -c conda-forge scipy=1.15.2  All seems to be ok, no error messages and a request to restart the kernel, which I do. However when I try to run a program that uses the "Gaussian" function from scipy.signal I get the following error message: from scipy.signal import gaussian ImportError: cannot import name 'gaussian' from 'scipy.signal' (C:\Users\USERNAME\AppData\Local\spyder-6\envs\spyder-runtime\Lib\site-packages\scipy\signal\__init__.py)  I've tried to instead install Scipy from Windows command prompt with pip but it's made no impact. Could anyone please advise as to what is occurring here. From what I've seen there had been issues previously that newer versions of Scipy had addressed, but the advice in those guides has not helped me. Thanks

Answer: Try from scipy.signal.windows import gaussian
Question: I have two address columns and I want to extract the last word from the first column and the first word from the second column. In the provided example there aren't two words in column 'Address2', but I want to build the code in such a way that it will work regardless of how the dataset will look like. Sometimes the address2 can be one word, something it will have 2, etc.. data = { 'Address1': ['3 Steel Street', '1 Arnprior Crescent', '40 Bargeddie Street Blackhill'], 'Address2': ['Saltmarket', 'Castlemilk', 'Blackhill'] } df = pd.DataFrame(data)  I have no problem with column 'Address1': df[['StringStart', 'LastWord']] = df['Address1'].str.rsplit(' ', n=1, expand=True)  The problem comes with column 'Address2' where if I apply the above code I an error: Columns must be same length as key I understand where the problem is coming from - I am trying to split one column which has one element into two columns. I am sure there is a way in which this can be handled to allow the split anyway and return Null if there isn't a word and a value if there is.

Answer: Using str.extract() might be better for several reasons: it handles all cases, offers precision with regular expressions, and eliminates the risk of value errors. import pandas as pd data = { 'Address1': ['3 Steel Street', '1 Arnprior Crescent', '40 Bargeddie Street Blackhill'], 'Address2': ['Saltmarket', 'Castlemilk East', 'Blackhill'] } df = pd.DataFrame(data) df[['StringStart', 'LastWord']] = df['Address1'].str.rsplit(' ', n=1, expand=True) df[['FirstWord_Address2', 'Remaining_Address2']] = ( df['Address2'].str.extract(r'^(\S+)\s*(.*)$') ) print(df)  Or: df[['Address1_Prefix', 'Address1_LastWord']] = df['Address1'].str.extract(r'^(.*\b)\s+(\S+)$') df[['Address2_FirstWord', 'Address2_Remaining']] = df['Address2'].str.extract(r'^(\S+)\s*(.*)$')  Output:  Address1 Address2 StringStart LastWord FirstWord_Address2 Remaining_Address2 0 3 Steel Street Saltmarket 3 Steel Street Saltmarket 1 1 Arnprior Crescent Castlemilk East 1 Arnprior Crescent Castlemilk East 2 40 Bargeddie Street Blackhill Blackhill 40 Bargeddie Street Blackhill Blackhill
Question: I want to build a page that has each author with their quote. I have tried, but everything i tried has failed. The function below is what causes me the issues. quotes = { "Arthur Ashe": "Start where you are, Use what you have, Do what you can.", "Steve Jobs": "Don’t watch the clock; do what it does. Keep going.", "Sam Levenson": "Don’t watch the clock; do what it does. Keep going.", " Robert Collier": "Success is the sum of small efforts, repeated day in and day out.", "Nelson Mandela": "It always seems impossible until it’s done.", "Mahatma Gandhi": "The future depends on what you do today.", "Zig Ziglar": "You don’t have to be great to start, but you have to start to be great.", "Dave": "Discipline is doing what needs to be done, even if you don’t want to do it.", "Suzy Kassem": "Doubt kills more dreams than failure ever will.", "Pablo Picasso": "Action is the foundational key to all success." } def mypage(request): messages = [quotes[item] for item in quotes] authors = [item for item in quotes] return render(request, "quotes/mypage.html", {"authors": authors, "quotes":messages}) 

Answer: You can pass the entire dictionary: def mypage(request): messages = [quotes[item] for item in quotes] authors = [item for item in quotes] return render(request, 'quotes/mypage.html', {'quotes': quotes}) and then in the template, enumerate over the .items() of the quotes, so: {% for author, quote in quotes.items %} {{author}} said: "{{ quote }}" {% endfor % }