eshangj/stackoverflow_q_and_a_sample huggingface dataset:
Question: I slowly started using python in excel. Somehow I managing the code, however this time I I have received my output data is delimited with ",". How to do it with py option excel. Based on the below pic, sum of aa,dd,ee is 18, bb,gg is 9 and so on.. I can do it sum individual but not sure how to deal with delimited text. Any help would appreciate. Thank you. 

Answer: Enter in E2 with =PY: df_nv = xl("A1:B10", headers=True) names = xl("D2:D5").iloc[:,0] nv = dict(zip(df_nv["Name"],df_nv["Value"])) def sum_split(a): return sum(map(lambda a: nv[a], a.split(","))) list(map(sum_split, names))    Siddharth Rout's answer has better checks - for example xl returns dataFrame only for multi-cell inputs.
Question: I am attempting to write a code that keeps track of the text for the links in the left handed gray box on this webpage. In this case the code should return Valykrie, The Acid Baby Here is the code I am trying to use: import requests from bs4 import BeautifulSoup url = 'https://www.mountainproject.com/area/109928429/aasgard-sentinel' page = requests.get(url) soup = BeautifulSoup(page.text, "html.parser") for link in soup.findAll('a', class_= 'new-indicator'): print(link)  It is not working (otherwise I wouldn't be here!) I'm pretty new to BeautifulSoup, and coding in general. No matter how much I inspect the page source I can't seem to figure out the inputs to the findAll to get it to return what I want!

Answer: Search for the table with a specific I’d, than the rows: import requests from bs4 import BeautifulSoup url = 'https://www.mountainproject.com/area/109928429/aasgard-sentinel' page = requests.get(url) soup = BeautifulSoup(page.text, "html.parser") table = soup.find(lambda tag: tag.name=='table' and tag.has_attr('id') and tag['id']=="left-nav-route-table") rows = table.findAll(lambda tag: tag.name=='a') for member in rows: print(member.text)  Output: Acid Baby Valkyrie, The  If you would also see the href link add print(member.get('href'))
Question: I'm trying to determine if a given positive integer ( N ) is a power of 3, i.e., if there exists a nonnegative integer ( x ) such that ( 3^x = N ). The challenge is that ( N ) can be extremely large, with up to ( 10^7 ) digits. Here's the logic I want to implement:  If ( N ) is less than or equal to 0, return -1. Use logarithmic calculations to determine if ( N ) is a power of 3. If it is a power of 3, print the value of ( x ); otherwise, print -1.  I've tried the following code, but I'm concerned about precision issues with large integers: import math def is_power_of_three(N): if N <= 0: return -1 log3_N = math.log10(N) / math.log10(3) if abs(log3_N - round(log3_N)) < 1e-10: return int(round(log3_N)) else: return -1 # Example usage: N = int(input("Enter a number: ")) print(is_power_of_three(N))  Question:  Is there a more efficient way to check if ( N ) is a power of 3, especially for very large values? How can I handle precision issues in Python when calculating logarithms for such large integers? Are there alternative methods to achieve the same goal? 

Answer: Any approach that involves repeated division is going to be slow with numbers that large. Instead, consider using the bit length of the number (effectively the ceiling of its base 2 logarithm) to approximate the corresponding power of three, then check to see if it is indeed equal: import math def what_power_of_3(n): if n <= 0: return -1 k = math.floor(n.bit_length() * (math.log(2) / math.log(3))) if n == 3**k: return k else: return -1  This is fast: the test requires only a single exponentiation, so for e.g. 3**10000000, it requires only a few seconds on my smartphone.  A brief sketch to show why this is correct: Because of the equality check, if n is not a power of 3, the answer will always be correct (because n == 3**k cannot be true). So it suffices to prove that this answer always finds the correct k when n == 3 ** k. Let k > 0, n = 3 ** k, and t = n.bit_length(). Then, 3 ** k < 2 ** t < 3 ** (k + 1) by definition of bit_length(). Thus k < t * log(2) / log(3) < k + 1, and so n.bit_length() * (math.log(2) / math.log(3)) is going to be a fractional value that lies strictly between k and k+1; thus, the floor of that value will be exactly k.
Question: Is there any advantage to one of these over the other? Also, is there any better code than these to achieve the goal? My intuition is that in number 2, since it has already checked for x or y, the check for y is more efficient?   if x or y: do some stuff if y: do some OTHER stuff    if x or y: do some stuff if y: do some OTHER stuff   

Answer: The second one is marginally more efficient, because it won't redundantly test for y again in the case where x and y are both False. In cases where x and y are often False (i.e. the first if is rarely entered), this results in a slight improvement in performance. However, it should be noted that this improvement is very slight, and there is a tradeoff to making this change: the code becomes less readable due to the extra nesting level of if y. If y is expensive to compute, it will be better to pre-compute (cache) the condition once rather than evaluating it twice. That will improve performance much more than micro-optimizing the arrangement of these if statements.
Question: I have a square cost matrix stored as a pandas DataFrame. Rows and columns represent positions [i, j], and I want to multiply all off-diagonal elements (where i != j) by a constant c, without using any for loops for performance reasons. Is there an efficient way to achieve this in pandas or do I have to switch to numpy and then back to pandas to perform this task? Example import pandas as pd # Sample DataFrame cost_matrix = pd.DataFrame([ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]) # Constant c = 4 # Desired output # 1 8 12 # 16 5 24 # 28 16 9 

Answer: Build a boolean mask with numpy.identity and update the underlying array in place: cost_matrix.values[np.identity(n=len(cost_matrix))==0] *= c  output:  0 1 2 0 1 8 12 1 16 5 24 2 28 32 9  Intermediate: np.identity(n=len(cost_matrix))==0 array([[False, True, True], [ True, False, True], [ True, True, False]])  NB. for .values to be a view of the underlying array, the DataFrame must have been constructed from an homogeneous block. If not, it should be converted to one using cost_matrix = cost_matrix.copy(). Alternative @PaulS suggested to modify all the values and restore the diagonal. I would use: d = np.diag(cost_matrix) cost_matrix *= c np.fill_diagonal(cost_matrix.values, d)  Timings The mask approach seems to be faster on small/medium size inputs, and the diagonal restoration faster on large inputs. (My previous timings were performed online and I don't reproduce the results with perfplot). NB. the timings below were computed with c=1 or c=-1 to avoid increasing the values exponentially during the timing.
Question: Why do these two base classes result in the child objects having different behavior? class Base: _instance: "Base" = None def __new__(cls) -> "Base": if cls._instance is None: cls._instance = super().__new__(cls) return cls._instance class A(Base): def foo(self): return "foo" class B(Base): def quz(self): return "quz" a = A() b = B() print(id(a)) print(id(b))  140035075937792 140035075948400  On the other hand from typing import Dict class Base: _instances: Dict[int, "Base"] = {} def __new__(cls) -> "Base": if 0 not in cls._instances: cls._instances[0] = super().__new__(cls) return cls._instances[0] class A(Base): def foo(self): return "foo" class B(Base): def quz(self): return "quz" a = A() b = B() print(id(a)) print(id(b))  140035075947296 140035075947296 

Answer: When a = A() is executed, the __new__ method is called with the class A as its argument. This sets the value of the class attribute A._instance. Likewise, b = B() sets the value of B._instance. In the first case, the original value of Base._instance, A.instance and B._instance is None, which is a non-mutable object, so changing this value in A or B does not affect the other two classes. In the second case, A._instance, B._instance and Base._instance point to the same dictionary. Since a dictionary is a mutable object, modifying this dictionary via one class affect all three classes.
Question: I would like to hint a function as a mapping between instances of a class or its children and a value. takes_mapping below is an example. However, I am getting a static typing error when I use the following: from collections.abc import Mapping class Parent: pass class Child(Parent): pass assert issubclass(Child, Parent) def takes_mapping(mapping: Mapping[Parent, int]): return child = Child() my_dict: dict[Child, int] = {child: 1} my_mapping: Mapping[Child, int] = {child: 1} takes_mapping(my_dict) # typing error... takes_mapping(my_mapping) # same basic error, involving invariance (see below)  Pyright generates the following error: Argument of type "dict[Child, int]" cannot be assigned to parameter "mapping" of type "Mapping[Parent, int]" in function "takes_mapping" "dict[Child, int]" is not assignable to "Mapping[Parent, int]" Type parameter "_KT@Mapping" is invariant, but "Child" is not the same as "Parent" reportArgumentType  How can I hint the argument to take mapping in such a way that the keys may be an instance of Parent or any of its children (without typing errors)? In my use case, we may introduce additional children of Parent and it would be nice we didn't have to couple the hint to the hierarchy, i.e., Union will not really express what's desired since it depends on the specific unioned types.

Answer: The problem is that if the function is typed as taking a Mapping[Parent, ...] the body of the function is expected to try to access that dict with Parent keys, and that's likely not going to work if you pass in a dict[Child, ...]. (It could work depending how you implement __hash__, and you can make an argument for why mypy should allow it, but you can see why it also might make sense for it to be skittish.) To fix this I think you want to use a TypeVar, which allows the type to be narrowed within the bound of Parent: from collections.abc import Mapping from typing import TypeVar class Parent: pass class Child(Parent): pass assert issubclass(Child, Parent) _Parent = TypeVar("_Parent", bound=Parent) def takes_mapping(mapping: Mapping[_Parent, int]): # do stuff with the mapping, with some potentially-narrowed type for the mapping keys return child = Child() my_dict: dict[Child, int] = {child: 1} my_mapping: Mapping[Child, int] = {child: 1} takes_mapping(my_dict) # ok (_Parent is bound to the Child type) takes_mapping(my_mapping) # ok (same)
Question: I've encountered situations where standard methods like __module__ and __class__ become unreliable due to inheritance hierarchies or metaclass-based class creation. I need a robust approach that can accurately identify the module and class, regardless of the complexity of the class structure. Here's an example of a potential issue: class BaseClass: @classmethod def mymethod(cls): print(cls.__module__, cls.__name__) class DerivedClass(BaseClass): pass DerivedClass.mymethod()  This will output __main__ DerivedClass, which might not be the desired result if you need to differentiate between the base class and derived class methods. I'm looking for a solution that can handle scenarios like these and provide accurate module and class information, even in the presence of inheritance, metaclasses, and other advanced Python constructs.

Answer: To get the module and class of the defining type, you can look at the qualified name. With this approach, the derived class could be defined in a separate module entirely. import logging class BaseClass: @classmethod def mymethod(cls): _, _, methodname, _ = logging.getLogger().findCaller() func = getattr(cls, methodname) print("module:", func.__module__) print("method name:", methodname) print("qualname:", func.__qualname__) print("defining class:", func.__qualname__.split(".")[0]) class DerivedClass(BaseClass): pass DerivedClass.mymethod()  Note that direct invocations of findCaller() were bugged until Python 3.11. A somewhat deviant approach which still works in older Python versions: class BaseClass: @classmethod def mymethod(cls): methodname = (lambda:0).__qualname__.split(".")[1] func = getattr(cls, methodname) print("module:", func.__module__) print("method name:", methodname) print("qualname:", func.__qualname__) print("defining class:", func.__qualname__.split(".")[0]) class DerivedClass(BaseClass): pass DerivedClass.mymethod()
Question: I'm working with sympy to obtain symbolic solutions of equation. This is my code: import sympy as sp # Define the symbolic variable x = sp.symbols('x') # Define f(x) f = ((x**2 - 2)**2 - 2)**2 - 2 # Solve the equation f(x) = 0 solutions = sp.solve(f, x) # Filter only the positive solutions positive_solutions = [sol for sol in solutions if sol.is_real and sol > 0] # Print the positive solutions print("The positive solutions of the equation f(x) = 0 are:") for sol in positive_solutions: print(sol)  I'm using SymPy to solve the equation  and I am able to obtain the positive solutions. However, the solutions are returned with nested square roots in a way that makes the inner roots appear on the left side, like this: sqrt(2 - sqrt(2 - sqrt(2))) sqrt(2 - sqrt(sqrt(2) + 2)) sqrt(sqrt(2 - sqrt(2)) + 2) sqrt(sqrt(sqrt(2) + 2) + 2)  I would like the nested square roots to be formatted in a way that they all appear to the right, like this: sqrt(2 - sqrt(2 - sqrt(2))) sqrt(2 - sqrt(sqrt(2) + 2)) sqrt(2 + sqrt(2 - sqrt(2))) sqrt(2 + sqrt(sqrt(2) + 2))  Is there a way to adjust the output formatting of the solutions so that the nested square roots appear as desired? Any help would be greatly appreciated! Thank you!

Answer: Using your current code (applying reverse lexicographic order) print("The positive solutions of the equation f(x) = 0 are:") for sol in positive_solutions: pretty_print(sol, order='rev-lex')  Resulting The positive solutions of the equation f(x) = 0 are: ________________ ╱ ________ ╲╱ 2 - ╲╱ 2 - √2 ________________ ╱ ________ ╲╱ 2 - ╲╱ 2 + √2 ________________ ╱ ________ ╲╱ 2 + ╲╱ 2 - √2 ________________ ╱ ________ ╲╱ 2 + ╲╱ 2 + √2  For further details sorting docs
Question: In the "Fluent Python" book by Luciano Ramalho (2nd ed.) he defines the concepts of container sequences and flat sequences:  A container sequence holds references to the objects it contains, which may be of any type, while a flat sequence stores the value of its contents in its own memory space, not as distinct Python objects.  Can we say that objects of type range and bytearray are flat sequences (I think that these objects can't contain references but not sure)? And is there a simple way to test whether a sequence is flat or not?  I just found this post that gives a quote from the 1st edition of the "Fluent Python" book (I personally only have the 2nd edition):  Flat sequences: str, bytes, bytearray, memoryview, and array.array hold items of one type.  So, it seems that bytearray is a flat sequence. In the 2nd edition (p.22) the author says only that  Flat sequences – Hold items of one simple type. Some examples: str, bytes, and array.array. Container sequences – Can hold items of different types, including nested containers. Some examples: list, tuple, and collections.deque. 

Answer: You have it right about bytearray. It references an internal mutable memory space to hold objects of just one simple type (bytes), so it's a flat sequence, by the author's definition. A range object is a bit more tricky though. I'd say it doesn't match either of the criteria given by the author, though it's certainly a sequence. Rather than containing any values however, it just stores the start, stop and step arguments it's constructed with, and calculates what values it should contain on the fly, as they're needed. It doesn't reference them, as Python objects or otherwise, until then. This means you can create a range of any size in constant time, since nothing but a few integers needs to be allocated. A final note: The definitions of "flat sequence" and "container sequence" are not widely used in the Python development community. They may be useful in the context of your book to explain some differences in container behavior, but they won't be understood by Python programmers more widely. In general, the details of how a given container stores its contents are not usually something you need to care about. Leaving implementation details unspecified is often a deliberate choice by Python's design team, so that if a more efficient implementation becomes available, the code can be switched to it. This happened not long ago with Python's dictionaries, which became ordered (by insertion order) as a side effect of an implementation change. Because many of the previous implementation's details were unspecified, there was no break in backwards compatibility, since all the documented interfaces still worked the same way.
Question: I have 12 Excel files. Each is based on a month of the year and therefore each filename ends in 'Month YYYY'. For example, the file for March of 2021 ends in 'March 2021.xlsx'. I want to read each Excel file, select certain columns, drop empty rows, then merge each file into one excel file as a named worksheet. However, I want to search the file's name, identify the month and then rename the second column to say that month. How do I add a code to have the month of each file be used as the 'new name' for the second column of each df? Here's an example using two months: File one: January 2021.xlsx    A B     1 x   3 x    File three: February 2021.xlsx    A B     3 x   5 x    I want to rename B to represent the month of the respective excel file and then merge to get:    A January February     1 x 0   3 x x   5 0 x    This is what I have done so far. #Store Excel files in a python list excel_files = list(Path(DATA_DIR).glob('*.xlsx'))  #Read each file into a dataframe dfs = [] for excel_file in excel_files: df = pd.read_excel(excel_file,sheet_name='Sheet1',header=5,usecols='A,F',skipfooter=8) df.dropna(how='any', axis=0, inplace = True) df.rename(columns={'old-name': 'new-name'}, inplace=True) dfs.append(df)  #Compile the list of dataframes to merge data_frames = [dfs[0], dfs[1],dfs[2] ... dfs[11]]  #Merge all dataframes df_merged = reduce(lambda left,right: pd.merge(left,right,on=['A'], how='outer'), data_frames).fillna(0)  I need help adding the code to have the month of each file be used as the 'new name' for the second column of each df?

Answer: I think your question is similar to this: Extract month, day and year from date regex an advanced way to do this would be using regex, which is laid out a little in that prior post. a simpler way to do this would be to split (or rsplit) the on (' '), assuming that there is a space in front of the month as well as after: excel_file = "first bit of names MONTH 2021.xlsx": for excel_file in excel_files: new-name = str(excel_file).rsplit(' ', 2)[-2] # creates a list [[first bit of names...], [MONTH], [2021.xlsx]] and takes the 2nd to last element df = pd.read_excel(excel_file,sheet_name='Sheet1',header=5,usecols='A,F',skipfooter=8) df.dropna(how='any', axis=0, inplace = True) df.rename(columns={'old-name': new-name}, inplace=True) dfs.append(df)  I think this answers the question, but you may have another problem getting the "old name" in the way you propose. Hope this helps *edited to match comment
Question: I am writing a Pandas dataframe as HTML using this code import pandas as pd df = pd.DataFrame({ "a": [1] }) print(df.style.to_html())  I ran it once and it produced this output <style type="text/css"> </style> <table id="T_f9297"> <thead> <tr> <th class="blank level0" >&nbsp;</th> <th id="T_f9297_level0_col0" class="col_heading level0 col0" >a</th> </tr> </thead> <tbody> <tr> <th id="T_f9297_level0_row0" class="row_heading level0 row0" >0</th> <td id="T_f9297_row0_col0" class="data row0 col0" >1</td> </tr> </tbody> </table>  But when I run the same program again a moment later it gives <style type="text/css"> </style> <table id="T_d628d"> <thead> <tr> <th class="blank level0" >&nbsp;</th> <th id="T_d628d_level0_col0" class="col_heading level0 col0" >a</th> </tr> </thead> <tbody> <tr> <th id="T_d628d_level0_row0" class="row_heading level0 row0" >0</th> <td id="T_d628d_row0_col0" class="data row0 col0" >1</td> </tr> </tbody> </table>  I would like to get the same output each time. That is, the T_f9297 and T_d628d identifiers shouldn't change from one run to the next. How can I get that? I believe that I could generate HTML without any CSS styling and without the identifiers, but I do want CSS (I just omitted it from my example) and I'm happy to have the identifiers, as long as I get the same output given the same input data. I am using Python 3.11.7 and Pandas 2.1.4.

Answer: pandas.io.formats.style.Styler.to_html has a table_uuid parameter:  table_uuid str, optional: Id attribute assigned to the <table> HTML element in the format: <table id="T_<table_uuid>" ..>  If not provided it generates a random uuid. If set it will use the uuid provided: print(df.style.to_html(table_uuid="my_table_id"))  Output: <style type="text/css"> </style> <table id="T_my_table_id"> <thead> <tr> <th class="blank level0" >&nbsp;</th> <th id="T_my_table_id_level0_col0" class="col_heading level0 col0" >a</th> </tr> </thead> <tbody> <tr> <th id="T_my_table_id_level0_row0" class="row_heading level0 row0" >0</th> <td id="T_my_table_id_row0_col0" class="data row0 col0" >1</td> </tr> </tbody> </table>