eshangj/stackoverflow_q_and_a_sample huggingface dataset:
Question: I have a function like this in Python (the capital letters can represent constants, functions, anything, but not function calls): def f(x): a = foo1(A, B, foo3(E, foo2(A, B))) b = foo3(a, E) return b  and I want to break it up into "atomic" operations like this: def f(x): tmp1 = foo2(A, B) tmp2 = foo3(E, tmp1) a = foo1(A, B, tmp2) b = foo3(a, E) return b  In other words, exactly one function call and one assignment per line. Is there a way I can implement this source code transformation in Python? A program that takes in the string representation of such a function and returns the transformed version. I know I need to use the AST representation, but I don't really know how to proceed.

Answer: I think this is what you want exactly: s = """def f(x): a = foo1(A, B, foo3(E, foo2(A, B))) b = foo3(a, E) return b""" import re result = [] reg = re.compile(r'^(\s+).*?(\w+\([^\(\)]*\))') count = 1 for l in s.splitlines(): r = reg.search(l) if not r: result.append(l) continue while r: indent = r[1] var = f'tmp_{count}' result.append(f'{indent}{var} = {r[2]}') count += 1 l = l[:r.start(2)] + var + l[r.end(2):] r = reg.search(l) else: if l.strip(): result.append(l) print('\n'.join(result))  and it will show: def f(x): tmp_1 = foo2(A, B) tmp_2 = foo3(E, tmp_1) tmp_3 = foo1(A, B, tmp_2) a = tmp_3 tmp_4 = foo3(a, E) b = tmp_4 return b
Question: I'm running a simple python test to read json and output to excel. I'm get the following error: "ValueError("All arrays must be of the same length")" JSON file example (test data) { "content": [ { "id": "test_id", "url": "http://www.google.com/", "path1": "dir1/dir1_subdir_1/hello.txt", "path2": "external://hello.txt", "revision": "613592", "ChangedDate": "2024-01-01T17:00:00+00:00", "ChangedRevision": "12345", "tags": [], "Revision": { "id": "2468", "revision": "987654", "revisionSource": "google", "url": "http://www.google.com" } } ], "requestId": "12345", "errors": [], "warnings": [], "information": [], "isValid": true }  I don't need the last 5 lines in the output file Python code import pandas as pd df = pd.read_json("simple.json",).to_excel("json_output.xlsx", index=False) 

Answer: Assuming you only need content from the json file, you can first use json.load and then use it to create dataframe: import pandas as pd import json with open('simple.json') as json_data: data = json.load(json_data) df = pd.DataFrame(data['content']) # output to excel df.to_excel("json_output.xlsx", index=False)
Question: I'm using Altair to create a layered chart with bars and lines. In this simple example, I'm trying to make the strokedash with different colors without success. I have tried using scale scheme for the strokedash, but it didn't change the colors. Can anyone help? import altair as alt import pandas as pd data = { 'Date': ['2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02'], 'Value': [50, 200, 300, 150, 250, 350, 200, 200, 200, 200, 200, 200, 300, 300, 300, 300], 'DESCR': ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'D', 'D', 'D', 'D'], 'Company': ['X', 'X', 'Y', 'Y', 'X', 'X', 'Y', 'Y', 'X', 'X', 'Y', 'Y', 'X', 'X', 'Y', 'Y'], } source = pd.DataFrame(data) source_aux = source[(source['DESCR']=='C') | (source['DESCR']=='D')] source = source[(source['DESCR']=='A') | (source['DESCR']=='B')] bars = alt.Chart().mark_bar().encode( x='Date:N', y='Value:Q', color=alt.Color('DESCR:N', legend=alt.Legend(title=None)) ) limits = alt.Chart(source_aux).mark_rule().encode( y='Value', strokeDash=alt.StrokeDash('DESCR', legend=alt.Legend(title=None)) ) chart = alt.layer( bars, limits, data=source ).facet( column='Company:N' ) chart 

Answer: strokeDash controls the dash of the line but has no impact on the color. To change the color of the line you can either use the stroke or color encoding.  import altair as alt import pandas as pd data = { 'Date': ['2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02', '2023-01-01', '2023-01-02'], 'Value': [50, 200, 300, 150, 250, 350, 200, 200, 200, 200, 200, 200, 300, 300, 300, 300], 'DESCR': ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'D', 'D', 'D', 'D'], 'Company': ['X', 'X', 'Y', 'Y', 'X', 'X', 'Y', 'Y', 'X', 'X', 'Y', 'Y', 'X', 'X', 'Y', 'Y'], } source = pd.DataFrame(data) source_aux = source[(source['DESCR']=='C') | (source['DESCR']=='D')] source = source[(source['DESCR']=='A') | (source['DESCR']=='B')] bars = alt.Chart().mark_bar().encode( x='Date:N', y='Value:Q', color=alt.Color('DESCR:N', legend=alt.Legend(title=None)) ) limits = alt.Chart(source_aux).mark_rule().encode( y='Value', strokeDash=alt.StrokeDash('DESCR', legend=alt.Legend(title=None)), stroke=alt.Stroke('DESCR', legend=alt.Legend(title=None)) ) chart = alt.layer( bars, limits, data=source ).facet( column='Company:N' ) chart
Question: I'm dealing with a list of integers that represent the pages in which a keyword was found. I would like to build block of code that converts this list into a string with a specific format that follow some simple rules. Single integers are converted into string. Consecutive integers are considered as intervals (left bound, hyphen, right bound) and then converted into string. Each conversion is comma separated. This is an example of input and desired output: input = [4, 5, 6, 7, 8, 9, 10, 22, 23, 26, 62, 63, 113, 137, 138, 139] desired_output = "4-10, 22, 23, 26, 62, 63, 113, 137-139"  I wrote this code: res = [4, 5, 6, 7, 8, 9, 10, 22, 23, 26, 62, 63, 113, 137, 138, 139] if len(res)>0: resStr = str(res[0]) isConsecutive = False for index in range(1, len(res)): diff = res[index] - res[index-1] if diff == 1: isConsecutive = True if index == len(res)-1: resStr = resStr + "-" + str(res[index]) continue else: if isConsecutive: isConsecutive = False resStr = resStr + "-" + str(res[index-1]) + ", " + str(res[index]) else: resStr = resStr + ", " + str(res[index]) print(res) print(resStr)  This code gives me as a result: 4-10, 22-23, 26, 62-63, 113, 137-139  It doesn't recognize that only two consecutive numbers have not be considered as an interval: "22, 23" and not "22-23", as well as "62, 63" and not "62-63". How can be solved this issue? Is there a simpler or more efficient way to perform the conversion?

Answer: Track first and last of a range of consecutive numbers, collect the string of the range when a number is non-consecutive. I used a sub-function to avoid repeated code: res = [4, 5, 6, 7, 8, 9, 10, 22, 23, 26, 62, 63, 113, 137, 138, 139] def span(numbers): if len(numbers) == 0: # empty case return '' result = [] def _process(first, last): if first != last: # range, but handle consecutive case result.append(f"{first}{'-' if last - first > 1 else ', '}{last}") else: result.append(str(first)) # first==last, single number first = last = numbers[0] # initialize range tracking to first number for item in numbers[1:]: # iterate remaining numbers if item <= last: # input not increasing raise ValueError('not consecutive numbers') if item == last + 1: # next number is consecutive last = item else: _process(first, last) # next number not consecutive first = last = item # reset for next range _process(first, last) # no more numbers, finish last range return ', '.join(result) print(res) print(span(res)) # Test cases print(span([])) print(span([1])) print(span([1,2])) print(span([1,2,3])) print(span([1,2,4])) print(span([1,3])) print(span([1,3,4])) print(span([1,3,4,5])) print(span([1,2,3,5])) print(span([1,2,4,5,6])) print(span([1,2,3,5,6])) print(span([1,2,3,5,6,7]))  Output: [4, 5, 6, 7, 8, 9, 10, 22, 23, 26, 62, 63, 113, 137, 138, 139] 4-10, 22, 23, 26, 62, 63, 113, 137-139 1 1, 2 1-3 1, 2, 4 1, 3 1, 3, 4 1, 3-5 1-3, 5 1, 2, 4-6 1-3, 5, 6 1-3, 5-7
Question: I have a dataframe that has user ID, code, and value. user code value 0001 P 10 0001 P 20 0001 N 10 0002 N 40 0002 N 30 0003 P 10  I am trying to add a new column that groups by User ID, filters for code = P and sums the value. However I want this value to be applied to every row. So for the example above, the output I'm looking for would be: user code value Sum_of_P 0001 P 10 30 0001 P 20 30 0001 N 10 30 0002 N 40 0 0002 N 30 0 0003 P 10 10  I tried doing df['Sum_of_P'] = df.loc[df['code'] == 'P', 'value'].groupby(df['user']).transform('sum'), but this applies only to the rows with code = P. Is there a way to have to have this apply to all rows?

Answer: Use a mask and where rather than loc: df['Sum_of_P'] = (df['value'].where(df['code'].eq('P'), 0) .groupby(df['user']).transform('sum') )  Variant with NaNs as masked values: df['Sum_of_P'] = (df['value'].where(df['code'].eq('P')) .groupby(df['user']).transform('sum') .convert_dtypes() )  If you want to use loc you should aggregate rather than transform, then map the values from the group: s = df.loc[df['code'] == 'P'].groupby('user')['value'].sum() df['Sum_of_P'] = df['user'].map(s).fillna(0).convert_dtypes()  Output:  user code value Sum_of_P 0 1 P 10 30 1 1 P 20 30 2 1 N 10 30 3 2 N 40 0 4 2 N 30 0 5 3 P 10 10
Question: Given a python list that contains nested lists of arbitrary levels of nesting, the goal is to return a completely flattened list i.e for the sample input [1, [2], [[[3]]], 1], the output should be [1, 2, 3, 1]. My solution: def flatten(lst): stack = [[lst, 0]] result = [] while stack: current_lst, start_index = stack[-1] for i in range(start_index, len(current_lst)): if isinstance(current_lst[i], list): # Update the start_index of current list # to the next element after the nested list stack[-1][1] = i + 1 # Add nested list to stack # and initialize its start_index to 0 stack.append([current_lst[i], 0]) # Pause current_lst traversal break # non list item # add item to result result.append(current_lst[i]) else: # no nested list # remove current list from stack stack.pop() return result  I would like to know the time and space complexity (auxiliary space) of my solution if correct. What I think Time Complexity: I believe the solution has a time complexity of O(m + n) where m is the total number of nested lists at all levels and n is the total number of atomic elements at all levels (non list elements) Space Complexity: I believe the space complexity is O(d), where d is the depth of the most nested list. This is because the stack tracks the current state of traversal, and its size is proportional to the nesting depth Is the solution correct? Is the time and space analysis correct?

Answer: Yes, the solution is correct. Yes, the time and space analysis are correct ... if you don't count the space used by result as auxiliary space, which is reasonable. Although note that result overallocates/reallocates, which you could regard as taking O(n) auxiliary space. You could optimize that by doing two passes over the whole input, one to count the atomic elements, then create result = [None] * n, and then another pass to fill it. Btw it's better to use iterators instead of your own list+index pairs (Attempt This Online!): def flatten(lst): stack = [iter(lst)] result = [] while stack: for item in stack[-1]: if isinstance(item, list): stack.append(iter(item)) break result.append(item) else: stack.pop() return result  Or with the mentioned space optimization (Attempt This Online!): def flatten(lst): def atoms(): stack = [iter(lst)] while stack: for item in stack[-1]: if isinstance(item, list): stack.append(iter(item)) break yield item else: stack.pop() n = sum(1 for _ in atoms()) result = [None] * n i = 0 for result[i] in atoms(): i += 1 return result
Question: I would like to take the product within a group in polars The following works but I suspect there is a more elegant/efficient way to perform this operation. Thank you import polars as pl import numpy as np D = pl.DataFrame({'g':['a','a','b','b'],'v':[1,2,3,4],'v2':[2,3,4,5]}) D.group_by('g').agg(pl.all().map_elements( lambda group: np.prod(group.to_numpy()),return_dtype=pl.Float64)) 

Answer: You could use Expr.product: D.group_by('g').agg(pl.all().product())  Output: ┌─────┬─────┬─────┐ │ g ┆ v ┆ v2 │ │ --- ┆ --- ┆ --- │ │ str ┆ i64 ┆ i64 │ ╞═════╪═════╪═════╡ │ b ┆ 12 ┆ 20 │ │ a ┆ 2 ┆ 6 │ └─────┴─────┴─────┘  If you want Floats: D.group_by('g').agg(pl.all().product().cast(pl.Float64)) ┌─────┬──────┬──────┐ │ g ┆ v ┆ v2 │ │ --- ┆ --- ┆ --- │ │ str ┆ f64 ┆ f64 │ ╞═════╪══════╪══════╡ │ b ┆ 12.0 ┆ 20.0 │ │ a ┆ 2.0 ┆ 6.0 │ └─────┴──────┴──────┘
Question: I have been trying to run the below code to calculate upper and lower confidence intervals using t distribution, but it keeps throwing the error in the subject. The piece of code is as below: def trans_threshold(Day): Tran_Cnt=Tran_Cnt_DF[['Sample',Day]].dropna() Tran_Cnt=Tran_Cnt.astype({'Sample':'str'}) Tran_Cnt.dtypes #Finding outliers in Materiality via IQR X_Tran = Tran_Cnt.drop('Sample', axis=1) Tran_arr1 = X_Tran.values #Finding the first quartile Tran_q1= np.quantile(Tran_arr1, 0.25) # finding the 3rd quartile Tran_q3 = np.quantile(Tran_arr1, 0.75) # finding the iqr region Tran_iqr = Tran_q3-Tran_q1 # finding upper and lower outliers Tran_upper_bound = Tran_q3+(1.5*Tran_iqr) Tran_lower_bound = Tran_q1-(1.5*Tran_iqr) # removing outliers Tran_arr2 = Tran_arr1[(Tran_arr1 >= Tran_lower_bound) & (Tran_arr1 <= Tran_upper_bound)] #Using t distribution for Materiality Limits Tran_Threshold_mat=st.t.interval(alpha=0.99999999999, df=len(Tran_arr2-1), loc=np.mean(Tran_arr2), scale=st.sem(Tran_arr2)) return Tran_Threshold_mat trn_lim_FullFeed_Mon = trans_threshold(Day) --------------------------------------------------------------------------- TypeError Traceback (most recent call last) Cell In[106], line 19 17 Tran_arr2 = Tran_arr1[(Tran_arr1 >= Tran_lower_bound) & (Tran_arr1 <= Tran_upper_bound)] 18 #Using t distribution for Materiality Limits ---> 19 Tran_Threshold_mat=st.t.interval(alpha=0.99999999999, df=len(Tran_arr2-1), 20 loc=np.mean(Tran_arr2), 21 scale=st.sem(Tran_arr2)) TypeError: rv_generic.interval() missing 1 required positional argument: 'confidence'  The issue seems to be with piece of code below. However, I have provided all parameters required to calculate confidence intervals, including degrees of freedom, but it still gives this error. Where am I going wrong and what needs to be done? Tran_Threshold_mat=st.t.interval(alpha=0.99999999999, df=len(Tran_arr2-1), loc=np.mean(Tran_arr2), scale=st.sem(Tran_arr2))  Also, the Tran_arr2 list looks like below: array([12617., 12000., 1123., 537., 8605., 4365., 11292., 12231., 7640., 9583., 9257., 13864., 14682., 11744., 10501., 8694., 5327., 10066., 13022., 11092., 7444., 11658., 14920., 12849., 14681., 5719., 11029., 3814., 14703., 5593., 9772., 8851., 9551., 15975., 6532., 13827., 8547.])  Hence, there is no issue, up until the last like of the code block which estimates confidence intervals using t distribution. I have used the below packages: import pandas as pd import numpy as np import scipy.stats as st import matplotlib.pyplot as plt import matplotlib.ticker as tkr import matplotlib.scale as mscale from matplotlib.ticker import FixedLocator, NullFormatter pd.options.display.float_format = '{:.0f}'.format pd.options.mode.chained_assignment = None 

Answer: Note that the signature of scipy.stats.t is interval(confidence, df, loc=0, scale=1). There is no alpha keyword, pass it as positional or relabel it to confidence.
Question: Say I have a 1D numpy-array given by np.array([1,2,3]). Is there a built-in command for calculating the relative difference between each element and display it in a 2D-array? The result would then be given by np.array([[0,-50,-100*2/3], [100,0,-100*1/3], [200,50,0]]) Alternatively I would have to use a for-loop.

Answer: Use numpy broadcasting: a = np.array([1,2,3]) out = (a[:, None]-a)/a*100  Output: array([[ 0. , -50. , -66.66666667], [100. , 0. , -33.33333333], [200. , 50. , 0. ]])
Question: I am currently trying out moviepy to burn subtitles on a video. However I keep getting the same error message no matter what I do. This is the code I am using: from moviepy import TextClip from moviepy.video.tools.subtitles import SubtitlesClip ... generator = lambda txt: TextClip( text = self.txt, font = self.font_path, font_size = 100, color= self.text_color, stroke_color="black", stroke_width=5, ) subtitles = SubtitlesClip(self.subtitles_path, generator)  where self.font_path is currently the string "/fonts/my_font.otf". Regardless of the font however, I keep getting the error message: ValueError: Invalid font <function _spyderpdb_code.<locals>.<lambda> at 0x7b89f4d60e00>, pillow failed to use it with error 'function' object has no attribute 'read' 

Answer: I was facing the same issue and I managed to solve it.  Corrected code from moviepy import TextClip from moviepy.video.tools.subtitles import SubtitlesClip ... generator = lambda txt: TextClip( self.font_path, text = self.txt, font_size = 100, color= self.text_color, stroke_color="black", stroke_width=5, ) subtitles = SubtitlesClip(self.subtitles_path, make_textclip=generator)   Explanation The error occurs on this portion of code in the VideoClip.py from MoviePy: try: _ = ImageFont.truetype(font) except Exception as e: raise ValueError( "Invalid font {}, pillow failed to use it with error {}".format(font, e) )  The thing is that font is interpreted as a your generator function. As said AKX:  pass your generator as make_textclip  But this did not fully solve the issue because you will face another error:  TypeError: multiple values for argument 'font'  To solve it you need to look at the TextClip class: class TextClip(ImageClip): ... @convert_path_to_string("filename") def __init__( self, font, text=None, filename=None, font_size=None, size=(None, None), margin=(None, None), color="black", bg_color=None, stroke_color=None, stroke_width=0, method="label", text_align="left", horizontal_align="center", vertical_align="center", interline=4, transparent=True, duration=None, ):  The first argument needed to initialise a TextClip object is the font as a positional argument and then the text as a named argument. Note You also don't need to pass the full path of the font based on PIL's truetype doctring: :param font: A filename or file-like object containing a TrueType font. If the file is not found in this filename, the loader may also search in other directories, such as: * The :file:`fonts/` directory on Windows, * :file:`/Library/Fonts/`, :file:`/System/Library/Fonts/` and :file:`~/Library/Fonts/` on macOS. * :file:`~/.local/share/fonts`, :file:`/usr/local/share/fonts`, and :file:`/usr/share/fonts` on Linux; or those specified by the ``XDG_DATA_HOME`` and ``XDG_DATA_DIRS`` environment variables for user-installed and system-wide fonts, respectively.  It will automatically search for the font in the correct folder according to your OS.
Question: Specifically: Write a function (in python, java or pseudocode) average(nums) that prints the mean of the numbers in list nums, and returns a tuple:  The tuple's first element is the index of the number furthest from the mean. The tuple's second element is the number furthest from the mean.  Can this be done with one loop? I know how to write the function with two loops, and using one or two loops are both O(n) time, so the code doesn't technically become more efficient with one loop. But I'm just curious if it can be done. A party trick, if you will. Some of my thought process: So computing the average as you iterate through the list is as simple as total += nums[i], then avg = total/(i+1). And I guess you could calculate which number is furthest from your current average, but by the time you get to the end of the list an extreme number could make your code incorrect , I think?

Answer:  Is it possible to find the average of a group of numbers and return the index of the number furthest from the average with one loop?  Yes. The number furthest from the mean will be either the largest or the smallest in the sample. With a single pass through the data, you can compute the average and track the indices of the largest and smallest elements seen so far. At the end, you'll know the indices of the overall largest and smallest. You can check which of these is further from the mean, and return the information for that one.
Question: Given a python list of lists containing numbers i.e lists = [ [1, 2], [2, 1], [3, 4] ], the problem is to return a list of all unique lists from the input list. A list is considered a duplicate if it can be generated from another list by reordering the items in the list. i.e [2, 1] is a duplicate of [1, 2]. Given the input [ [1, 2], [2, 1], [3, 4] ], the output should be [ [1, 2], [3, 4]] . Any reordering of [ [1, 2], [3, 4]]  is also correct i.e [1, 2], [4, 3]], My approach is to first sort all lists in the input list, convert the lists to tuples, use a set data structure to filter out duplicate tuples, and finally convert the unique tuples back to lists. The time complexity for sorting all lists is O(m*nlogn) where m is the number of lists and n is the size of each list(assuming same size lists). Converting the lists to tuples takes O(mn) time, creating a set from the tuples takes O(mn), converting the set of unique tuples back to lists also takes O(mn) making the total time complexity O(mnlogn + mn + mn + mn) = O(mnlogn). Can we do any better than O(mnlogn)? Code: def find_unique(lists): sorted_lists = [ sorted(lst) for lst in lists] tuples = [tuple(lst) for lst in sorted_lists] unique_tuples = set(tuples) return list(unique_tuples) 

Answer: You can get an O(m*n) solution as long as the "key" you are using is O(m*n). This can be accomplished in two ways. If the inner lists can't contain duplicates, then a set of frozen sets is an elegant solution. Note, frozenset(mylist) is O(n): def unique(lists): seen = set() result = [] for sub in lists: key = frozenset(sub) if key not in seen: result.append(sub) seen.add(key) return result  The above returns the first seen "unique" list that is actually in the input. If any order of a unique list is valid, even an order not seen in the original input (I presume this is the case because that is what your solution does) then perhaps more tersely: def unique(lists): return list(map(list, set(map(frozenset, lists))))  If the inner lists can contain duplicates, then the above won't work, but you can use a collections.Counter which can act like a multiset, then use a frozent-set of the items in the counter: from collections import Counter def unique(lists): seen = set() result = [] for sub in lists: key = frozenset(Counter(sub).items()) if key not in seen: result.append(sub) seen.add(key) return result  Note, if n is smallish, I bet the sorted solution is faster.