The full dataset viewer is not available (click to read why). Only showing a preview of the rows.
The dataset generation failed because of a cast error
Error code: DatasetGenerationCastError
Exception: DatasetGenerationCastError
Message: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 2 new columns ({'Name', 'CategoryID'}) and 2 missing columns ({'errors', 'data'}).
This happened while the json dataset builder was generating data using
hf://datasets/antoinelb7/alloprof/data/questions/categories.json (at revision 0faa90fee1ad1a6e3e461d7be49abf71488e6687)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2011, in _prepare_split_single
writer.write_table(table)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/arrow_writer.py", line 585, in write_table
pa_table = table_cast(pa_table, self._schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2302, in table_cast
return cast_table_to_schema(table, schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2256, in cast_table_to_schema
raise CastError(
datasets.table.CastError: Couldn't cast
Name: string
CategoryID: int64
to
{'errors': [{'extensions': {'code': Value(dtype='string', id=None)}, 'locations': [{'column': Value(dtype='int64', id=None), 'line': Value(dtype='int64', id=None)}], 'message': Value(dtype='string', id=None), 'path': Sequence(feature=Value(dtype='string', id=None), length=-1, id=None)}], 'data': {'file': {'breadcrumbs': {'children': [{'routerLink': Value(dtype='string', id=None), 'title': Value(dtype='string', id=None)}], 'current': {'routerLink': Value(dtype='string', id=None), 'title': Value(dtype='string', id=None)}, 'parents': [{'routerLink': Value(dtype='string', id=None), 'title': Value(dtype='string', id=None)}], 'root': {'routerLink': Value(dtype='string', id=None), 'title': Value(dtype='string', id=None)}}, 'code': Value(dtype='string', id=None), 'image': Value(dtype='null', id=None), 'imageStyles': Sequence(feature=Value(dtype='null', id=None), length=-1, id=None), 'introduction': Value(dtype='string', id=None), 'lang': Value(dtype='string', id=None), 'levels': Sequence(feature=Value(dtype='int64', id=None), length=-1, id=None), 'metatags': [{'attributes': {'content': Value(dtype='string', id=None), 'href': Value(dtype='string', id=None), 'name': Value(dtype='string', id=None), 'property': Value(dtype='string', id=None), 'rel': Value(dtype='string', id=None)}, 'tag': Value(dtype='string', id=None)}], 'navigation': [{'anchor': Value(dtype='string', id=None), 'level': Value(dtype='string', id=None), 'text': Value(dtype='string', id=None), 'type': Value(dtype='null', i
...
), 'type': Value(dtype='string', id=None), 'uuid': Value(dtype='string', id=None), 'verticalScrollbar': Value(dtype='bool', id=None), 'video': {'code': Value(dtype='string', id=None), 'description': Value(dtype='null', id=None), 'image': Value(dtype='null', id=None), 'imageStyles': Sequence(feature=Value(dtype='null', id=None), length=-1, id=None), 'tags': Sequence(feature=Value(dtype='string', id=None), length=-1, id=None), 'title': Value(dtype='string', id=None), 'uuid': Value(dtype='string', id=None), 'videoId': Value(dtype='string', id=None)}, 'vocabularies': [{'audio': Value(dtype='string', id=None), 'image': {'alt': Value(dtype='string', id=None), 'url': Value(dtype='string', id=None)}, 'imageStyles': [{'link': Value(dtype='string', id=None), 'style': Value(dtype='string', id=None)}], 'name': Value(dtype='string', id=None), 'translation': Value(dtype='string', id=None), 'type': Value(dtype='string', id=None), 'uuid': Value(dtype='string', id=None)}], 'width': Value(dtype='int64', id=None)}], 'slug': Value(dtype='string', id=None), 'title': Value(dtype='string', id=None), 'type': Value(dtype='string', id=None), 'uuid': Value(dtype='string', id=None)}], 'slug': Value(dtype='string', id=None), 'subtype': Value(dtype='string', id=None), 'tags': Sequence(feature=Value(dtype='string', id=None), length=-1, id=None), 'title': Value(dtype='string', id=None), 'topic': Value(dtype='string', id=None), 'type': Value(dtype='string', id=None), 'uuid': Value(dtype='string', id=None)}}}
because column names don't match
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1577, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1191, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1027, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1122, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1882, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2013, in _prepare_split_single
raise DatasetGenerationCastError.from_cast_error(
datasets.exceptions.DatasetGenerationCastError: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 2 new columns ({'Name', 'CategoryID'}) and 2 missing columns ({'errors', 'data'}).
This happened while the json dataset builder was generating data using
hf://datasets/antoinelb7/alloprof/data/questions/categories.json (at revision 0faa90fee1ad1a6e3e461d7be49abf71488e6687)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)Need help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
data
dict | errors
null |
|---|---|
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"text": "<p>L’<strong>usinage</strong>, aussi nommé <em>façonnage</em>, est l’action de modifier un matériau pour lui donner la configuration voulue à l’aide d’outils et de machines-outils.</p>\r\n",
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"text": "<p>Lors de l’usinage, on utilise différents outils et diverses techniques afin de façonner les pièces telles qu’elles sont décrites par les <a data-entity-substitution=\"canonical\" data-entity-type=\"node\" data-entity-uuid=\"ccf8ea9f-1f07-4d9e-b72c-587180861b92\" href=\"/fr/eleves/bv/sciences/le-dessin-de-fabrication-s1409\">dessins de fabrication</a> présentés dans la <a data-entity-substitution=\"canonical\" data-entity-type=\"node\" data-entity-uuid=\"29e34767-0f97-41d4-87f4-4265ab076e3d\" href=\"/fr/eleves/bv/sciences/la-gamme-de-fabrication-s1452#gamme-fabrication\">gamme de fabrication</a>.</p>\r\n\r\n<p>L’usinage se déroule généralement de la façon suivante.</p>\r\n\r\n<ol>\r\n\t<li>\r\n\t<p>L’ébauchage</p>\r\n\t</li>\r\n\t<li>\r\n\t<p>Le choix des outils et des techniques</p>\r\n\t</li>\r\n\t<li>\r\n\t<p>Le façonnage du matériau</p>\r\n\t</li>\r\n</ol>\r\n\r\n<p>D’abord, l’ébauchage consiste à couper grossièrement le matériau afin d’en retirer les parties superflues et de ne conserver que la portion nécessaire à la fabrication.<br/>\r\n<br/>\r\nEnsuite, les techniques, les instruments et les outils sont choisis en fonction du type de matériau à façonner. Par exemple, puisque les métaux sont généralement plus durs que le bois, on utilise des forets et des lames de scie de nature différente pour travailler ces matériaux.<br/>\r\n<br/>\r\nFinalement, le façonnement de la pièce permet de donner la configuration souhaitée aux matériaux.</p>\r\n",
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"text": "<p>Le <strong>cambrage</strong>, parfois appelé <em>pliage</em>, est une technique qui consiste à plier un matériau de façon permanente afin de modifier sa forme.</p>\r\n",
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"text": "<p>Les matériaux les plus susceptibles de faire l’objet d’un cambrage sont les métaux et les thermoplastiques puisqu’ils ont une bonne <a data-entity-substitution=\"canonical\" data-entity-type=\"node\" data-entity-uuid=\"c5a1d7bf-2a8a-4d19-8196-901c3db23830\" href=\"/fr/eleves/bv/sciences/les-proprietes-des-materiaux-s1447#la-malleabilite\">malléabilité</a>. Le cambrage provoque une déformation plastique du matériau usiné; le pli y reste marqué de façon permanente.</p>\r\n",
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"text": "<p>Pour réaliser un pliage, on peut utiliser une presse à plier. Les feuilles de métal y sont pliées par la pression exercée entre le poinçon (la partie supérieure de la presse) et la matrice (la partie inférieure de la presse) qui ont des formes en V.</p>\r\n",
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"text": "<p>Pour réaliser un taraudage, on fixe un taraud sur un porte-taraud (aussi nommé <em>tourne-à-gauche</em>) et on le visse à l’intérieur du matériau à tarauder. Cette opération peut également être réalisée en fixant un taraud au mandrin d’une perceuse. Le taraud utilisé doit être choisi en fonction du type de filet souhaité et de la nature du matériau à tarauder.</p>\r\n",
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"text": "<p><strong>a)</strong><br/>\r\n<br/>\r\nTo answer question a), use the formula. To calculate the kinetic energy, first <a data-entity-substitution=\"canonical\" data-entity-type=\"node\" data-entity-uuid=\"f0f45def-20d2-4c55-a3ee-191cf2f8bde6\" href=\"/en/students/vl/mathematics/units-for-measuring-mass-and-their-conversion-m1394#converting-units-of-mass\">convert the mass</a> of the ball into kilograms, since this is the unit of measurement of |m| for the formula for kinetic energy. |58\\ \\text{g}=0.058\\ \\text{kg}|<br/>\r\n||\\begin{align} E_k &= \\frac{1}{2}\\color{blue}{m}v^2 \\\\ E_k &= \\frac{1}{2}\\color{blue}{(0.058)}v^2 \\\\ E_k &=0.029 v^2 \\end{align}||</p>\r\n\r\n<p>Now, we want to know the value of |E_k| when |v=198\\ \\text{km/h}|. Before replacing |v| with the actual speed in the equation of the function, first <a href=\"#\">convert the km/h to m/s</a>. ||\\begin{align} 198\\ \\frac{\\text{km}}{h} &= 198\\ 000\\ \\frac{\\text{m}}{h} \\times \\frac{1\\ \\text{h}}{3\\ 600\\ \\text{s}} \\\\ &= 55\\ \\frac{\\text{m}}{s} \\end{align}||</p>\r\n\r\n<p>Finally, calculate |E_k| when |v=55|.||\\begin{align} E_k &= 0.029\\color{green}{v}^2 \\\\ &= 0.029\\color{green}{(55)}^2 \\\\&= 87.725\\ \\text{joules} \\end{align}||</p>\r\n\r\n<p><strong>Answer</strong>: |E_k=87.725\\ \\text{joules}|</p>\r\n\r\n<hr/>\r\n<p><br/>\r\n<strong>b)</strong><br/>\r\n<br/>\r\nBe careful: here the equation is not the same as in question a), because the mass of the ball is not the same. In this question, |m=44\\ \\text{g}=0.044\\ \\text{kg}|. The formula is therefore:||\\begin{align} E_k &= \\frac{1}{2}\\color{blue}{m}v^2 \\\\ E_k &= \\frac{1}{2}\\color{blue}{(0.044)}v^2 \\\\ E_k &=0.022 v^2 \\end{align}||</p>\r\n\r\n<p>Now we want to know the value of |v| when |E_k=87.725\\ \\text{joules}|. Replace |E_k| with |87.725| and isolate |v|.||\\begin{align} \\color{brown}{E_k} &= 0.022v^2 \\\\ \\color{brown}{87.725} &= 0.022v^2 \\\\ \\frac{87.725}{\\color{red}{0.022}} &= \\frac{0.022v^2 }{\\color{red}{0.022}} \\\\ 3\\ 987.5 &= v^2 \\\\ \\sqrt{3\\ 987.5} &= \\sqrt{v^2} \\\\ 63.15\\ \\text{m/s} &\\approx v \\end{align}||</p>\r\n\r\n<p>All that is left to do is convert the answer from m/s to km/h. ||\\begin{align} 63.15\\ \\frac{\\text{m}}{\\text{s}} \\times \\frac{3\\ 600\\ \\text{s}}{1\\ \\text{h}} &= 227\\ 340\\ \\frac{\\text{m}}{\\text{h}} \\\\ &= 227.34\\ \\frac{\\text{km}}{\\text{h}} \\end{align}||</p>\r\n\r\n<p><strong>Answer:</strong> The golf ball is moving at a speed of |227.34\\ \\text{km/h}.|<br/>\r\n<br/>\r\nThe problem is represented in the following Cartesian plane.</p>\r\n",
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"alt": "This graph shows the kinetic energy of a tennis ball and a golf ball as a function of their speed.",
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"text": "<p>Only the first quadrant is represented because neither the speed nor the energy can be negative.</p>\r\n\r\n<p>Notice that the kinetic energy of the tennis ball increases more rapidly since its mass is greater, which means that the parameter |a| of the function is greater. This is why a golf ball must be moving with greater speed than a tennis ball to have the same kinetic energy, confirming our results.</p>\r\n",
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"text": "<p>The kinetic energy of an object, denoted |E_k,| is the energy that an object possesses due to its motion. The formula for calculating the kinetic energy of an object as a function of its speed is a second-degree polynomial function. ||\\begin{align} E_k = \\frac{1}{2}&mv^2 \\\\\\\\ \\text{where} \\quad E_k &: \\text{kinetic energy (J)}\\\\ m\\ &: \\text{mass of the object (kg)} \\\\ v\\ \\ &: \\text{speed of the object (m/s)}\\end{align}||</p>\r\n\r\n<p><strong>a)</strong> What is the kinetic energy of a tennis ball that weighs |58| grams and moves at a speed of |198\\ \\text{km/h}|?<br/>\r\n<br/>\r\n<strong>b)</strong> What is the speed in |\\text{km/h}| of a golf ball weighing |44| grams if it has the same kinetic energy as the tennis ball in question <strong>a)</strong>?</p>\r\n",
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"text": "<p>It is very important to define the variables.<br/>\r\n<br/>\r\n|x|: number of months since the share was listed on the stock market<br/>\r\n|y|: share value in |$|</p>\r\n\r\n<p>The statement from the word problem confirms that the function passes through the point |(6,4)| which corresponds to the minimum value of the function.<br/>\r\n<br/>\r\n<strong>a)</strong><br/>\r\n<br/>\r\nThe problem says that the function passes through the point |(0,6).|<br/>\r\n<br/>\r\nTo answer the question, first determine the equation of the function. The minimum of the function is known, that is, its vertex |(h,k)=(6,4).| Additionally, a second point |(0,6)| is also known. Use the vertex form |y=a(x-h)^2+k| to determine the equation.</p>\r\n\r\n<p>Find the value of the parameter |a| by substituting in the known values in the equation. ||\\begin{align}y&=a(x-h)^2+k \\\\6&=a(0-6)^2+4\\\\6&=a(-6)^2+4\\\\6&=36a + 4\\\\2 &= 36a\\\\ \\displaystyle \\frac{1}{18}&=a \\end{align}|| So, the equation of the function is |y=\\displaystyle \\frac{1}{18}(x-6)^2+4.| To answer the question, simply replace |x| with |4| to find the value of |y.| ||\\begin{align}y&=\\displaystyle \\frac{1}{18}(4-6)^2 + 4\\\\y&= \\displaystyle \\frac{1}{18}(-2)^2+4\\\\y&=\\displaystyle \\frac{1}{18} (4) + 4\\\\y&= \\displaystyle \\frac{2}{9} + 4\\\\y&=\\displaystyle \\frac{38}{9} \\approx 4.22\\end{align}|| <strong>Answer: </strong>Four months after being listed, the stock is valued at |\\$ 4.22| per share.</p>\r\n\r\n<hr/>\r\n<p><strong>b)</strong><br/>\r\n<br/>\r\nTo determine after how many months the stock is valued at |\\$ 5.00,| replace |y| with |5|, and then isolate |x.| ||5=\\displaystyle \\frac{1}{18}(x-6)^2+4||<br/>\r\n<br/>\r\nThe left side of the equation must be equal to 0. ||0=\\displaystyle \\frac{1}{18}(x-6)^2-1||<br/>\r\n<br/>\r\nThe formula |x_{1,2} = h \\pm \\displaystyle \\sqrt{-\\frac{k}{a}}| can be used to find the missing value.</p>\r\n\r\n<p>In this case, |a=\\dfrac{1}{18},h=6,| and |k=-1.| ||\\begin{align}x_{1,2} &= \\displaystyle 6 \\pm \\sqrt{-\\frac{-1}{\\frac{1}{18}}}\\\\x_{1,2} &= 6 \\pm \\sqrt{18}\\end{align}||<br/>\r\n<br/>\r\nThen, separate the formula into two parts. ||\\begin{align}x_1 &= 6 + \\sqrt{18} \\approx 10.24\\\\x_2 &= 6 - \\sqrt{18} \\approx 1.76 \\end{align}||<br/>\r\n<br/>\r\nSo, to answer the question, disregard the first value found and only keep the second value.<br/>\r\n<br/>\r\n<strong>Answer:</strong> The value of the share is |\\$ 5.00| a little over a month after being listed.</p>\r\n",
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"text": "<p>A stock listed on the stock exchange reaches a minimum value of |\\$ 4.00| six months after being listed on the stock exchange. The function that represents the decrease in the value of the stock during the first six months after being listed is a quadratic polynomial function.<br/>\r\n<br/>\r\n<strong>a)</strong> If the stock was valued at |\\$ 6.00| when it was first listed, what is its value four months later?<br/>\r\n<br/>\r\n<strong>b)</strong> At what point in the first six months did the stock reach a value of |\\$ 5.00|?</p>\r\n",
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"text": "<p>In the previous example, there were only equations and not inequalities. A final question using inequalities could be: \"During what time interval was the stock's value less than |\\$ 5.00|?\"</p>\r\n\r\n<p>To find out how to solve this kind of question, consult the following concept sheet: <a data-entity-substitution=\"canonical\" data-entity-type=\"node\" data-entity-uuid=\"ff32c663-e5d6-4291-aa9e-08ae96a16293\" href=\"/en/students/vl/mathematics/solving-a-second-degree-equation-or-inequality-m1397#solving-a-second-degree-polynomial\">Solving a Single-variable Second-degree Polynomial Inequality.</a></p>\r\n",
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"text": "<p>It is very important to define the variables.<br/>\r\n<br/>\r\n|x|: time of day (in hours)<br/>\r\n|y|: amount of water in the tank</p>\r\n\r\n<p>It is possible to find two points belonging to the function. These are the points |(12, 25\\ 000)| and |(20,\\ 0).|<br/>\r\n<br/>\r\nThe point |(12, 25\\ 000)| corresponds to the maximum value of the function.<br/>\r\n<br/>\r\n<strong>a)</strong><br/>\r\n<br/>\r\nTo <a href=\"#\">find the equation of the function</a> modeling this situation, use the vertex form of the quadratic function. The point |(12, 25\\ 000)| being the vertex, the vertex form is best for finding the equation.</p>\r\n\r\n<p>The vertex form of the quadratic function is |y=a(x-h)^2+k.| In this situation, |(h,k)=(12, 25\\ 000)| and |(x,y)=(20,0).| Then replace these parameters and variables in the equation. ||y=a(x-h)^2+k \\rightarrow 0=a(20-12)^2+25\\ 000||<br/>\r\nNow isolate the value of |a.| || \\begin{align}0&=a(20-12)^2 + 25\\ 000 \\\\ 0&= a(8)^2 + 25\\ 000 \\\\ 0&= 64a + 25\\ 000 \\\\ -25\\ 000 &= 64a \\\\ -390.625&= a \\end{align}||<br/>\r\nSo, the equation of the function in vertex form is |y=-390.625(x-12)^2 + 25\\ 000.|<br/>\r\nAll that is left to do is to convert from the vertex form to the general form using algebraic manipulations. ||\\begin{align}y&=-390.625(x-12)^2+25\\ 000 \\\\ y&= -390.625(x-12)(x-12) +25\\ 000 \\\\ y&= -390.625(x^2-24x+144) + 25\\ 000 \\\\ y&= -390.625x^2 + 9\\ 375x - 56\\ 250 + 25\\ 000 \\\\ y&= -390.625x^2 + 9\\ 375x - 31\\ 250 \\end{align}||<br/>\r\n<strong>Answer:</strong> The general form is therefore |y=-390.625x^2 + 9\\ 375x - 31\\ 250.|</p>\r\n\r\n<hr/>\r\n<p><br/>\r\n<strong>b)</strong><br/>\r\n<br/>\r\nTo determine at what times the tank contains |15\\ 000\\ \\text{L},| replace |y| with |15\\ 000| in the equation. ||15\\ 000=-390.625x^2+9\\ 375x-31\\ 250||<br/>\r\nTo <a data-entity-substitution=\"canonical\" data-entity-type=\"node\" data-entity-uuid=\"ff32c663-e5d6-4291-aa9e-08ae96a16293\" href=\"/en/students/vl/mathematics/solving-a-second-degree-equation-or-inequality-m1397\">solve</a> such an equation, make the left side equal to 0. ||\\begin{align} 15\\ 000 &= -390.625x^2 + 9\\ 375x - 31\\ 250 \\\\ \\Rightarrow \\quad 0 &=-390.625x^2 + 9\\ 375x-46\\ 250 \\end{align}||<br/>\r\nNow, in order to find the solutions of this equation, use the quadratic formula: |\\displaystyle x_{1,2} = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}.|<br/>\r\nIn this case, |a=-390.625, b=9\\ 375,| and |c=-46\\ 250|. ||\\begin{align}\\displaystyle x_{1,2} &= \\frac{-9\\ 375 \\pm \\sqrt{9\\ 375^2 - 4\\times -390.625 \\times -46\\ 250}}{2 \\times -390.625} \\\\ \\displaystyle x_{1,2} &= \\frac{-9\\ 375 \\pm \\sqrt{15\\ 625\\ 000}}{-781.25}\\end{align}||<br/>\r\nThen, separate the formula into two parts. ||\\begin{align}\\displaystyle x_1 &= \\frac{-9\\ 375 + \\sqrt{15\\ 625\\ 000}}{-781.25} \\approx 6.94\\\\ \\displaystyle x_2 &= \\frac{-9\\ 375 - \\sqrt{15\\ 625\\ 000}}{-781.25} \\approx 17.06 \\end{align}||<br/>\r\nTo give the answer to the question, convert the two values |x_1| and |x_2| into hours and minutes.</p>\r\n\r\n<ul>\r\n\t<li>\r\n\t<p>|x_1| corresponds to |6| hours and |56| minutes. Essentially, |0.94 \\times 60 = 56.4| minutes, which can be rounded to |56| minutes.</p>\r\n\t</li>\r\n\t<li>\r\n\t<p>|x_2| corresponds to |17| hours and |4| minutes. Essentially, |0.06 \\times 60 = 3.6| minutes, which can be rounded to |4| minutes.</p>\r\n\t</li>\r\n</ul>\r\n\r\n<p><strong>Answer:</strong> The water level of the tank will be |15\\ 000\\ \\text{L}| at 6:56 a.m. and 5:04 p.m (17:04).</p>\r\n",
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"text": "<ol>\r\n\t<li><strong>1.</strong> Un <strong>accident </strong>de voiture.</li>\r\n\t<li>Oups! Un petit <strong>accident </strong>: l'assiette est tombée. </li>\r\n\t<li>Le spectacle a été retardé, mais cela n'a été qu’un petit <strong>incident </strong>dans cette magnifique soirée.</li>\r\n</ol>\r\n",
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