My Dialog

office_hours_20241129_alexis.ipynb

{"cells": [{"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nAre you there?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nYes, I'm here and ready to help! What would you like to learn about today? It would be helpful to know a bit about your background and interests so I can tailor our discussion appropriately."}, {"cell_type": "code", "metadata": {}, "source": "samples=\"\"\"1000\n2000\n3000\n\n4000\n\n5000\n6000\n\n7000\n8000\n9000\n\n10000\"\"\"", "outputs": [], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "lines = samples.splitlines()\nlines", "outputs": [{"data": {"text/plain": ["['1000',\n", " '2000',\n", " '3000',\n", " '',\n", " '4000',\n", " '',\n", " '5000',\n", " '6000',\n", " '',\n", " '7000',\n", " '8000',\n", " '9000',\n", " '',\n", " '10000']"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "import itertools\nfrom fastcore.utils import *", "outputs": [], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "groups = list(itertools.groupby(lines,key=lambda line:line==\"\"))", "outputs": [], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "groups", "outputs": [{"data": {"text/plain": ["[(False, <itertools._grouper at 0x7b1d6aff5f00>),\n", " (True, <itertools._grouper at 0x7b1d6aff5ff0>),\n", " (False, <itertools._grouper at 0x7b1d6aff5fc0>),\n", " (True, <itertools._grouper at 0x7b1d6aff5f90>),\n", " (False, <itertools._grouper at 0x7b1d6aff5ba0>),\n", " (True, <itertools._grouper at 0x7b1d6aff5c60>),\n", " (False, <itertools._grouper at 0x7b1d6aff6080>),\n", " (True, <itertools._grouper at 0x7b1d6aff60b0>),\n", " (False, <itertools._grouper at 0x7b1d6aff60e0>)]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "xxs = [it for (p,it) in groups if p==False]\nxxs", "outputs": [{"data": {"text/plain": ["[<itertools._grouper at 0x7b1d6aff5f00>,\n", " <itertools._grouper at 0x7b1d6aff5fc0>,\n", " <itertools._grouper at 0x7b1d6aff5ba0>,\n", " <itertools._grouper at 0x7b1d6aff6080>,\n", " <itertools._grouper at 0x7b1d6aff60e0>]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[list(xs) for xs in xxs]", "outputs": [{"data": {"text/plain": ["[[], [], [], [], []]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nWhy are the individual groupers not producing lists when I call list on them?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nThe grouper objects from `itertools.groupby()` are iterator objects that can only be consumed once. It looks like they've already been consumed by the time you tried to convert them to lists. This is a common gotcha with `groupby()`.\n\nTo fix this, you need to convert the grouper objects to lists immediately when you create them. Would you like me to show you how to modify your code to make it work?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nWait, before showing me that, please explain when they are being consumed?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nIn your code sequence, the grouper objects are actually being consumed when you first inspect the `groups` variable. When you did:\n\n```python\ngroups<outputs><output type=\"execute_result\">[(False, <itertools._grouper at 0x7523f69d5870>),\n (True, <itertools._grouper at 0x7523f69d54b0>),\n ...]</output></outputs>\n```\n\nEven though this looks like it's just displaying the objects, the Python interactive environment actually iterates through the groupers to display this representation, consuming them in the process. That's why by the time you try to use them later, they're already exhausted.\n\nWould you like me to explain more about how Python iterators work, or shall we look at how to fix this issue?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nI know how iterators work. This is why your answer confuses me. You say \"the Python interactive environment actually iterates through the groupers to display this represetnation\". But the only way to consume them, is to call iter on them and then call next on them repeatedl. And if the Python interactive environment had in fact done so, then it would have shown the resulting values not the itertools._grouper placeholder. Where am I wrong here?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nAh, you make an excellent point, and I was wrong. Looking at the output more carefully, you're absolutely right - we're just seeing the repr of the grouper objects themselves, not their contents. They haven't been consumed at this point.\n\nLet me revise my understanding: We create the groups, and then create `xxs` by filtering for `p==False`, but at no point are we actually consuming the iterators. \n\nCould you help me understand what's happening here? Because you're right that we should be able to convert these groupers to lists if they haven't been consumed yet."}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nI am baffled. Pleaes show me an example of groupby wher I am able to access the group values."}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nLet's create a simple example to understand how groupby works:\n\n```python\nfrom itertools import groupby\ndata = [1,1,1,2,2,3,3,3,1]\ngroups = groupby(data)\nfor k,g in groups:\n    print(f\"key {k}:\", list(g))\n```\n\nThis should show us the groups clearly. Would you like to try this and then we can compare it to your original scenario to spot what's different?"}, {"cell_type": "code", "metadata": {}, "source": "from itertools import groupby\ndata = [1,1,1,2,2,3,3,3,1]\ngroups = groupby(data)\nfor k,g in groups:\n    print(f\"key {k}:\", list(g))", "outputs": [{"name": "stdout", "output_type": "stream", "text": ["key 1: [1, 1, 1]\n", "key 2: [2, 2]\n", "key 3: [3, 3, 3]\n", "key 1: [1]\n"], "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "from itertools import groupby\ndata = [1,1,1,0,2,2,0,3,3,3,0,1]\ngroups = groupby(data,key=lambda x:x==0)\nfor k,g in groups:\n    print(f\"key {k}:\", list(g))", "outputs": [{"name": "stdout", "output_type": "stream", "text": ["key False: [1, 1, 1]\n", "key True: [0]\n", "key False: [2, 2]\n", "key True: [0]\n", "key False: [3, 3, 3]\n", "key True: [0]\n", "key False: [1]\n"], "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "groups2 = list(itertools.groupby(data,key=lambda x:x==0))\ngroups2", "outputs": [{"data": {"text/plain": ["[(False, <itertools._grouper at 0x7b1d6aff6770>),\n", " (True, <itertools._grouper at 0x7b1d6aff61a0>),\n", " (False, <itertools._grouper at 0x7b1d6aff5d80>),\n", " (True, <itertools._grouper at 0x7b1d6aff6170>),\n", " (False, <itertools._grouper at 0x7b1d6aff5cf0>),\n", " (True, <itertools._grouper at 0x7b1d6aff6530>),\n", " (False, <itertools._grouper at 0x7b1d6aff6710>)]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[(k,list(v)) for (k,v) in groups2]", "outputs": [{"data": {"text/plain": ["[(False, []),\n", " (True, []),\n", " (False, []),\n", " (True, []),\n", " (False, []),\n", " (True, []),\n", " (False, [])]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nPlease consider the example above, which shows the same puzzling issue in the case of integers not strings. Thoughts?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nAh! Now I see the issue more clearly. When we store the groupby results in a list (groups2), and then try to access them later, we get empty lists. But when we process the groupby results immediately in the for loop, we get the actual grouped values.\n\nThis suggests that groupby creates iterators that can only be used once and must be consumed immediately. When we store the results in a list, we're storing the grouper objects, but by the time we try to use them later, something about their internal state means they no longer have access to the original data.\n\nWould you like to try an experiment where we create the lists immediately when creating groups2, rather than storing the grouper objects?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nRather, I will try an experiment where I never convert the value returned by groupby as a list. Watch me work!"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nOK! I'm watching and eager to see what you discover about how groupby's return value behaves without converting it to a list first."}, {"cell_type": "markdown", "metadata": {}, "source": "I'll create a generator not a list of the subgroups"}, {"cell_type": "code", "metadata": {}, "source": "groups3 = (v for (k,v) in itertools.groupby(lines,key=lambda line:line!=\"\") if k==True)\ngroups3", "outputs": [{"data": {"text/plain": ["<generator object <genexpr> at 0x7b1d6b8bdb60>"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[list(v) for v in groups3]", "outputs": [{"data": {"text/plain": ["[['1000', '2000', '3000'],\n", " ['4000'],\n", " ['5000', '6000'],\n", " ['7000', '8000', '9000'],\n", " ['10000']]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nWhat seems to me to be happening is the following:\n\n- `itertools.groupby` returns an iterator, which yields individual items which are themselves iterators.\n- However, these \"child\" iterators are not independent of the parent iterator.\n- Here is one way they interact: after iterating through the parent iterator, it is impossible to iterate through the child iterators. \n- In other words, if you want to materialize the sequences represented by the child iterators, you need to do so for every child iterator completely, before even getting the handle of the next child iterator.\n\nwdyt of this theory?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nYes, that's an excellent analysis! Your theory explains perfectly why:\n\n1. Converting the groupby result to a list first fails - because by the time we try to access the child iterators, the parent has already advanced past all the data\n2. Your generator expression approach works - because we're processing each child iterator completely (via `list(v)`) before moving to the next group in the parent iterator\n3. The immediate for-loop approach worked - for the same reason as #2\n\nThis is a really interesting example of how iterators in Python can have dependencies on their parent iterator's state. It's quite different from, say, a list comprehension, where each result is independent.\n\nWould you like to test this theory further with some specific examples to confirm the parent-child iterator relationship?"}, {"cell_type": "markdown", "metadata": {}, "source": "The fact that consuming the parent iterator implicitly consumes the child iterators, indicates to me that it is more helpful to think of `groupby` as returning one \"underlying\" iterator, which ranges over all the elements in the sublists, but which can only be consumed by consuming the parent and child iterators in a particular order."}, {"cell_type": "code", "metadata": {}, "source": "groups4 = (list(v) for (k,v) in itertools.groupby(lines,key=lambda line:line!=\"\") if k==True)\ngroups4", "outputs": [{"data": {"text/plain": ["<generator object <genexpr> at 0x7b1d6b8bde00>"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "list(groups4)", "outputs": [{"data": {"text/plain": ["[['1000', '2000', '3000'],\n", " ['4000'],\n", " ['5000', '6000'],\n", " ['7000', '8000', '9000'],\n", " ['10000']]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "groups5 = [list(v) for (k,v) in itertools.groupby(lines,key=lambda line:line!=\"\") if k==True]\ngroups5", "outputs": [{"data": {"text/plain": ["[['1000', '2000', '3000'],\n", " ['4000'],\n", " ['5000', '6000'],\n", " ['7000', '8000', '9000'],\n", " ['10000']]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "The above two examples work, because in both cases `list(v)` is called _as_ the parent iterator is consumed.\n\nThe following example will fail, because it consumes the parent iterator fully and then tries to access the child iterators:"}, {"cell_type": "code", "metadata": {}, "source": "list6 = list(groups6)", "outputs": [], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "list6", "outputs": [{"data": {"text/plain": ["[<itertools._grouper at 0x7b1d6aff6410>,\n", " <itertools._grouper at 0x7b1d6aff6fe0>,\n", " <itertools._grouper at 0x7b1d6aff65f0>,\n", " <itertools._grouper at 0x7b1d6aff6a10>,\n", " <itertools._grouper at 0x7b1d6aff5e40>]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[list(v) for v in list6]", "outputs": [{"data": {"text/plain": ["[[], [], [], [], []]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}], "metadata": {"kernelspec": {"display_name": "python3", "language": "python", "name": "python3"}}, "nbformat": 4, "nbformat_minor": 4}