{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "
Gyakorló feladatok megoldásai
\n", "\n", "\n", "

**Ha valaki az igazságot és a törvényszerűséget keresi, nem tehet különbséget kicsiny és nagy problémák között.
Aki a kicsiny dolgokban nem veszi komolyan az igazságot, abban az emberben nagy dolgokkal kapcsolatban sem bízhatunk meg.**\n", "
*Albert Einstein*

\n" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "hide_input": false, "run_control": { "frozen": false, "marked": false, "read_only": false } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Populating the interactive namespace from numpy and matplotlib\n" ] } ], "source": [ "%pylab inline " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Több megoldást is közzéteszünk, mindegyik helyes, és tökéletesen elvégzi a kiadott feladatot.\n", "


Először mindig futtasuk le a \"%pylab inline\" parancsot" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Generálj egy egész számokból álló 1000 tagú véletlen számsorozatot (0-100-ig vannak elemek). Majd írasd ki a 100.-tól a 200.-ig elmeig 7-sével az értékeket. Ehhez használd a \"randint()\" függvényt." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (A)" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[44, 98, 45, 74, 15, 73, 48, 64, 70, 51, 75, 2, 16, 40, 7]" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "A=[] # Létrehozunk egy üres listát, amit majd később lehet növelni\n", "for i in range(0,1000): # Elindítunk egy ciklust, ami 1000-szer fut le\n", " B=randint(0,100) # randint választ egy számot 0 és 100 között\n", " A.append(B) # hozzáírjuk az A listához a B számot\n", "A[100:200:7] # kiírjuk a lista elemeit 100 és 200 között 7-sével" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (B)" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "array([40, 89, 22, 18, 45, 7, 30, 8, 71, 52, 20, 43, 68, 72, 88])" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "B=randint(0,100,1000) # Generálunk egy 1000 tagú listát a randint-tel\n", "C=arange(100,200,7) # megalkotjuk azoknak a számoknak a listáját, amilyen indexű elmeet kiíratni szeretnénk\n", "B[C] #kiírjuk a megfelelő indexű számokat" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (C)" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "array([69, 66, 83, 73, 96, 60, 44, 20, 12, 19, 20, 17, 83, 52, 20])" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "B=randint(0,100,1000) # Generálunk egy 1000 tagú listát a randint-tel\n", "B[100:200:7] #Lekérjük a megfelelő elemeket a listából" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (D) (1 sorban megoldva)" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "array([38, 54, 68, 29, 17, 71, 53, 70, 35, 1, 14, 47, 41, 71, 58])" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "randint(0,100,1000)[100:200:7] #A generált lista megfelelő elemeét hívjuk meg\n", "#Magyarázat: A randint egy listát ad, aminek hívatkozhatunk az elemeire" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Írjunk egy olyan programot, ami egy alsó és egy felső határ között kirajzol egy $log10(a*x)+b$ függvényt. (Itt a és b is a felhasználó által bevitt paraméter)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (A)" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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uSe6e1KFDh+D3oBak78jnwmc+Z9rSTG47LZGXrh6qEBCRsBHMqCEDXgBS3P2x\ngzSbAVwVGD10IrDH3bPMrHmggxkzaw6cCayqptprxQertnHe3+ezLbeAF38+hF+d0ZNoPTBORMJI\nMJeGhgNXAivNbFlg2V1AHIC7TwRmAmcDqcA+4OpAu07AtMA19AbAv939g2qrvgaVlJbx8Idr+ce8\nNAZ0bc0zV/yAmDZNQ12WiEi1C2bU0HzgkL8CB0YL3VzJ8jRgwBFXFyI5eYXc+uoSFqTt4ooT4/jD\nuX1p3CA61GWJiNQI3Vl8gMWbdvGLKUvYs7+Yx34ygIsGdw11SSIiNUpBEODu/GvBJh54N5nOrZsy\n9aah9O3SKtRliYjUOAUBgaGh01YydWkmp/XuyGM/Gai7hEUkYkR8EGzZtY8b/rWYlG25/Or0ntx6\nag9NIykiESWig2DeuhxufXVp+ZSSPxvCmN4dQ12SiEiti8ggcHee/XQDD3+4ll6dWvKPK3/Ace2a\nh7osEZGQiLggyC8s4bdvLmfmym38cEAX/vqjfjRrFHGHQUTkvyLqG3DjjnzG/2sRqdl7ufvsPlw3\nUk8NFRGJmCD4ZG02v3x1KVFRxsvXDGNEYvtQlyQiUieEfRC4OxM/TeN/P1xDr04tee6qJGLbNgt1\nWSIidUZYB8G+ohLueHMF767I4pz+nXn4x/3VHyAicoCw/VbM+GYf418uvz/gjrG9uGl0d/UHiIhU\nIiyDYGHaTm6asoTi0jLdHyAichhhFwRTFm7i3rdXE9euGc9dlUT3Di1CXZKISJ0WNkFQXFrGn95Z\nzSsLNnNKrw48edkgWjXR84JERA4nLIJgV34Rv5iymAVpu7hhdDfuOKu3ZhETEQlSvQ+CddvzuPaf\nX7M9t5C/XTKACwdp/gARkaqo10Hwccp2bnttGU0bRfP6+BMZFHdMqEsSEal36mUQuDvPfZbGg++v\n4fgurXjuqiQ6t9Z8wiIiR6LeBUFhSSl3T1vFm4szOKdfZx65eABNG2k+YRGRI1WvgmDn3kJuemUJ\nX23cxW2nJTLh9ETdJCYicpTqTRB82ymcnVvIk5cN4rwBXUJdkohIWKgXQfDpuhxumbKEJo2ief2G\nkxgY2ybUJYmIhI06HwT/+nIj972TTM9OLXnhZ0l0aaNOYRGR6hR1uAZmFmtmc80s2cxWm9ltlbQx\nM3vSzFLNbIWZDa6wbqyZrQ2s+12whZWWOffNWM0f3l7NKT078MaNJykERERqQDBnBCXA7e6+xMxa\nAovNbJY/dpMeAAAFMklEQVS7J1doMw5IDPwZBjwLDDOzaOBp4AwgA/jazGYcsO33lLlz/cuLmLMm\nm2tHJHDX2X10p7CISA05bBC4exaQFXidZ2YpQAxQ8cv8fOBld3dggZm1MbPOQDyQ6u5pAGb2WqDt\nIYNgQ04+e9bl8OcLTuCKE487gt0SEZFgVamPwMzigUHAwgNWxQBbKvycEVhW2fJhh/ucopIyXvz5\nEEb17FCV8kRE5Agcto/gW2bWAngLmODuudVdiJmNN7NFZraobYMihYCISC0JKgjMrCHlITDF3adW\n0iQTiK3wc9fAsoMt/x53n+TuSe6e1LmjJpYXEaktwYwaMuAFIMXdHztIsxnAVYHRQycCewJ9C18D\niWaWYGaNgEsDbUVEpI4Ipo9gOHAlsNLMlgWW3QXEAbj7RGAmcDaQCuwDrg6sKzGzW4APgWhgsruv\nrtY9EBGRoxLMqKH5wCHHbgZGC918kHUzKQ8KERGpg4LuLBYRkfCkIBARiXAKAhGRCKcgEBGJcAoC\nEZEIZ+UDfuoWM8sD1oa6jjqiPbAj1EXUAToO39Gx+I6OxXd6uXvLI9mwrs5HsNbdk0JdRF1gZot0\nLHQcKtKx+I6OxXfMbNGRbqtLQyIiEU5BICIS4epqEEwKdQF1iI5FOR2H7+hYfEfH4jtHfCzqZGex\niIjUnrp6RiAiIrUkZEFgZpPNLNvMVh1kvZnZk4FJ71eY2eDarrG2BHEsfho4BivN7AszG1DbNdaW\nwx2LCu2GmFmJmf24tmqrbcEcCzM7xcyWmdlqM/u0NuurTUH8H2ltZu+Y2fLAsbi6tmusDWYWa2Zz\nzSw5sJ+3VdKmyt+doTwjeAkYe4j144DEwJ/xwLO1UFOovMShj0U6MNrd+wEPEN7XRV/i0McCM4sG\n/gp8VBsFhdBLHOJYmFkb4BngPHc/Hri4luoKhZc49L+Lm4Fkdx8AnAI8GpgDJdyUALe7e1/gROBm\nM+t7QJsqf3eGLAjcfR6w6xBNzgde9nILgDZm1rl2qqtdhzsW7v6Fu38T+HEB5TO9haUg/l0A3Er5\njHnZNV9R6ARxLC4Hprr75kD7sD0eQRwLB1oGJtJqEWhbUhu11SZ3z3L3JYHXeUAK5XPDV1Tl7866\n3EdQ2cT3B+5wJLoWeD/URYSKmcUAFxLeZ4jB6gkcY2afmNliM7sq1AWF0FNAH2ArsBK4zd3LQltS\nzTKzeGAQsPCAVVX+7qyrdxZLJcxsDOVBMCLUtYTQ48Cd7l5W/stfRGsA/AA4DWgKfGlmC9x9XWjL\nComzgGXAqUB3YJaZfebuuaEtq2aYWQvKz4onVMc+1uUgCHri+0hgZv2B54Fx7r4z1PWEUBLwWiAE\n2gNnm1mJu08PbVkhkQHsdPd8IN/M5gEDgEgMgquBhwKzJaaaWTrQG/gqtGVVPzNrSHkITHH3qZU0\nqfJ3Z12+NDQDuCrQA34isMfds0JdVCiYWRwwFbgyQn/b+y93T3D3eHePB94EfhGhIQDwNjDCzBqY\nWTNgGOXXjCPRZsrPjDCzTkAvIC2kFdWAQB/IC0CKuz92kGZV/u4M2RmBmb1Kee9+ezPLAO4FGgK4\n+0TK5zk+G0gF9lGe+GEpiGPxR6Ad8EzgN+GScH3QVhDHImIc7li4e4qZfQCsAMqA5939kMNu66sg\n/l08ALxkZispn2P9TncPx6eSDgeuBFaa2bLAsruAODjy707dWSwiEuHq8qUhERGpBQoCEZEIpyAQ\nEYlwCgIRkQinIBARiXAKAhGRCKcgEBGJcAoCEZEI938Lps/A4o6kxwAAAABJRU5ErkJggg==\n", "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "mini=1.0 #bekérjük a kezdő értéket\n", "maxi=2.0 #bekérjük a végső értéket\n", "a=1 # bekérjük az \"a\" értéket\n", "b=2 # bekérjük a \"b\" értéket\n", "xlim(mini, maxi) # megadjuk az ábrázolási tartományt (x tengelyen)\n", "x=arange(mini,maxi, 0.01) # 0,01-s lépésközzel megadom az x tengely értékeit\n", "plot(x,(log10(a*x)+b)) # x-et és az egyenletet ábrázoljuk" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (B)" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 7, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "mini=1.0 #bekérjük a kezdő értéket\n", "maxi=2.0 #bekérjük a végső értéket\n", "a=1 # bekérjük az \"a\" értéket\n", "b=2 # bekérjük a \"b\" értéket\n", "x=linspace(mini, maxi,1000) # Generálunk 1000 számot a tartományban egyenlő lépésközzel\n", "plot(x,log10(a*x)+b) # x-et és az egyenletet ábrázoljuk" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (C)" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "mini=1.0 #bekérjük a kezdő értéket\n", "maxi=2.0 #bekérjük a végső értéket\n", "a=1 # bekérjük az \"a\" értéket\n", "b=2 # bekérjük a \"b\" értéket\n", "plot(linspace(mini,maxi, 1000),log10(a*x)+b) # Beírható a generálás közvetlen a plot-ba" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (D)" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 9, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "mini=1.0 #bekérjük a kezdő értéket\n", "maxi=2.0 #bekérjük a végső értéket\n", "a=1 # bekérjük az \"a\" értéket\n", "b=2 # bekérjük a \"b\" értéket\n", "def func(x,a,b): # létrehozok egy függvényt, ami elvégzi a kért műveletet\n", "# A függvény megkapja az \"x\", \"a\" és \"b\" értékeket!\n", " y=log10(a*x)+b # kiszámoljuk az y-t\n", " return y # a függvény visszatér az ny értékkel\n", "x=linspace(mini,maxi, 1000) \n", "plot(x,func(x,a,b)) #ábárolzjuk az x-et és a függvényt" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (E)" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 10, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "mini=1.0 #bekérjük a kezdő értéket\n", "maxi=2.0 #bekérjük a végső értéket\n", "a=1 # bekérjük az \"a\" értéket\n", "b=2 # bekérjük a \"b\" értéket\n", "def func(x,a,b):\n", " y=log10(a*x)+b\n", " return y\n", "x=linspace(mini,maxi, 1000)\n", "adat=(a,b) #létrehozok egy \"tuple\" listát az a,b paraméterekből\n", "plot(x,func(x,*adat)) # a paraméterekkel hívjuk mega függvényt" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Írjunk olyan programot, ahol egy 2 dimenziós rácson haladunk előre (0,0-ból indulva). X vagy Y koordináta változhat, 0 vagy +1 értékkel. Hasznájuk a random.choice függvényt a megoldáshoz. Tegyünk meg 100 lépést. Minden lépést mentsünk egy 100 x 2-es tömbbe, és ábrázoljuk a mozgást." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (A)\n", "\n", "- Létrehozunk egy tömböt, melynek van 100 sor és 2 oszlopa (csupa nullával töltjük fel)\n", "- Végigmegyünk a tömb elemein (ciklussal)\n", "- Először az adott sorba beírjuk az előző sor értékeit, azaz megmondjuk, hogy honnan fogunk továbblépni (emlékezzünk mindenhova alapból 0 van beírva!) !! Ezért kell 1-től indítani a cilust és nem nullától !!\n", "- Kiválasztjuk a lépés irányát és nagyságát\n", "- Csinálunk egy elágazást, ahol megadjuk, hogy merre lépünk\n", "- Az adott lépést elvégezzük a megadott irányban (itt a 0. oszlop az x irány és az 1. oszlop az y irányt jelenti)\n", "- Ábrázoljuk a mozgást" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "b=zeros((100,2))\n", "for i in arange(1,100):\n", " b[i,:]=b[i-1,:]\n", " irany=random.choice(['X','Y'])\n", " lepes=random.choice([0,1])\n", " if irany == \"X\":\n", " b[i,0]=b[i,0]+lepes\n", " else:\n", " b[i,1]=b[i,1]+lepes\n", "plot(b[:,0], b[:,1], \"o\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (B)\n", "\n", "- Létrehozok egy 2 elemű vektort, 0-0 elmekkel\n", "- Létrehozunk egy tömböt, melynek van 100 sor és 2 oszlopa (csupa nullával töltjük fel)\n", "- Végigmegyünk a tömb elemein (ciklussal)\n", "- Kiválasztjuk a lépés irányát\n", "- Csinálunk egy elágazást, ahol megadjuk, hogy merre lépünk\n", "- A vektor kiválasztott elemének helyére véletlenül választunk lépésnagyságot és ezzel megnöveljük a nagyságát. (itt a 0. oszlop az x irány és az 1. oszlop az y irányt jelenti)\n", "- Beírjuk a tömb adott sorába az aktuális kordinétapárost\n", "- Ábrázoljuk a mozgást" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "run_control": { "frozen": false, "read_only": false }, "scrolled": true }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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5hBYMUVgEoF7bLejaQQs6AEgvtAUdt1wAIBIEOgBEgnvoBdq6/4gOnzw/93rD6qU6sGN9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\n", "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "a=[0,0]\n", "b=zeros((100,2))\n", "for i in arange(0,100):\n", " irany=random.choice(['X','Y'])\n", " if irany == \"X\":\n", " a[0]=a[0]+random.choice([0,1])\n", " else:\n", " a[1]=a[1]+random.choice([0,1])\n", " b[i,:]=a[:]\n", "plot(b[:,0], b[:,1], \"o\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (C)\n", "\n", "- Létrehozunk egy tömböt, melynek van 100 sor és 2 oszlopa (csupa nullával töltjük fel)\n", "- Végigmegyünk a tömb elemein (ciklussal)\n", "- Először az adott sorba beírjuk az előző sor értékeit, azaz megmondjuk, hogy honnan fogunk továbblépni (emlékezzünk mindenhova alapból 0 van beírva!) !! Ezért kell 1-től indítani a cilust és nem nullától !!\n", "- Kiválasztjuk a lépés irányát\n", "- Csinálunk egy elágazást, ahol megadjuk, hogy merre lépünk\n", "- A megadott irányban választunk egy lépést, és el is végezzük (itt a 0. oszlop az x irány és az 1. oszlop az y irányt jelenti)\n", "- Ábrázoljuk a mozgást" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "b=zeros((100,2))\n", "for i in arange(1,100):\n", " b[i,:]=b[i-1,:]\n", " irany=random.choice(['X','Y'])\n", " if irany == \"X\":\n", " b[i,0]=b[i,0]+random.choice([0,1])\n", " else:\n", " b[i,1]=b[i,1]+random.choice([0,1])\n", "plot(b[:,0], b[:,1], \"o\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (D)\n", "\n", "- Létrehozunk egy tömböt, melynek van 100 sor és 2 oszlopa (csupa nullával töltjük fel)\n", "- Végigmegyünk a tömb elemein (ciklussal)\n", "- Először az adott sorba beírjuk az előző sor értékeit, azaz megmondjuk, hogy honnan fogunk továbblépni (emlékezzünk mindenhova alapból 0 van beírva!) !! Ezért kell 1-től indítani a cilust és nem nullától !!\n", "- Kiválasztjuk a lépés irányát, és nagysagat egyben, úgy hogy felismerjük, hogy az irány is lehet 0-1, és így 2-szer kell ugyanaból választani (itt a 0. oszlop az x irány és az 1. oszlop az y irányt jelenti)\n", "- A mátrix megadott eleméhez hozzáadjuk a választott lépés nagyságát (itt szintén a 0. oszlop az x irány és az 1. oszlop az y irányt jelenti)\n", "- Ábrázoljuk a mozgást" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 14, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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BGbooLAIQh9wHOoVFAGKR+0suFBYBiEXuA53CIgCxyH2gU1gEIBa5D3QKiwDEIvc3RSks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\n", "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "xy=zeros((100,2))\n", "for i in range(1,100):\n", " xy[i]=xy[i-1]\n", " irany,lepes=random.choice([0,1],2)\n", " xy[i,irany]+=lepes\n", "plot(xy[:,0], xy[:,1], \"o\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (E)\n", "\n", "- Létrehozok egy 2 elemű vektort, 0-0 elmekkel\n", "- Létrehozunk egy tömböt, melynek van 100 sor és 2 oszlopa (csupa nullával töltjük fel)\n", "- Végigmegyünk a tömb elemein (ciklussal)\n", "- Módósítjuk a \"a\" vektor elemét a választott helyen a választott mértékben\n", "- A mátrix megadott sorába beírjuk az előző sor plusz az \"a\" vektor értékét\n", "- Ábrázoljuk a mozgást" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 15, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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Vwxs0c1hbciip+aqscgngc8BzmfkH6556GNhd3N8NPDT+8jSKsvNWerlSRWqPKpHLtcC/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\nQ5eklrChS1JL2NAlqSX+PypOMLvNG4cNAAAAAElFTkSuQmCC\n", "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "a=[0,0]\n", "xy=zeros((100,2))\n", "for i in range(1,100):\n", " a[random.choice([0,1])]=random.choice([0,1])\n", " xy[i,:]=xy[i-1,:]+a\n", "plot(xy[:,0], xy[:,1], \"o\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (F)" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 16, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "x,y=cumsum(list(map(eval,random.choice(('[0,0]','[0,1]','[1,0]'),100))),axis=0).T\n", "plot(x,y)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Aranymetszés aránya (aranyarány) előállítása Fibonacci-számokkal. A Fibonacci-sorozat egymást követő tagjainak hányadosából képzett sorozat (1/1, 2/1, 3/2, 5/3, …) határértéke éppen az aranymetszés arány. Határozzuk meg, hanyadik tagtól vállik a különbség 1e16-nál kisebbé (azaz 0.00000000e+00-at ír a python). Az aranymetszés pontos értéke: $\\frac{1+\\sqrt {5}}{2}$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (A)" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "run_control": { "frozen": false, "read_only": false }, "scrolled": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2 -0.61803398875\n", "3 0.38196601125\n", "4 -0.11803398875\n", "5 0.0486326779168\n", "6 -0.0180339887499\n", "7 0.00696601125011\n", "8 -0.00264937336528\n", "9 0.00101363029772\n", "10 -0.000386929926365\n", "11 0.000147829431923\n", "12 -5.64606600073e-05\n", "13 2.15668056607e-05\n", "14 -8.23767693348e-06\n", "15 3.14652861966e-06\n", "16 -1.20186464891e-06\n", "17 4.59071787029e-07\n", "18 -1.75349769593e-07\n", "19 6.69776591966e-08\n", "20 -2.55831884566e-08\n", "21 9.77190839357e-09\n", "22 -3.73253694619e-09\n", "23 1.42570222295e-09\n", "24 -5.44569944694e-10\n", "25 2.08007167046e-10\n", "26 -7.94517784897e-11\n", "27 3.03477243335e-11\n", "28 -1.15918386001e-11\n", "29 4.42756942221e-12\n", "30 -1.69131375571e-12\n", "31 6.45927755727e-13\n", "32 -2.46691556072e-13\n", "33 9.41469124882e-14\n", "34 -3.59712259979e-14\n", "35 1.37667655054e-14\n", "36 -5.3290705182e-15\n", "37 1.99840144433e-15\n", "38 -8.881784197e-16\n", "39 2.22044604925e-16\n", "40 -2.22044604925e-16\n", "41 0.0\n", "42 0.0\n", "43 0.0\n", "44 0.0\n", "45 0.0\n", "46 0.0\n", "47 0.0\n", "48 0.0\n", "49 0.0\n", "50 0.0\n", "51 0.0\n" ] } ], "source": [ "fiblist = [0,1] # Megadjuk a sorozat első kettő elemét\n", "for i in range(50): # létrehozunk egy ciklust, ami kiszámolja a sorozatot\n", " #a sorozathoz hozzáírjuk az előző két tag összegét:\n", " fiblist.append(fiblist[i] + fiblist[i+1]) \n", "\n", "hanyados=[] # létrehozok egy üres listát\n", "for i in range(2,len(fiblist)):\n", " # az arány listába beírom a sorozat tagjainak hányadoását\n", " hanyados.append((fiblist[i]) / (fiblist[i-1]))\n", " print(i,hanyados[i-2]-(1+sqrt(5))/2) #kiíratom a hanyados és az aranymetszés különbségét\n", "# Vegyük észre, hogy a ciklus 2-től indul (így nincs zéróval való osztás), de a\n", "# hányadost ugyanúgy nullától kell indexelni, ezért van az \"i-2\" benne" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás (B)" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "it: 41\n" ] } ], "source": [ "fiblist = [0,1] # megadjuk az első 2 tagot\n", "for i in range(2,100):\n", " #a sorozathoz hozzáírjuk az előző két tag összegét:\n", " fiblist.append(fiblist[-1] + fiblist[-2]) \n", " hanyados = (fiblist[-1]) / (fiblist[-2]) # kiszámoljuk a hányadost\n", " if hanyados == (1+sqrt(5))/2: # Megvizsgáljuk mikor egyezik meg a hányados az aranymetszéssel\n", " print('it:',i) \n", " break # Megállítjuk a program futását" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Teszteljük a random.choice() függvény eloszlását véletlenszerűségét. Készíts programot, ami rendre 20, 50, 100, 200, 500, 1000 elmet választ az [1,2,3,4] tömbből." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás: 5 külön kódcellában" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "(array([ 1., 0., 0., 7., 0., 0., 5., 0., 0., 7.]),\n", " array([ 1. , 1.3, 1.6, 1.9, 2.2, 2.5, 2.8, 3.1, 3.4, 3.7, 4. ]),\n", " )" ] }, "execution_count": 19, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "hist(random.choice([1,2,3,4],20)) # generálok egy 20 elemű véletlen listát, és hisztogrammal ábrázolom" ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "(array([ 18., 0., 0., 9., 0., 0., 16., 0., 0., 7.]),\n", " array([ 1. , 1.3, 1.6, 1.9, 2.2, 2.5, 2.8, 3.1, 3.4, 3.7, 4. ]),\n", " )" ] }, "execution_count": 20, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "hist(random.choice([1,2,3,4],50))" ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "text/plain": [ "(array([ 23., 0., 0., 26., 0., 0., 24., 0., 0., 27.]),\n", " array([ 1. , 1.3, 1.6, 1.9, 2.2, 2.5, 2.8, 3.1, 3.4, 3.7, 4. ]),\n", " )" ] }, "execution_count": 21, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "hist(random.choice([1,2,3,4],1000))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "### Megoldás több ciklussal" ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "j=1 # legyen j változó 1\n", "for n in [20,50,100,200,1000]: # menjen egy ciklus végig a az elemszámok listáján\n", " a=[] # legyen \"a\" egy üres vektor\n", " for i in range(n): # egy ciklus végigmegy az adott elemszámig\n", " a.append(random.choice([1, 2, 3, 4])) # folyamatosan hozzáírok az \"a\" vektorhoz\n", " subplot(2,3,j) # suboplotban beállítom, hogy hanyadik részképet ábrázolom\n", " hist(a,normed=1) # elkészítem a hisztogramot\n", " j=j+1 # megnövelem j-t, hogy másik részképet válasszak" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás 1 ciklussal" ] }, { "cell_type": "code", "execution_count": 25, "metadata": { "run_control": { "frozen": false, "read_only": false } }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n=[20,50,100,200,1000] # létrehozom az elemszámok listáját\n", "a=random.choice([1,2,3,4],1000) # generálok egy 1000 elemű véletlen listát\n", "for j in range(0,len(n)): # egy ciklus anyiszor ismétlődik, aminnyi elemt tartalmaz az \"n\" lista\n", " subplot(2,3,j+1) # használjuk a subplot parancsot, és mindig új részegységben rajzolunk\n", " hist(a[0:n[j]]) # az adott darabszámú részlistáról csinálunk hisztogramot" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Megoldás elegánsan" ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "run_control": { "frozen": false, "read_only": false }, "scrolled": false }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "for n in [20,50,100,500,1000]: # egy ciklus végigmegy az elemszámok listáján\n", " figure() #mindig új ábrát fog nyitni\n", " # hasonlóan működik, mint a subplot, de itt nem egy képet osztok fel kis részekre,\n", " # hanem magát egy teljesen új képet használok (mintha egy új cellát használnék az ábrázolásra)\n", " hist(random.choice([1, 2, 3, 4],n) ,normed=True) #hisztogram készítése" ] } ], "metadata": { "hide_input": false, "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.2" }, "latex_envs": { "bibliofile": "biblio.bib", "cite_by": "key", "current_citInitial": 1, "eqLabelWithNumbers": false, "eqNumInitial": 0 }, "nav_menu": {}, "toc": { 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