This function M will count groups of a value. It outputs a list of group occurrences in this format: [index of group, length of group]. It uses a flag m to keep track of in/out of group. If it’s out, it starts a new group occurrence, if it’s in, it increments the group length. It … Read more
Though I am not compleately sure what your question demands, but as far as i comphreanded it, True and False are being printed to your file, which is not the desired output? Well that’s because re.search returns a Match Object, For example: >>> search_string = ‘piiig’ >>> output = re.search(‘iii’, search_string) >>> output <_sre.SRE_Match object … Read more
blob=cv2.dnn.blobFromImage(cv2.resize(image,(300,300),1.0,(300,300),(104.0, 177.0, 123.0)) here is one closing bracket missing 0 solved I am writing a simple face recognition program I faced Syntax Error [closed]
You can try this out: a = [[22,3,3], [5,3,7],[1,6,3]] b = [[2,4,7], [6,4,8],[3,66,13]] , [[2,23,6], [5,13,7],[11,6,34]] , [[22,53,6], [54,3,7],[11,6,33]] for i in range(len(a)): for j in range(len(a[i])): for x in range(len(b)): for y in range(len(b[x])): for z in range(len(b[y])): if (a[i][j]==b[x][y][z]): print(‘true’) else: print(‘false’) As it prints true or false it is going to give … Read more
Try This: def orangecap(match_details): players_data = {} for k, v in match_details.iteritems(): for player_name, score in v.iteritems(): prev_player = player_name if prev_player == player_name: score = players_data.get(player_name, 0) + score players_data[player_name] = score high_score_player = max(players_data, key=lambda i: players_data[i]) print (str(high_score_player), players_data[high_score_player]) 6 solved python program using dictionary
You can try time.sleep() pygame.time.wait() pyplot.pause() .after() asyncio.sleep() For more detail check 1st answer by ‘Trooper Z’ here solved How to delay a python script? [closed]
If you are trying to delete this lines from the file, try this: import os def removeUrl(url): url_file = “url.txt” fileIn = open(url_file, ‘r’) fileOut = open(“temp.txt”, ‘w’) for line in fileIn: if url not in line: fileOut.write(line) fileIn.close() fileOut.close() os.remove(url_file) os.rename(“temp.txt”, url_file) removeUrl(“www.youtube.com”) solved python – history file
raw = [{‘Name’: ‘Erica’,’Value’:12},{‘Name’:’Sam’,’Value’:8},{‘Name’:’Joe’,’Value’:60}] raw.sort(key=lambda d: d[‘Value’], reverse=True) result = [] for i in range(2): result.append(raw[i][‘Name’]) print(result) # => [‘Joe’, ‘Erica’] print(result) Try this. 2 solved Dictionary Sorting based on lower dictionary value
Once you define the function, you can subsequently invoke it: Func(5) solved Calling upon a defined function with users input
a and b are divisors of n if a * b = n. You can, without loss of generality, set b >= a. Therefore a * a is the upper limit; i.e. you only need to consider up to and including the square root of n. Once you have an a, you can trivially infer … Read more
cakes = int(input(‘How many cakes? ==> ‘)) donuts = int(input(‘How many dozens of donuts? ==> ‘)) cookies = int(input(‘How many dozen cookies? ==> ‘)) cake_eggs = 2 cake_butter = .5 cake_sugar = 1 cake_flour = 1.5 cookie_eggs = 2 cookie_butter = 2.5 cookie_sugar = 2 cookie_flour = 8 donuts_eggs = 3 donuts_butter = .25 donuts_sugar … Read more
<< and >> are the Binary Left Shift and Binary Right Shift respectively. The left operands value is moved left by the number of bits specified by the right operand. Example, Code: temp = 14 << 2 The variable temp has a value of 56 because 14 (00001110 in binary) shifted left two bits equals … Read more
There is a popular math theorem – Fermat’s Great (Last) Theorem – that states for any integer n > 2 there does not exist non-zero integers a, b, c such that a^n + b^n = c^n. 2 solved consider x^i+y^i=z^i, x
>>> mylist = [[‘dog’, ‘animal’],[‘daisy’,’flower’],[‘cat’,’animal’]] >>> x = mylist[2][1] >>> print x animal 4 solved How to Retrieve a Specific value from a Specific List inside a Larger List?
Write response = input(‘Would like to …?’).lower() if response == ‘y’ or response == ‘yes’: … Or if response in [‘y’, ‘yes’]: … Note that it’s not necessary to call str — your object is already a string. solved Test for answers in an IF block [duplicate]