If you have a string: a=”hi how \\are you” you can remove it by doing: a.replace(‘\\’,”) >’hi how are you’ If you have a specific context where you are having trouble, I recommend posting a bit more detail. 2 solved I want to remove ‘\’ from a string in python [duplicate]
It computes the Cartesian product over any number of iterables. Source So if you have two lists like [1,2] and [3,4], the Cartesian product is (1,3),(1,4),(2,3),(2,4) 10 solved What does product() do in python? [closed]
You have two independent if statements. That the second such statement has an else suite doesn’t matter here; that else suite is picked on the basis of the condition attached to the second if test; whatever happened in the first if statement doesn’t matter. If you want the two x tests to not be independent, … Read more
No, you do not need a main loop if you use the cmd.Cmd class included with Python: #! /usr/bin/env python3 import cmd class App(cmd.Cmd): prompt=”Type in a command: ” def do_a(self, arg): print(‘I am doing whatever action “a” is.’) def do_b(self, arg): print(‘I am doing whatever action “b” is.’) if __name__ == ‘__main__’: App().cmdloop() The … Read more
From what I can tell, it seems like you want to create a list of a bunch of random features. The way I would personally go about this is by using the random method Choice Choice allows us to pick a string from a list, then we use a function called .append that lets us … Read more
try this: In [16]: df.ix[df.groupby(df[‘datetime’].dt.date)[‘production’].transform(‘nunique’) < 44 * 4 * 24, ‘datetime’].dt.date.unique() Out[16]: array([datetime.date(2015, 12, 7)], dtype=object) this will give you all rows for the “problematic” days: df[df.groupby(df[‘datetime’].dt.date)[‘production’].transform(‘nunique’) < 44 * 4 * 24] PS there is a good reason why people asked you for a good reproducible sample data sets – with the one … Read more
You have to split(‘.’) your string to convert to a list, then reverse it using [::-1] and join it again adding the . mystring = “192.168.1.1” print ‘.’.join(mystring.split(‘.’)[::-1]) Output: 1.1.168.192 1 solved How to print string array in reverse order in python
You can use replace for this: for index, item in enumerate(my_list): my_list[index] = item.replace(‘[‘, ”).replace(‘]’, ”) 2 solved need regular expression to remove brackets [closed]
Since pythons list comprehensions are turing complete and require no line breaks, any program can be written as a python oneliner. If you enforce arbitrary restrictions (like “order of the statements” – what does that even mean? Execution order? First apperarance in sourcecode?), then the answer is: you can eliminate some linebreaks, but not all. … Read more
This should work: text_file = open(“write_it.txt”, “w”) while 1: word = input(“Please add to a text file: “) if not word: break text_file.write(word) text_file.close() solved How can I write to the textfile with “while”?
Here’s some “python magic” name = [‘Elon Reeve Musk’] f”{”.join(filter(str.isupper, name[0]))}@company.com”.lower() >>> [email protected] Whether this is better than your method is debatable. Most of the time a few lines of easy to read code is better than a one line hack. My recommendation would be name = [‘Elon Reeve Musk’] initials=””.join(word[0] for word in name[0].split()) … Read more
Assuming you have a pandas dataframe, one option is to use pandas.crosstab to return another dataframe: import pandas as pd df = pd.read_csv(‘file.csv’) res = pd.crosstab(df[‘X’], df[‘Y’]) print(res) Y 0 1 X 0 3 7 1 1 3 A collections.Counter solution is also possible if a dictionary result is required: res = Counter(zip(df[‘X’].values, df[‘Y’].values)) 4 … Read more
Hoo boy, this was a big XY problem. You want to sort by the third character of the second element of each tuple. Use the sorted built-in with its kwarg key. lst = [(‘abc’, ‘bcd’), (‘abc’, ‘zza’)] result = sorted(lst, key=lambda t: t[1][2]) key here accepts a function that is given each element in turn, … Read more
this is where itertools.groupby may come in handy: from itertools import groupby a = [“a”, “a”, “b”, “b”, “a”, “a”, “c”, “c”] res = [key for key, _group in groupby(a)] print(res) # [‘a’, ‘b’, ‘a’, ‘c’] this is a version where you could ‘scale’ down the unique keys (but are guaranteed to have at leas … Read more
You can take advantage of CSS selector span[id$=lblResultsRaceName], which finds all spans that’s id ends with lblResultsRaceName and ‘td > span’, which finds all spans that have direct parent <td>: This code snippet will go through all racing result and prints all races: import requests from bs4 import BeautifulSoup url = “https://www.thedogs.com.au/Racing/Results.aspx?SearchDate=3-Jun-2018” def get_info(session,link): session.headers[‘User-Agent’] … Read more