One way to do this without itertools is to use Python’s sum function to concatenate lists. >>> L = [ [1], [2], [3], [4], [5], [6], [7] ] >>> L_extend = [ sum(L[0:i+1], []) for i in range(len(L)) ] 2 solved Append cumulative list of lists python
Why not use a dict? collections.defaultdict would work too: d = defaultdict(list) for score, student in l: d[score] += student solved How to reorganize a list of tuples?
EDIT: As pointed by @Petesh, the issue comes from stompy(external library), which hasn’t been ported to Python3. If you check the source code, you can find this: except socket.timeout, exc: which is invalid syntax for python3+ If you run your Lambdas in python3.6/3.7 environment, the syntax is invalid. The issue might go away if you … Read more
something like this? orig = ‘hello’ # table = {‘h’: ‘L’, ‘e’: ‘O’, ‘l’: ‘A’, ‘o’: ‘W’ } # table_tr = dict( (ord(a), ord(b)) for a,b in table.items() ) table_tr = str.maketrans(‘helo’, ‘LOAW’) res = orig.translate(table_tr) print(res) (this is for python3; for python2 you need to import string and use string.maketrans) solved Need to change … Read more
dictionary = {‘key’: [‘value’]} print(dictionary.get(‘key’)[0]) solved Get raw key value
Python dictionaries do not maintain order, so you cannot guarantee the order you entered the data is the same order it will be in when emitted from json.dumps(). If you need an ordered dictionary, check out collections.OrderedDict. 1 solved Python JSON union dictionay list
It splits a word into two parts: stem and end. There are three cases: The word ends with ss (or even more s): stem <- word and end <- “” The word ends with a single s: stem <- word without “s” and end <- “s” The word does not end with s: stem <- … Read more
You can’t have what you describe. You could have this: dct = {} dct[‘word1’] = 23 dct[‘word2’] = 12 dct[‘word1’] = 7 dct[‘word2′] = 2 But at the end all you’d end up with is this: {‘word1’: 7, ‘word2’: 2} Keys in a dictionary cannot be repeated. If your code is actually set up like … Read more
Here is one way if len(set(line.split()).intersection(keywords)) > 2: This splits the line into words with line.split() first . Then uses sets intersection function to find the common elements of 2 sets 3 solved Finding lines that include two or more words from a list
You are looking at whole tuples with unicode strings; the u” is normal when showing you a tuple with unicode values inside: >>> print u’Hello World!’ Hello World! >>> print (u’Hello World’,) (u’Hello World’,) You want to format each row: print u’ {:<15} {:<8} {:<6}’.format(*row) See the str.format() documentation, specifically the Format Syntax reference; the … Read more
A simple merge using df.merge does this: df1.merge(df2, on=[‘A’]) A B_x B_y 0 a 1 3 1 a 1 4 2 a 2 3 3 a 2 4 4 b 1 3 5 b 2 3 2 solved Merging two dataframes in python pandas [duplicate]
The converting function that you are referring to does the same – it weights the R,G and B channel values of each pixel, and takes the sum. Since OpenCV uses the BGR colorspace on reading images, your conversion function will be something like this- def rgbToGray(img): grayImg = 0.0722*img(:,:,1) + 0.7152*img(:,:,2) + 0.2126*img(:,:,3) return grayImg … Read more
I’ve never actually seen PyFunction_SetClosure used, but I would expect to use it for modifying an existing Python function object to change the closure variables. It probably isn’t suitable for implementing a generating a new function with a closure from scratch, because that would not be implemented in terms of PyFunctionObject (since these are specifically … Read more
You can try this code below: import pandas as pd df = pd.read_csv(r’your csv file path’) res = df[(df.apply(lambda s: s.eq(‘CA’).any(), axis=1))] print(res) 0 solved How to use python pandas to find a specific string in various rows
You can use in to test membership: >>> for attr in attrs: … if (‘href’, ‘http://fls-na.amazon.com’) in attr: … print attr … [(‘rel’, ‘dns-prefetch’), (‘href’, ‘http://fls-na.amazon.com’)] solved Find an element in a list of tuples in python [closed]