@Monso, as per your provided inputs, expected outputs in problem & in comments, I’ve written the code to solve your problem. In my case: Input directory: C:\Users\pc-user\Desktop\Input Output directory: C:\Users\pc-user\Desktop\Output And I have 3 file A.txt, B.txt & C.txt inside Input directory with contents as follows. You’ve to use your own paths as you’ve mentioned … Read more
First of all, there seems to be a mistake in your code. Both if and else conditions set your mat1[i,j] to 1… Supposing your code is actually: for i in range(0,4): for j in range(0,4): if edges[i,j]==1 or (edges[i,0]==1 and edges[0,j]==1): mat1[i,j]=1 Then a Tensorflow solution is: import tensorflow as tf edges = tf.constant([[0, 0, … Read more
If you’re printing only 4 decimals and you don’t want the number rounded that means you want the decimals truncated and regular formatting methods don’t provide facilities for that. You can easily achieve that if you pre-process your number as: def truncate_number(number, decimals): factor = 10.0 ** decimals return int(number * factor) / factor num … Read more
It seems that in the documentation of xlwings, it is necessary to have an empty row and column at the bottom and to the right. if not it will overwrite it http://docs.xlwings.org/en/stable/api.html#xlwings.xlwings.ret solved xlwings udf function erase next cell on excel
This looks like an excel datetime format. This is called a serial date. To convert from that serial date you can do this: data[‘Date’].apply(lambda x: datetime.fromtimestamp( (x – 25569) *86400.0)) Which outputs: >>> data[‘Date’].apply(lambda x: datetime.fromtimestamp( (x – 25569) *86400.0)) 0 2013-02-25 10:00:00.288 1 2013-02-26 10:00:00.288 2 2013-02-27 10:00:00.288 3 2013-02-28 10:00:00.288 To assign it … Read more
You can use a regex to look for words that follow the word python, for example >>> import re >>> re.findall(r’Python (\w+)’, s) [‘is’, ‘has’, ‘features’, ‘interpreters’, ‘code’, ‘is’, ‘Software’] Since this list may contain duplicates, you could create a set if you want a collection of unique words >>> set(re.findall(r’Python (\w+)’, s)) {‘Software’, ‘is’, … Read more
You could find all substrings of length 4+, and then down select from those to find only the shortest possible combinations that contain one of each letter: s=”AAGTCCTAG” def get_shortest(s): l, b = len(s), set(‘ATCG’) options = [s[i:j+1] for i in range(l) for j in range(i,l) if (j+1)-i > 3] return [i for i in … Read more
At the end of your function, return total_value_in_a_round At the end of OneRound, return the score to the calling program by combining the last two lines: return userscore() In your main program, set up a loop to play 5 times, and keep a running sum of the score: total_score = 0 for round in range(5): … Read more
Try and see! Here is a request for imagery of the O2 Arena on the river in London for an image from January 2017: curl “https://api.nasa.gov/planetary/earth/imagery/?lon=0&lat=51.5&date=2017-01-01&cloud_score=True&api_key=DEMO_KEY” Here is the result: { “cloud_score”: 0.047324414226919846, “date”: “2017-01-17T10:52:32”, “id”: “LC8_L1T_TOA/LC82010242017017LGN00”, “resource”: { “dataset”: “LC8_L1T_TOA”, “planet”: “earth” }, “service_version”: “v1”, “url”: “https://earthengine.googleapis.com/api/thumb?thumbid=a286185b3fda28fa900a3ce43b3aad8c&token=206c7f1b6d4f847d0d16646461013150” If you paste the URL at the … Read more
I think you need to do something like this # Print out predictions y = regressor.predict(input_fn=lambda: input_fn(prediction_set)) # .predict() returns an iterator; convert to a list and print predictions predictions = list(itertools.islice(y, 6)) print(“Predictions: {}”.format(str(predictions))) I found it from the boston tutorial: https://github.com/tensorflow/tensorflow/blob/master/tensorflow/examples/tutorials/input_fn/boston.py 0 solved Race Condition when reading from generator
Not sure why your question is drawing so much bad attention. permutations are what you need: from itertools import permutations def is_perm(letters,word): for p in permutations(letters): if ”.join(p) == word: return True return False letters = [“a”, “c”, “t”] word = ‘cat’ print is_perm(letters,word) Letters may of course be any list of strings, not just … Read more
You may achieve this using itertool.chain() with list comprehension expression as: >>> a=[[1,2,3],[4,5,6]] >>> b=[[5,8,9],[2,7,10]] # v join the matched sub-lists # v v your condition >>> [i + list(chain(*[j[1:] for j in b if i[1]==j[0]])) for i in a] [[1, 2, 3, 7, 10], [4, 5, 6, 8, 9]] solved How to merge two … Read more
If you are using raw queries, then you just need to merge the data. Something like this should work, def merge_order_data_and_detail(orders, details): “””Group details by order_id and merge it in orders.””” # create dictionary key:order_id value:[order_detail_data] dic = {} for d in details: if d[‘order_id’] not in dic: dic[d[‘order_id’]] = [] dic[d[‘order_id’]].append(d) # iterate orders … Read more
This is the approach you may look into: def thesum(the_numbers): the_numbers = [i for i in the_numbers if i % 3 == 0 or i % 5 == 0] return print(sum(the_numbers)) thesum(range(100)) 0 solved Sum in range that meet criteria [closed]
In here: for entry in my_dict_1.keys(): my_dict_1[entry] = metric You are assigning the value of metric to all your dict (my_dict_1) items. Reason why ‘it just adds the last arg passed’. Without cleaning up your code, here’s the patch to fix your issue: for i, metric in enumerate(my_metrics): … my_dict_1[i] = metric 4 solved Python … Read more