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Python: what is the difference between (1,2,3) and [1,2,3], and when should I use each?


By : pixelbart
Date : July 28 2020, 10:00 AM
like below fixes the issue From the Python FAQ:
code :


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Huge difference between python and fortran difference for a small program


By : Joel Elkins
Date : March 29 2020, 07:55 AM
like below fixes the issue In general Python is an interpreted language while Fortran is a compiled one. Therefore you have some overhead in Python. But it shouldn't take that long.
One thing that can be improved in the python version is to replace the for loop by an index operation.
code :
#create flag filled with zeros with same shape as data
flag=numpy.zeros(data.shape)
#get bool array stating where data>=threshold
barray=data>=threshold
#everywhere where barray==True put a 1 in flag
flag[barray]=1
#create flag filled with zeros with same shape as data
flag=numpy.zeros(data.shape)
#combine the two operations without temporary barray
flag[data>=threshold]=1

Python: Find difference of dictionary's key[value] with other values and return the id which has the minimum difference


By : tanjila
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You could use a one-liner where you iterate over all the 'Time' values, calculate the abs() difference, and take the min() difference. Then assign it to dummy_w['T_id']:
code :
min_diff = min((abs(dummy_w['Time']-d['Time']),d['t_id']) for d in [dummy_R001,dummy_R002,dummy_R003,dummy_R004,dummy_R005])
# (20000, 'y')

dummy_w['T_id'] = min_diff[1]
# {'T_id': 'y', 'Time': 1006120000}

Find the difference (set difference) between two dataframes in python


By : Den Ash
Date : March 29 2020, 07:55 AM
it should still fix some issue I have two dataframes: df1 and df2. I want to eliminate all occurrences of df2 rows in df1. Basically, this is the set difference operator but for dataframes. , try:
code :
df1[~df1.isin(df2)]

A,B,C,D

What's the difference between the new StanfordNLP native Python package and the python wrapper to Core-NLP?


By : Taqi Rizvi Ysr
Date : March 29 2020, 07:55 AM
I hope this helps you . The two systems are completely distinct. The Python-native neural pipeline roughly corresponds to Universal Dependency Parsing from Scratch, with the Tensorflow parser used there reproduced in PyTorch. It provides a fully neural pipeline for many languages from sentence splitting through dependency parsing, exploiting UD resources, but doesn't (at present) support other things such as NER, coreference, relation extraction, open IE, and hand-written pattern matching, and is only trained on UD resources. CoreNLP, which you can use through this or other Python wrappers, does provide all of these other components for a handful of languages, and some models, including English, are trained on much more data. It has the advantages and disadvantages of many pre-neural components (fast tokenizer! purely heuristic sentence-splitting). Most likely, if you're working with formal English text, you'll currently still do better with CoreNLP. In a bunch of other circumstances, you'll do better with the Python stanfordnlp.

How to calculate time difference between two difference values continously in pandas using python


By : sidleal
Date : March 29 2020, 07:55 AM
To fix the issue you can do Filter out rows with x3==0 and groupby with both columns with GroupBy.transform and GroupBy.first for reepat first value per all values of group, so possible subtract by original column with converting to hours:
code :
df['time_diff']= pd.to_datetime(df['date'] + " " + df['time'],
                                format='%d/%m/%Y %H:%M:%S', dayfirst=True)

mask = df['x3'].ne(0)
df['Duration'] = df[mask].groupby(['date','x3'])['time_diff'].transform('first')
df['Duration'] = df['time_diff'].sub(df['Duration']).dt.total_seconds().div(3600)
print (df)
         date      time  x3 Expected           time_diff  Duration
0   10/3/2018   6:00:00   0      NaN 2018-03-10 06:00:00       NaN
1   10/3/2018   7:00:00   5        0 2018-03-10 07:00:00       0.0
2   10/3/2018   8:00:00   0      NaN 2018-03-10 08:00:00       NaN
3   10/3/2018   9:00:00   7        0 2018-03-10 09:00:00       0.0
4   10/3/2018  10:00:00   0      NaN 2018-03-10 10:00:00       NaN
5   10/3/2018  11:00:00   0      NaN 2018-03-10 11:00:00       NaN
6   10/3/2018  12:00:00   0      NaN 2018-03-10 12:00:00       NaN
7   10/3/2018  13:45:00   0      NaN 2018-03-10 13:45:00       NaN
8   10/3/2018  15:00:00   0      NaN 2018-03-10 15:00:00       NaN
9   10/3/2018  16:00:00   0      NaN 2018-03-10 16:00:00       NaN
10  10/3/2018  17:00:00   0      NaN 2018-03-10 17:00:00       NaN
11  10/3/2018  18:00:00   0      NaN 2018-03-10 18:00:00       NaN
12  10/3/2018  19:00:00   5     12hr 2018-03-10 19:00:00      12.0
13  10/3/2018  20:00:00   0      NaN 2018-03-10 20:00:00       NaN
14  10/3/2018  21:30:00   7  12.30hr 2018-03-10 21:30:00      12.5
15  10/4/2018   6:00:00   0      NaN 2018-04-10 06:00:00       NaN
16  10/4/2018   7:00:00   0      NaN 2018-04-10 07:00:00       NaN
17  10/4/2018   8:00:00   5        0 2018-04-10 08:00:00       0.0
18  10/4/2018   9:00:00   7        0 2018-04-10 09:00:00       0.0
19  10/4/2018  11:00:00   5      3hr 2018-04-10 11:00:00       3.0
20  10/4/2018  12:00:00   5      4hr 2018-04-10 12:00:00       4.0
21  10/4/2018  13:00:00   5      5hr 2018-04-10 13:00:00       5.0
22  10/4/2018  16:00:00   0      NaN 2018-04-10 16:00:00       NaN
23  10/4/2018  17:00:00   0      NaN 2018-04-10 17:00:00       NaN
24  10/4/2018  18:00:00   7     11hr 2018-04-10 18:00:00       9.0
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