CODEWITHSUNDEEP
arraysnumpyrowsubstitutionreplace
Alice
Alice

Reputation: 191

Numpy rows substitution

I am new to Numpy and I am not an expert programmer at all...
This is my issue:

I have array a and array b (b < a).

I want to substitute some rows of a with all the rows of b (in order).

The rows in a to be substituted have one value in common with b rows.

for example:

a is:

1 a 2

3 b 4

0 z 0

5 c 6

0 y 0

b is

1 z 1

1 y 1

In this case I will want to substitute rows 3 and 5 in a with 1 and 2 in b. The arrays are very big and, as in the example, there are some character types (so I set the array to dtype=object).

Upvotes: 1

Views: 131

Answers (1)

askewchan
askewchan

Reputation: 46530

First, I would recommend changing your dtype from object to [('a',int),('letter','S1'),('b',int)]. What this does is allow you to have columns of different dtypes (note the length of that dtype list is the number of columns). Then you have:

In [63]: a = np.array(
   ....:         [(1,'a',2),(3,'b',4),(0,'z',0),(5,'c',6),(0,'y',0)],
   ....:         dtype=[('a',int),('letter','S1'),('b',int)])

In [64]: a
Out[64]: 
array([(1, 'a', 2), (3, 'b', 4), (0, 'z', 0), (5, 'c', 6), (0, 'y', 0)], 
      dtype=[('a', '<i8'), ('letter', 'S1'), ('b', '<i8')])

In [65]: a['a']
Out[65]: array([1, 3, 0, 5, 0])

In [66]: a['b']
Out[66]: array([2, 4, 0, 6, 0])

In [67]: a['letter']
Out[67]: 
array(['a', 'b', 'z', 'c', 'y'], 
      dtype='|S1')

Then you can use the same dtype for b:

In [69]: b = np.array([(1,'z',1),(1,'y',1)], dtype=a.dtype)

In [70]: b
Out[70]: 
array([(1, 'z', 1), (1, 'y', 1)], 
      dtype=[('a', '<i8'), ('letter', 'S1'), ('b', '<i8')])

In [71]: b['letter']
Out[71]: 
array(['z', 'y'], 
      dtype='|S1')

Now, you can simply replace the parts of a with the parts of b where the letters match:

In [73]: a
Out[73]: 
array([(1, 'a', 2), (3, 'b', 4), (0, 'z', 0), (5, 'c', 6), (0, 'y', 0)], 
      dtype=[('a', '<i8'), ('letter', 'S1'), ('b', '<i8')])

In [74]: b
Out[74]: 
array([(1, 'z', 1), (1, 'y', 1)], 
      dtype=[('a', '<i8'), ('letter', 'S1'), ('b', '<i8')])

In [75]: for row in b:
   ....:     a[a['letter']==row['letter']] = row
   ....:     

In [76]: a
Out[76]: 
array([(1, 'a', 2), (3, 'b', 4), (1, 'z', 1), (5, 'c', 6), (1, 'y', 1)], 
      dtype=[('a', '<i8'), ('letter', 'S1'), ('b', '<i8')])

Upvotes: 1

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