How to join multiple rows sequentially in a numpy array?

Question

I've a data like this,

data = [array(['a', 'b', 'c']), 
        array([['d', 'e', 'f'], ['g', 'h', 'i']]), 
        array([['j', 'k', 'l'], ['m', 'n', 'o'], ['p', 'q', 'r']])]

I want to join values in inner list sequentially. This is the desired output that I need.

[['a', 'b', 'c'], ['d g', 'e h', 'f i'], ['j m p', 'k n q', 'l o r']]

I tried using using multiple loops and join but it's not giving me required output.

for i in data:
    for j in i:
        print(" ".join(j))

I'm not really sure on how to achieve this? Need an efficient and faster approach because my original data is really large.

That's not a NumPy array, though? You're not going to get any performance improvement if all you do is loop over a NumPy array, by the way. — AMC
– AMC, Commented Feb 16, 2020 at 20:18
@AMC actually it's a numpy array but I posted it as normal list. — user_12
– user_12, Commented Feb 16, 2020 at 20:21
@AMC I was given this data. I didn't prepare it this way. What would you suggest? I have large data in which I want this desired output. It's a list of numpy arrays. — user_12
– user_12, Commented Feb 16, 2020 at 20:22
Is it a list of NumPy ndarrays, or is it a multidimensional ndarray? I'm trying to convert the data you shared properly. — AMC
– AMC, Commented Feb 16, 2020 at 20:23

hpaulj · Accepted Answer · 2020-02-16 21:02:52Z

3

In [300]: [np.array([' '.join(ij) for ij in zip(*np.atleast_2d(row))]) for row in data]        
Out[300]: 
[array(['a', 'b', 'c'], dtype='<U1'),
 array(['d g', 'e h', 'f i'], dtype='<U3'),
 array(['j m p', 'k n q', 'l o r'], dtype='<U5')]

The first array is 1d, thus requiring the atleast_2d for consistency. Most the rest is just iterative application of string join.

answered Feb 16, 2020 at 21:02

hpaulj

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1 Comment

Andy L. Over a year ago

very nice use of np.atleast_2d :) +1

Georgina Skibinski · Accepted Answer · 2020-02-16 19:42:02Z

2

Try:

res=list(map(lambda x: list(map(' '.join, zip(*x))) if isinstance(x[0], list) else x, data))

Outputs:

[['a', 'b', 'c'], ['d g', 'e h', 'f i'], ['j m p', 'k n q', 'l o r']]

edited Feb 16, 2020 at 19:42

answered Feb 16, 2020 at 19:32

Georgina Skibinski

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4 Comments

user_12 Over a year ago

@GrzegorzSkibinski Thank you but I'm getting this error TypeError: sequence item 0: expected str instance, Timestamp found

Georgina Skibinski Over a year ago

Try:

res=list(map(lambda x: list(map(lambda z: ' '.join([str(el) for el in z]), zip(*x))) if isinstance(x[0], list) else x, data))

user_12 Over a year ago

@GrzegorzSkibinski I tried but it's not working. I mean the error is gone but it returns input data as output. It's not joining

Georgina Skibinski Over a year ago

Ok, since it's not a list, but numpy try:

res=list(map(lambda x: list(map(lambda z: ' '.join([str(el) for el in z]), zip(*x))) if isinstance(x[0], np.ndarray) else x, data))

Andy L. · Accepted Answer · 2020-02-16 21:12:46Z

1

Since you say there are also timestamp values, I cast to string for every sub-array.

out = [[' '.join(tup) for tup in zip(*arr.astype(str))] if arr.ndim > 1 else arr.tolist() 
               for arr in data]

Out[89]: [['a', 'b', 'c'], ['d g', 'e h', 'f i'], ['j m p', 'k n q', 'l o r']]

answered Feb 16, 2020 at 21:12

Andy L.

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