Converting strings Series to numeric one
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0
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I'm doing
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
Where x has strings of hex colors, and I'd like to map them to RGB arrays (3 values each). After that, X hasdtype='object, and X.values is a numpy array of numpy arrays.
My final goal is making it an 3 * n numpy array and use it with sklearn.cluster.KMeans. What is the best way to achieving this?
pandas
asked Nov 19 at 14:24
galah92
884816
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up vote
0
down vote
favorite
I'm doing
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
Where x has strings of hex colors, and I'd like to map them to RGB arrays (3 values each). After that, X hasdtype='object, and X.values is a numpy array of numpy arrays.
My final goal is making it an 3 * n numpy array and use it with sklearn.cluster.KMeans. What is the best way to achieving this?
pandas
asked Nov 19 at 14:24
galah92
884816
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm doing
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
Where x has strings of hex colors, and I'd like to map them to RGB arrays (3 values each). After that, X hasdtype='object, and X.values is a numpy array of numpy arrays.
My final goal is making it an 3 * n numpy array and use it with sklearn.cluster.KMeans. What is the best way to achieving this?
pandas
asked Nov 19 at 14:24
galah92
884816
I'm doing
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
Where x has strings of hex colors, and I'd like to map them to RGB arrays (3 values each). After that, X hasdtype='object, and X.values is a numpy array of numpy arrays.
My final goal is making it an 3 * n numpy array and use it with sklearn.cluster.KMeans. What is the best way to achieving this?
pandas
pandas
asked Nov 19 at 14:24
galah92
884816
asked Nov 19 at 14:24
galah92
884816
edited Nov 19 at 14:35
asked Nov 19 at 14:24
galah92
884816
asked Nov 19 at 14:24
galah92
884816
asked Nov 19 at 14:24
galah92
884816
884816
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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0
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After creating X, you can split up the data into 3 columns like this
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
data[['R','G','B']] = pd.DataFrame(X.values.tolist(), index=X.index)
so that
data[['R','G','B']]
has the result in three columns for further processing
answered Nov 19 at 15:14
576i
2,1831033
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
After creating X, you can split up the data into 3 columns like this
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
data[['R','G','B']] = pd.DataFrame(X.values.tolist(), index=X.index)
so that
data[['R','G','B']]
has the result in three columns for further processing
answered Nov 19 at 15:14
576i
2,1831033
add a comment |
up vote
0
down vote
After creating X, you can split up the data into 3 columns like this
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
data[['R','G','B']] = pd.DataFrame(X.values.tolist(), index=X.index)
so that
data[['R','G','B']]
has the result in three columns for further processing
answered Nov 19 at 15:14
576i
2,1831033
add a comment |
up vote
0
down vote
up vote
0
down vote
After creating X, you can split up the data into 3 columns like this
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
data[['R','G','B']] = pd.DataFrame(X.values.tolist(), index=X.index)
so that
data[['R','G','B']]
has the result in three columns for further processing
answered Nov 19 at 15:14
576i
2,1831033
After creating X, you can split up the data into 3 columns like this
X = data['x'].apply(lambda h: [int(h[i:i + 2], 16) for i in (0, 2 ,4)])
data[['R','G','B']] = pd.DataFrame(X.values.tolist(), index=X.index)
so that
data[['R','G','B']]
has the result in three columns for further processing
answered Nov 19 at 15:14
576i
2,1831033
answered Nov 19 at 15:14
576i
2,1831033
answered Nov 19 at 15:14
576i
2,1831033
answered Nov 19 at 15:14
576i
2,1831033
2,1831033
add a comment |
add a comment |
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