ここでは、アレイ内のこれらのキーのそれぞれの位置をバックトレースするnp.searchsorted
に基づいてベクトル化一つだし、次に置き換えると(ただし、それを助けることができなかった)、ここではほぼ性差別的関数名を言い訳してください -
def replace_with_dict(ar, dic):
# Extract out keys and values
k = np.array(list(dic.keys()))
v = np.array(list(dic.values()))
# Get argsort indices
sidx = k.argsort()
# Drop the magic bomb with searchsorted to get the corresponding
# places for a in keys (using sorter since a is not necessarily sorted).
# Then trace it back to original order with indexing into sidx
# Finally index into values for desired output.
return v[sidx[np.searchsorted(k,ar,sorter=sidx)]]
サンプル実行 -
In [82]: dic ={334:0, 4:22, 8:31, 12:16, 16:17, 24:27, 28:18, 32:21, 36:1}
...:
...: np.random.seed(0)
...: a = np.random.choice(dic.keys(), 20)
...:
In [83]: a
Out[83]:
array([ 28, 16, 32, 32, 334, 32, 28, 4, 8, 334, 12, 36, 36,
24, 12, 334, 334, 36, 24, 28])
In [84]: replace_with_dict(a, dic)
Out[84]:
array([18, 17, 21, 21, 0, 21, 18, 22, 31, 0, 16, 1, 1, 27, 16, 0, 0,
1, 27, 18])
改善
大きなアレイの
速い一方がソート値とキー配列であり、次いで、sorter
なしsearchsorted
を使用するように希望 -
def replace_with_dict2(ar, dic):
# Extract out keys and values
k = np.array(list(dic.keys()))
v = np.array(list(dic.values()))
# Get argsort indices
sidx = k.argsort()
ks = k[sidx]
vs = v[sidx]
return vs[np.searchsorted(ks,ar)]
ランタイム試験 -
In [91]: dic ={334:0, 4:22, 8:31, 12:16, 16:17, 24:27, 28:18, 32:21, 36:1}
...:
...: np.random.seed(0)
...: a = np.random.choice(dic.keys(), 20000)
In [92]: out1 = replace_with_dict(a, dic)
...: out2 = replace_with_dict2(a, dic)
...: print np.allclose(out1, out2)
True
In [93]: %timeit replace_with_dict(a, dic)
1000 loops, best of 3: 453 µs per loop
In [95]: %timeit replace_with_dict2(a, dic)
1000 loops, best of 3: 341 µs per loop
汎用ケースすべての配列要素が辞書にない場合
入力配列のすべての要素が辞書に存在することが保証されていない場合、リストされているとおりにもう少し作業が必要です以下 -
def replace_with_dict2_generic(ar, dic, assume_all_present=True):
# Extract out keys and values
k = np.array(list(dic.keys()))
v = np.array(list(dic.values()))
# Get argsort indices
sidx = k.argsort()
ks = k[sidx]
vs = v[sidx]
idx = np.searchsorted(ks,ar)
if assume_all_present==0:
idx[idx==len(vs)] = 0
mask = ks[idx] == ar
return np.where(mask, vs[idx], ar)
else:
return vs[idx]
サンプル実行 -
In [163]: dic ={334:0, 4:22, 8:31, 12:16, 16:17, 24:27, 28:18, 32:21, 36:1}
...:
...: np.random.seed(0)
...: a = np.random.choice(dic.keys(), (20))
...: a[-1] = 400
In [165]: a
Out[165]:
array([ 28, 16, 32, 32, 334, 32, 28, 4, 8, 334, 12, 36, 36,
24, 12, 334, 334, 36, 24, 400])
In [166]: replace_with_dict2_generic(a, dic, assume_all_present=False)
Out[166]:
array([ 18, 17, 21, 21, 0, 21, 18, 22, 31, 0, 16, 1, 1,
27, 16, 0, 0, 1, 27, 400])
これまでに試したコードを投稿できますか? –
'df ['your_col']。map(df2.set_index( 'code'))' –