在 Python 中解碼十六進制腳本

Running decodescriptond1a076722d11ddfd0fadb96d6d960b8b2ad6c6369570cd00df70a801a6be9e44正在返回：

{
 "asm": "0 30440220278ee027abaf4720f580ce469139d75e330e120f91ff52b2c73955a712b96f3902204f11b7b1541dee5497cffd50291347b796e44ab1b6dac845d43a7185fdda184901 3045022100c3602c2872248e73f543fbf21ddc9fa3b73052261f619452fd5254852c9ea55102202c1323830f38a6fe149e0eccb18f92eba962165733285812deb96bcb8482de5d01 30450221008caaadf9b887828d02911ff22df335606c372f968d5c2ea0fa2f8c1a083fe5f502204c55eb85c0f06e10d488d7643a5c571bbabdeb935acc917836d44e755db6b1f601 532102939a016f9a4cfc4c73ba395c2baf16382a4de078dadf2bafaaac6846d786ed392102c7cf0bd62fb500d8f6ed39a46c7e72dc1b5185754e1a48d25a22195f113b35d8210306179c491a25d6db330a3a3e3ba0788b3ccea45aebfa2a06fc114787aadc01532103b8a79bd499afbc0f4b68e4c0e175176a013b1e2143da6ec95772add5c6728d6d2103bec7b18285d7af85b4f2c0936a106b746eaad14071f41788257fdffeb2e68e9a2103e4fda639ffcf2d996dc372db5eda18857afdd21054009d65ddd17ce96d5feed22103f5d1065abd2a8b5adc6ecda48712d2a572a24130b786f538629ff834b298379857ae",
 "type": "nonstandard",
 "p2sh": "32QZxW8qHSufifvr5UzXpzGG6jdcgnTCSF",
 "segwit": {
   "asm": "0 59d0d03149e611e8f8e879ab7c1133c99091ec09758ed29a9111341e424d9891",
   "hex": "002059d0d03149e611e8f8e879ab7c1133c99091ec09758ed29a9111341e424d9891",
   "reqSigs": 1,
   "type": "witness_v0_scripthash",
   "addresses": [
     "bc1qt8gdqv2fucg7378g0x4hcyfnexgfrmqfwk8d9x53zy6pusjdnzgsdvlm7s"
   ],
   "p2sh-segwit": "36sk1CLrrv8HXgE1d1CKkNkwnK2tWKeWSL"
 }
}

原始十六進制：

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

Blockstream explorer 以 OP_CODEs 的形式返回腳本，我的問題是，是否有任何 python 庫可以將其解碼為相同的狀態？

我知道至少有兩個：Python-bitcoinlib和bitcoin_tools

引用自：https://bitcoin.stackexchange.com/questions/91715