在 Pandas 中分解和解包列
Explode and unpack column in Pandas
假设我有一个具有以下结构的数据,如何展开包含列表的列然后解压缩展开的列?
来源:
d = {
"_id" : "5f2",
"connId" : 128,
"hospitalList" : [
{
"hospitalId" : 29,
"boardId" : 1019,
"siteId" : 1
},
{
"hospitalId" : 3091,
"boardId" : 2163,
"siteId" : 382
},
{
"hospitalId" : 28,
"boardId" : 1017,
"siteId" : 5
}]
}
代码:
root = pd.json_normalize(d)
nested_cols = [i for i in root.columns if isinstance(root[i][0], list)]
l = [root.drop(nested_cols,1),]
for i in nested_cols:
l.append(pd.json_normalize(d, record_path=i))
output = pd.concat(l, axis=1)
print(output)
实际结果:
_id connId hospitalId boardId siteId
0 5f2 128.0 29 1019 1
1 NaN NaN 3091 2163 382
2 NaN NaN 28 1017 5
预期结果:
_id connId hospitalId boardId siteId
0 5f2 128.0 29 1019 1
1 5f2 128.0 3091 2163 382
2 5f2 128.0 28 1017 5
这会输出你想要的。
root = pd.json_normalize(d)
nested_cols = [i for i in root.columns if isinstance(root[i][0], list)]
l = [root.drop(nested_cols,1),]
for i in nested_cols:
l.append(pd.json_normalize(d, record_path=i))
output = pd.concat(l, axis=1)
output.fillna(method='ffill', inplace=True)
不过,很遗憾,我不知道您将在何种情况下使用该代码,and/or如果您必须进行调整。
假设我有一个具有以下结构的数据,如何展开包含列表的列然后解压缩展开的列?
来源:
d = {
"_id" : "5f2",
"connId" : 128,
"hospitalList" : [
{
"hospitalId" : 29,
"boardId" : 1019,
"siteId" : 1
},
{
"hospitalId" : 3091,
"boardId" : 2163,
"siteId" : 382
},
{
"hospitalId" : 28,
"boardId" : 1017,
"siteId" : 5
}]
}
代码:
root = pd.json_normalize(d)
nested_cols = [i for i in root.columns if isinstance(root[i][0], list)]
l = [root.drop(nested_cols,1),]
for i in nested_cols:
l.append(pd.json_normalize(d, record_path=i))
output = pd.concat(l, axis=1)
print(output)
实际结果:
_id connId hospitalId boardId siteId
0 5f2 128.0 29 1019 1
1 NaN NaN 3091 2163 382
2 NaN NaN 28 1017 5
预期结果:
_id connId hospitalId boardId siteId
0 5f2 128.0 29 1019 1
1 5f2 128.0 3091 2163 382
2 5f2 128.0 28 1017 5
这会输出你想要的。
root = pd.json_normalize(d)
nested_cols = [i for i in root.columns if isinstance(root[i][0], list)]
l = [root.drop(nested_cols,1),]
for i in nested_cols:
l.append(pd.json_normalize(d, record_path=i))
output = pd.concat(l, axis=1)
output.fillna(method='ffill', inplace=True)
不过,很遗憾,我不知道您将在何种情况下使用该代码,and/or如果您必须进行调整。