熊猫地图返回列表

Question

我需要它来合并“target_cw_id”上的“companies”数据集和“locations”数据集以及“relations”数据集中的“source_cw_id”数据集

公司

 row_id     cw_id   cik     company_name    source_type     source_id
0   1   1   20.0    MOTHER COMPANY  filers  35791
1   2   2   1750.0  FATHER COMPANY  filers  40788
2   3   3   1800.0  LITTLE SISTER   filers  60238
3   4   4   1800.0  MIDDLE SISTER   filers  60238
4   5   5   2132.0  BABY BROTHER    filers  8286
5   6   6   543.0   NAUGHTY COUSIN  filers  8286
6   7   7   4546.0  BIG BROTHER     filers  8286

关系

    relation_id     target_cw_id    source_cw_id    relation_type   relation_origin     origin_id   year
0   1   3   1   NaN     relationships   2507504     2010
1   2   4   1   NaN     relationships   824847  2005
2   3   5   2   NaN     relationships   841281  2006
3   4   6   2   NaN     relationships   864758  2007
4   5   7   2   NaN     relationships   1288382     2008

地点

    cw_id   country_code
0   1   US
1   2   AT
2   3   US
3   4   US
5   5   SU
6   6   US
7   7   US

这按预期工作，但我想让它减少冗余

merged = pd.merge(left=relations, right=companies, left_on="source_cw_id", right_on="cw_id", how="left")
merged = pd.merge(left=merged, right=companies, left_on="target_cw_id", right_on="cw_id", how="left",  suffixes=('_source', '_target'))
merged = pd.merge(left=merged, right=locations, left_on="source_cw_id", right_on="cw_id", how="left")
merged = pd.merge(left=merged, right=locations, left_on="target_cw_id", right_on="cw_id", how="left",  suffixes=('_source', '_target'))

所以我正在尝试map 和lambda

merged = pd.DataFrame()

dfs = [relations, merged, merged, merged]
dfs2 = [companies, companies, locations, locations]
ids = ["source_cw_id","target_cw_id","source_cw_id","target_cw_id"]

merged = map(lambda x, y, z: pd.merge(left=x, right=y, left_on=z, right_on="cw_id", how="left",suffixes=('_source','_target')), dfs,dfs2,ids)

但是，第一次迭代返回一个列表而不是数据框，然后我得到一个

KeyError "target_cw_id"

这些是我期望在最终文件中的列名：

[u'relation_id', u'source_cw_id', u'target_cw_id', u'relation_type',
       u'relation_origin', u'origin_id', u'year', u'row_id_source',
       u'cw_id_source', u'cik_source', u'company_name_source',
       u'source_type_source', u'source_id_source', u'row_id_target',
       u'cw_id_target', u'cik_target', u'company_name_target',
       u'source_type_target', u'source_id_target', u'cw_id_source',
       u'country_code_source', u'cw_id_target', u'country_code_target']

任何想法表示赞赏！

Answer 1

首先，您使用 map 错误 (docs)

merged = map(
    lambda x, y, z: pd.merge(left=x, right=y, left_on=z, right_on="cw_id", how="left",suffixes=('_source','_target')),
    dfs,dfs2,ids
)

第一个参数是一个函数，你说得对。第二个是可迭代的，你错了。

您执行了 4 次合并，我希望有 4 个项目可以迭代。我会做以下事情。

def mymerge(x, y, z):
    kwargs = dict(left=x, right=y, left_on=z,
                  right_on="cw_id", how="left",
                  suffixes=('_source', '_targee'))
    return pd.merge(**kwargs)

# Initialize merged
merged = relations.copy()

dfs = [merged, merged, merged, merged]
dfs2 = [companies, companies, locations, locations]
ids = ["source_cw_id", "target_cw_id", "source_cw_id", "target_cw_id"]

myiterable = zip(dfs, dfs2, ids)

# Instead of map
for args in myiterables:
    merged = mymerge(*args)

---

请记住，我无法在您的数据上对此进行测试，因为您没有提供任何示例数据或期望输出。

MaxU 建议阅读 https://stackoverflow.com/help/mcve 。我也是。