如何减少这个熊猫数据框连接代码

编程问答 2022-05-02

问题描述

我有一个以下元组列表形式的模板，我将使用数据帧连接实例化它。

rule = [('#1','X','Y'),('#2','Z'),('#3','Z','Y')]

我还有一个模板的每个组件的实例作为字典。

rComp_substitution =

{('#1','Y'):           pred  subj  obj
                   0  nationality  BART  USA,'Z'):            pred  subj      obj
                   0  placeOfBirth  BART  NEWYORK
                   1     hasFather  BART   HOMMER,'Y'):           pred     subj  obj
                   0    locatedIn  NEWYORK  USA
                   1  nationality   HOMMER  USA }

每个组件对应的实例是一个pandas数据框，有三列。对于 ('#1','Y')，#1 对应于 pred，X 对应于 subj，而 Y 对应于 obj。

比如先实例化('#1','Z').

我们可以检查 ('#1','Y') 和 ('#2','Z') 的公共变量。

并用一个键连接每个数据帧的公共变量 X(subj) 以获得 ('#1','Z') 的实例').

下面是我的代码。

depth = 0    
# step1 check common variable
current_subj = rule[depth][1] #['X']
current_obj = rule[depth][2] #['Y']
next_subj = rule[depth+1][1] #['X']
next_obj = rule[depth+1][2] #['Z']
if current_subj == next_subj or current_subj == next_obj:
    comVar = current_subj
elif current_obj == next_subj or current_obj == next_obj:
    comVar = current_obj

# step2 Create currnt_rComp with common variable for joining dataframes
current_rComp = rComp_substitution[rule[depth]]
unified_rComp = []
for col in current_rComp.itertuples(index=False):
    if comVar == current_subj:
        unified_rComp.append([col.subj,[list(col)]])
    elif comVar == current_obj:
        unified_rComp.append([col.obj,[list(col)]])
current_rComp = pd.DataFrame(unified_rComp,columns=['comVar','triples'])

# step3 Create next_rComp with common variable for joining dataframes
next_rComp = rComp_substitution[rule[depth+1]]
unified_rComp = []
for col in next_rComp.itertuples(index=False):
    if comVar == next_subj:
        unified_rComp.append([col.subj,[list(col)]])
    elif comVar == next_obj:
        unified_rComp.append([col.obj,[list(col)]])
next_rComp = pd.DataFrame(unified_rComp,'triples'])

# step4 Join currnt_rComp and next_rComp with common variable as key
partial_proof_path = pd.merge(current_rComp,next_rComp,how='inner',on='comVar')
print(partial_proof_path)

这个代码输出是

  comVar                   triples_x                        triples_y
0   BART  [[nationality,BART,USA]]  [[placeOfBirth,NEWYORK]]
1   BART  [[nationality,USA]]      [[hasFather,HOMMER]]

我认为这段代码太长了。有没有办法用更简单的代码做同样的事情？

解决方法

输入数据：

rComp_substitution = {('#1','X','Y'): pd.DataFrame({'pred': ['nationality'],'subj': ['BART'],'obj': ['USA']}),('#2','Z'): pd.DataFrame({'pred': ['placeOfBirth','hasFather'],'subj': ['BART','BART'],'obj': ['NEWYORK','HOMMER']}),('#3','Z','Y'): pd.DataFrame({'pred': ['locatedIn','nationality'],'subj': ['NEWYORK','HOMMER'],'obj': ['USA','USA']})}

rules = list(rComp_substitution.keys())

主要功能：

def merge_from_common_key(rule0,rule1):
  # Load dataframes
  df0 = rComp_substitution[rule0]
  df1 = rComp_substitution[rule1]

  # Rename ["pred","subj","obj"] by ruleN
  df0.columns = rule0
  df1.columns = rule1

  # Find the common key(s) and merge the two dataframes
  key = df0.columns.intersection(df1.columns).tolist()
  df = pd.merge(df0,df1,on=key)

  # Build the new dataframe
  return pd.DataFrame({"common": df["X"].values.tolist(),"left": df[list(rules[0])].values.tolist(),"right": df[list(rules[1])].values.tolist()})

用法：

>>> merge_from_common_key(rules[0],rules[1])

  common                      left                          right
0   BART  [nationality,BART,USA]  [placeOfBirth,NEWYORK]
1   BART  [nationality,USA]      [hasFather,HOMMER]
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