熊猫apply导致UnboundLocalError

如何解决熊猫apply导致UnboundLocalError

我有一个包含2列[客户ID，群集]的数据框（df_cluster）。大约有13个群集，我正在尝试使用python中的apply（）为每个群集分配一个名称。我过去使用过相同的功能，但效果很好，但是现在出现“ UnboundLocalError”错误。

如果我做错了任何事情，请告诉我。我对apply（）的理解是，它沿轴传递了函数（在这种情况下，将为每一行传递函数cluster_name）

这是代码

def cluster_name(df):
    if df['cluster'] == 1:
        value = 'A'
    elif df['cluster'] == 2:
        value = 'B'    
    elif df['cluster'] == 3:
        value = 'C'
    elif df['cluster'] == 4:
        value = 'D'
    elif df['cluster'] == 5:
        value = 'E'
    elif df['cluster'] == 6:
        value = 'F'
    elif df['cluster'] == 7:
        value = 'G'
    return value

df_cluster['cluster_name'] = df_cluster.apply(cluster_name,axis = 1)

错误

UnboundLocalError                         Traceback (most recent call last)
<ipython-input-16-b64f3fdc1260> in <module>
     16     return value
     17 
---> 18 df_cluster['cluster_name'] = df_cluster.apply(cluster_name,axis = 1)
     19 df_cluster['cluster_name'].value_counts()

/opt/cloudera/parcels/Anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/frame.py in apply(self,func,axis,broadcast,raw,reduce,result_type,args,**kwds)
   6926             kwds=kwds,6927         )
-> 6928         return op.get_result()
   6929 
   6930     def applymap(self,func):

/opt/cloudera/parcels/Anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/apply.py in get_result(self)
    184             return self.apply_raw()
    185 
--> 186         return self.apply_standard()
    187 
    188     def apply_empty_result(self):

/opt/cloudera/parcels/Anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/apply.py in apply_standard(self)
    290 
    291         # compute the result using the series generator
--> 292         self.apply_series_generator()
    293 
    294         # wrap results

/opt/cloudera/parcels/Anaconda/envs/py36/lib/python3.6/site-packages/pandas/core/apply.py in apply_series_generator(self)
    319             try:
    320                 for i,v in enumerate(series_gen):
--> 321                     results[i] = self.f(v)
    322                     keys.append(v.name)
    323             except Exception as e:

<ipython-input-16-b64f3fdc1260> in cluster_name(df)
     14     elif df['cluster'] == 7:
     15         value = 'G'
---> 16     return value
     17 
     18 df_cluster['cluster_name'] = df_cluster.apply(cluster_name,axis = 1)

UnboundLocalError: ("local variable 'value' referenced before assignment",'occurred at index 0')
'''

解决方法

似乎您的问题已在评论中得到了解答，因此我将提出一种面向大熊猫的方法来解决您的问题。将apply(axis=1)与DataFrame一起使用非常慢，几乎不需要（与遍历数据帧中的行相同），因此更好的方法是使用矢量化方法。最简单的方法是在字典中定义集群-> cluster_name映射，并使用map方法：

df = pd.DataFrame(
    {"cluster": [1,2,3,4,5,6,7]}
)

# repeat this dataframe 10000 times
df = pd.concat([df] * 10000)

应用方法：

def mapping_func(row):
    if row['cluster'] == 1:
        value = 'A'
    elif row['cluster'] == 2:
        value = 'B'    
    elif row['cluster'] == 3:
        value = 'C'
    elif row['cluster'] == 4:
        value = 'D'
    elif row['cluster'] == 5:
        value = 'E'
    elif row['cluster'] == 6:
        value = 'F'
    elif row['cluster'] == 7:
        value = 'G'
    else:
        # This is a "catch-all" in case none of the values in the column are 1-7
        value = "Z"
        
    return value

%timeit df.apply(mapping_func,axis=1)
# 1.32 s ± 91.3 ms per loop (mean ± std. dev. of 7 runs,1 loop each)

.map方法

mapping_dict = {
    1: "A",2: "B",3: "C",4: "D",5: "E",6: "F",7: "G"
}

# the `fillna` is our "catch-all" statement.
#  essentially if `map` encounters a value not in the dictionary
#  it will place a NaN there. So I fill those NaNs with "Z" to
#  be consistent with the above example
%timeit df["cluster"].map(mapping_dict).fillna("Z")
# 4.87 ms ± 195 µs per loop (mean ± std. dev. of 7 runs,100 loops each)

我们可以看到，使用字典方法的map比apply快得多，同时还避免了较长的if/elif语句链。

,

您的函数中缺少else：

def cluster_name(df):
    if df['cluster'] == 1:
        value = 'A'
    elif df['cluster'] == 2:
        value = 'B'    
    elif df['cluster'] == 3:
        value = 'C'
    elif df['cluster'] == 4:
        value = 'D'
    elif df['cluster'] == 5:
        value = 'E'
    elif df['cluster'] == 6:
        value = 'F'
    elif df['cluster'] == 7:
        value = 'G'
    else:
        value = ...
    return value

否则，如果value不在值{1、2，...，7}中，则将不会设置df['cluster']，这将导致异常。

,

• 手动创建if-else函数会被高估，并且可能会遗漏条件。
• 由于您要分配字母作为'cluster_name'，因此请使用string.ascii_uppercase获取所有字母的list，并将zip赋予{{ 1}}
  • 从压缩值和.map中创建一个'cluster'，以创建dict列。
• 此实现使用列中的唯一值来创建地图，因此'cluster_name'不会有问题。
  • 在发生错误的情况下，这是由于当列中的某个值不符合您的"local variable 'value' referenced before assignment"条件时执行return value，这意味着未分配if-else在功能中。

value