pandas 尝试使用函数中定义的数据帧名称时发生意外的 NameError
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Unexpected NameError occurs when trying to use a dataframe name defined within a function
提问by Jason
Could somebody explain why the following code produces a NameError?
有人可以解释为什么下面的代码会产生NameError?
def nonull(df, col, name):
name = df[pd.notnull(df[col])]
print name[col].count(), df[col].count()
return name
nonull(sve, 'DOC_mg/L', 'sveDOC')
sveDOC.count()
NameError: name 'sveDOC' is not defined
711 711
The dataframeseems to be created as the printstatement works, so I don't understand why when I try to use sveDOC(which was nameinside the function) it produces an error.
在dataframe似乎为要创建print声明的作品，所以我不明白为什么当我尝试使用sveDOC（这是name在函数内部）会产生错误。
Here's an example of what I'd like to do within the function:
这是我想在函数中执行的操作的示例：
import pandas as pd
d = {'one' : pd.Series([1., 1., 1., 1.], index=['a', 'b', 'c', 'd']),
'two' : pd.Series([1., 2., 3., 4.], index=['a', 'b', 'c', 'd'])}
pd.DataFrame(d)
df = pd.DataFrame(d)
df1 = df
df = df * 2
print df.head(), df1.head()
one two
a 2 2
b 2 4
c 2 6
d 2 8
one two
a 1 1
b 1 2
c 1 3
d 1 4
回答by jonrsharpe
Python names do notwork the way you seem to think. Here's what your code actually does:
Python 名称并不像您认为的那样工作。这是您的代码实际执行的操作：
def nonull(df, col, name):
name = df # rebind the name 'name' to the object referenced by 'df'
name = df[pd.notnull(name[col])] # rebind the name 'name' again
print name[col].count(), df[col].count()
return name # return the instance
nonull(sve, 'DOC_mg/L', 'sveDOC') # call the function and ignore the return value
The function never actually uses the 'sveDOC'argument. Here's what you should actually do:
该函数从不实际使用该'sveDOC'参数。以下是您实际应该做的：
def nonull(df, col):
name = df[pd.notnull(df[col])]
print name[col].count(), df[col].count()
return name
sveDOC = nonull(sve, 'DOC_mg/L')
sveDOC.count()
Your conception of Python's use of names and references is completely wrong.
您对 Python 使用名称和引用的概念是完全错误的。
pd.DataFrame(d) # creates a new DataFrame but doesn't do anything with it
# (what was the point of this line?)
df = pd.DataFrame(d) # assigns a second new DataFrame to the name 'df'
df1 = df # assigns the name `df1` to the same object that 'df' refers to
# - note that this does *not* create a copy
df = df * 2 # create a new DataFrame based on the one referenced by 'df'
# (and 'df1'!)and assign to the name 'df'
To demonstrate this:
为了证明这一点：
df1 = pd.DataFrame(d)
df2 = df1
df1 is df2
Out[5]: True # still the same object
df2 = df2 * 2
df1 is df2
Out[7]: False #?now different
If you want to create a copy of a DataFrame, do so explicitly:
如果要创建 a 的副本DataFrame，请明确执行以下操作：
df2 = copy(df1)
You can either do this outside nonulland pass the copy, or do it inside nonulland returnthe modified copy.
您可以在外部执行此操作nonull并传递副本，也可以在内部执行nonull并return修改后的副本。
