我正在溝通，pysimplegui但我有UnboundLocalError我的一項職能。這是function：def load_file(file): file_details = file.split('.') if file_details[1] == "csv": df = pd.read_csv(file) elif file_details[1] == "xlsx": df = pd.read_excel(file) elif file_details[1] != "csv" and file_details[1] != "xlsx": sg.popup("Unsupported file type") else: sg.popup("Your file does not meet the requirements", "Please check the file requirements for more info") return df我的代碼可能有什么問題？
2 回答
繁星淼淼
TA貢獻1775條經(jīng)驗 獲得超11個贊
df如果詳細(xì)信息是“csv”或“xlsx”，您只需創(chuàng)建一個名為 的對象。如果他們有另一種類型，則不會df創(chuàng)建。因此，您不能在函數(shù)結(jié)束時返回 df。每當(dāng)您嘗試時，它都會崩潰。
有兩種可能性：
僅當(dāng) df 存在時才返回它，方法是將它從 if-elif 塊中返回。
如果文件類型無效，則創(chuàng)建一個沒有任何值的 df。
選項1：
def load_file(file):
file_details = file.split('.')
if file_details[1] == "csv":
df = pd.read_csv(file)
return df
elif file_details[1] == "xlsx":
df = pd.read_excel(file)
return df
elif file_details[1] != "csv" and file_details[1] != "xlsx":
sg.popup("Unsupported file type")
else:
sg.popup("Your file does not meet the requirements", "Please check the file requirements for more info")
選項 2
def load_file(file):
file_details = file.split('.')
if file_details[1] == "csv":
df = pd.read_csv(file)
elif file_details[1] == "xlsx":
df = pd.read_excel(file)
elif file_details[1] != "csv" and file_details[1] != "xlsx":
sg.popup("Unsupported file type")
df = None
else:
sg.popup("Your file does not meet the requirements", "Please check the file requirements for more info")
df = None
return df
None對于這兩個選項，如果類型不受支持，函數(shù)將返回。
函數(shù)式編程
TA貢獻1807條經(jīng)驗 獲得超9個贊
您應(yīng)該在任何條件之前初始化空變量，如下所示
df = ''
def load_file(file): file_details = file.split('.')
**df = ''**
if file_details[1] == "csv":
df = pd.read_csv(file)
elif file_details[1] == "xlsx":
df = pd.read_excel(file)
elif file_details[1] != "csv" and file_details[1] != "xlsx":
sg.popup("Unsupported file type")
else:
sg.popup("Your file does not meet the requirements", "Please check the file requirements for more info")
return df
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