我在 pandas 中有一個像這樣格式化的數(shù)據(jù)框。(df)School ID Column 1 Column 2 Column 3School 1 8100 8200 School 2 9-999 School 3 9300 9500 School 4 7700 7800 School 5 8999....我希望能夠輸入一個值，例如如果該值是直接命中第1列（字符串類型），例如輸入數(shù)字：9-999，它將返回Input Number: 9-999School ID Column 1School 2 9-999但如果輸入的數(shù)字介于第 2 列和第 3 列（浮點數(shù)）中的數(shù)字之間，我想返回關(guān)聯(lián)的學(xué)校 ID，如下所示。Input Number: 8110School ID Column 2 Column 3School 5 8100 8200現(xiàn)在我有這個代碼：def find(num) :d1=df.loc[df['Column 1']==num]if len(d1)>0 : return d1[['School ID','Column 1']]else : return df.loc[(num>= df['Column 2']) & (num<= df['Column 3'])][['School ID','Column 2','Column 3']]但我收到一條錯誤消息：“返回”外部函數(shù)
1 回答
搖曳的薔薇
TA貢獻(xiàn)1793條經(jīng)驗 獲得超6個贊
s = '9-999'
q = df[df['Column_1']==str(s)]
if len(q):
print(q)
else:
m = df[['Column_2', 'Column_3']].apply(lambda x: x['Column_2'] <= s <= x['Column_3'], axis=1)
print(df[m])
印刷：
School_ID Column_1 Column_2 Column_3
1 School 2 9-999 0.0 0.0
為了s = 8110：
School_ID Column_1 Column_2 Column_3
0 School 1 0 8100.0 8200.0
編輯：為了獲得一致的數(shù)據(jù)類型，您可以將 Column2 和 Column3 轉(zhuǎn)換為浮點數(shù)：
s = '8110'
q = df[df['Column_1']==str(s)]
df['Column_2'] = df['Column_2'].astype(float)
df['Column_3'] = df['Column_3'].astype(float)
if len(q):
print(q)
else:
s = float(s)
m = df[['Column_2', 'Column_3']].apply(lambda x: x['Column_2'] <= s <= x['Column_3'], axis=1)
print(df[m])
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