例如，我有一個數據框 (df_1)，其中一列包含一些文本數據。第二個數據框 (df_2) 包含一些數字。如何檢查文本是否包含第二個數據框中的數字？df_1 Note0 The code to this is 10031 The code to this is 1004df_2 Code_Number0 10061 1003所以我想檢查 df_1 [Note] 中的條目是否包含 df_2 [Code_Number] 中的條目我嘗試使用以下代碼：df_1[df_1['Note'].str.contains(df_2['Code_Number'])]并且我知道我不能使用連接，因為我沒有加入的密鑰。應用過濾后我正在尋找的最終結果是： Note 0 The code to this is 1003
3 回答
慕仙森
TA貢獻1827條經驗 獲得超8個贊
試試這個，看看它是否涵蓋了您的用例：使用itertools 的產品和基于條件的過濾器獲取兩列的交叉笛卡爾：
from itertools import product
m = [ left for left, right
in product(df.Note,df1.Code_Number)
if str(right) in left]
pd.DataFrame(m,columns=['Note'])
Note
0 The code to this is 1003
拉莫斯之舞
TA貢獻1820條經驗 獲得超10個贊
做這個：
df_1.loc[df_1['Note'].apply(lambda x: any(str(number) in x for number in df_2['Code_Number']))]
qq_花開花謝_0
TA貢獻1835條經驗 獲得超7個贊
Firstly, you have to create 1 column in your df1 where the notes are with a list of numbers that are present in the Notes and then Compare the List column of numbers with the List column of the df2 where the numbers are present(both should be in list format)
#Extract Numbers from Notes
a_string = "0abcadda1 11 def 23 10007"
numbers = [int(word) for word in a_string.split() if word.isdigit()]
print(numbers)
list_test = "103,23"
#Finding common element from both lists the list
L1 = [2,3,4]
L2 = [1,2]
[i for i in L1 if i in L2]
S1 = set(L1)
S2 = set(L2)
print(S1.intersection(S2))
#If you want to find out the common element
def common_data(list1, list2):
result = False
# traverse in the 1st list
for x in list1:
# traverse in the 2nd list
for y in list2:
# if one common
if x == y:
result = True
return result
return result
# driver code
a = [1, 2, 3, 4, 5]
b = [5, 6, 7, 8, 9]
print(common_data(a, b))
a = [1, 2, 3, 4, 5]
b = [6, 7, 8, 9]
print(common_data(a, b))
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