我有一个如下图所示的DF:
DF =
id token argument1 argument2
1 Tza Tuvia Tza Moscow
2 perugia umbria perugia
3 associated the associated press Nelson
我现在想比较列 argumentX 和 token 的值，并相应地为新列 ARG 选择值。
DF =
id token argument1 argument2 ARG
1 Tza Tuvia Tza Moscow ARG1
2 perugia umbria perugia ARG2
3 associated the associated press Nelson ARG1
这是我尝试过的:
conditions = [
(DF["token"] == (DF["Argument1"])),
DF["token"] == (DF["Argument2"])]
choices = ["ARG1", "ARG2"]
DF["ARG"] = np.select(conditions, choices, default=nan)
这只会比较整个字符串，如果它们相同则匹配。 .isin、.contains 等构造或使用 DF["ARG_cat"] = DF.apply(lambda row: row['token '] 行 ['argument2'],axis=1) 无效。有什么想法吗？
最佳答案
使用str.contains使用正则表达式 - join token 中的所有值 | for regex OR 用于检查带单词边界的子字符串:
pat = '|'.join(r"\b{}\b".format(re.escape(x)) for x in DF["token"])
conditions = [ DF["argument1"].str.contains(pat), DF["argument2"].str.contains(pat)]
choices = ["ARG1", "ARG2"]
DF["ARG"] = np.select(conditions, choices, default=np.nan)
print (DF)
id token argument1 argument2 ARG
0 1 Tza Tuvia Tza Moscow ARG1
1 2 perugia umbria perugia ARG2
2 3 associated the associated ress Nelson ARG1
编辑:
如果要比较每一行:
d = {'id': [1, 2, 3],
'token': ["Tza","perugia","israel"],
"argument1": ["Tuvia Tza","umbria","Tuvia Tza"],
"argument2": ["israel","perugia","israel"]}
DF = pd.DataFrame(data=d)
print (DF)
id token argument1 argument2
0 1 Tza Tuvia Tza israel
1 2 perugia umbria perugia
2 3 israel Tuvia Tza israel
conditions = [[x[0] in x[1] for x in zip(DF['token'], DF['argument1'])],
[x[0] in x[1] for x in zip(DF['token'], DF['argument2'])]]
choices = ["ARG1", "ARG2"]
DF["ARG"] = np.select(conditions, choices, default=np.nan)
print (DF)
id token argument1 argument2 ARG
0 1 Tza Tuvia Tza israel ARG1
1 2 perugia umbria perugia ARG2
2 3 israel Tuvia Tza israel ARG2
关于python - 比较 DF 中两列的(子)字符串，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51836554/