Pandas 如何比较和标记最短的字符串行

我的目标是通过使用

transform（）

或

apply（）

函数来检测基于

res

列的连续行是否相等

我的数据帧：

data = [[111, 123, "aa", 0], 
        [111, 124, "bb", 1], 
        [111, 125, "bb", 2],
        [111, 126, "cc", 0],
        [111, 127, "dd", 1],
        [111, 128, "cc", 2],
        [222, 133, "xx", 1],
        [222, 134, "yy", 2],
        [222, 135, "zz", 0], 
        [222, 136, "zz", 1],] 
df = pd.DataFrame(data, columns = ["uuid", "foo_id", "res", "num"]) 

111, 123, "aa", 0, 0 
111, 124, "bb", 1, 1 
111, 125, "bb", 2, 1
111, 126, "cc", 0, 0
111, 127, "dd", 1, 0
111, 128, "cc", 2, 0
222, 133, "xx", 1, 0
222, 134, "yy", 2, 0
222, 135, "zz", 0, 1
222, 136, "zz", 1, 1

df['flag'] = df.groupby('uuid')['res'].tranform(lambda x:  1 if x == x.shift(-1) else 0)

*ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().*

我在找什么：

data = [[111, 123, "aa", 0], 
        [111, 124, "bb", 1], 
        [111, 125, "bb", 2],
        [111, 126, "cc", 0],
        [111, 127, "dd", 1],
        [111, 128, "cc", 2],
        [222, 133, "xx", 1],
        [222, 134, "yy", 2],
        [222, 135, "zz", 0], 
        [222, 136, "zz", 1],] 
df = pd.DataFrame(data, columns = ["uuid", "foo_id", "res", "num"]) 

111, 123, "aa", 0, 0 
111, 124, "bb", 1, 1 
111, 125, "bb", 2, 1
111, 126, "cc", 0, 0
111, 127, "dd", 1, 0
111, 128, "cc", 2, 0
222, 133, "xx", 1, 0
222, 134, "yy", 2, 0
222, 135, "zz", 0, 1
222, 136, "zz", 1, 1

df['flag'] = df.groupby('uuid')['res'].tranform(lambda x:  1 if x == x.shift(-1) else 0)

*ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().*

我用过：

data = [[111, 123, "aa", 0], 
        [111, 124, "bb", 1], 
        [111, 125, "bb", 2],
        [111, 126, "cc", 0],
        [111, 127, "dd", 1],
        [111, 128, "cc", 2],
        [222, 133, "xx", 1],
        [222, 134, "yy", 2],
        [222, 135, "zz", 0], 
        [222, 136, "zz", 1],] 
df = pd.DataFrame(data, columns = ["uuid", "foo_id", "res", "num"]) 

111, 123, "aa", 0, 0 
111, 124, "bb", 1, 1 
111, 125, "bb", 2, 1
111, 126, "cc", 0, 0
111, 127, "dd", 1, 0
111, 128, "cc", 2, 0
222, 133, "xx", 1, 0
222, 134, "yy", 2, 0
222, 135, "zz", 0, 1
222, 136, "zz", 1, 1

df['flag'] = df.groupby('uuid')['res'].tranform(lambda x:  1 if x == x.shift(-1) else 0)

*ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().*

返回：

data = [[111, 123, "aa", 0], 
        [111, 124, "bb", 1], 
        [111, 125, "bb", 2],
        [111, 126, "cc", 0],
        [111, 127, "dd", 1],
        [111, 128, "cc", 2],
        [222, 133, "xx", 1],
        [222, 134, "yy", 2],
        [222, 135, "zz", 0], 
        [222, 136, "zz", 1],] 
df = pd.DataFrame(data, columns = ["uuid", "foo_id", "res", "num"]) 

111, 123, "aa", 0, 0 
111, 124, "bb", 1, 1 
111, 125, "bb", 2, 1
111, 126, "cc", 0, 0
111, 127, "dd", 1, 0
111, 128, "cc", 2, 0
222, 133, "xx", 1, 0
222, 134, "yy", 2, 0
222, 135, "zz", 0, 1
222, 136, "zz", 1, 1

df['flag'] = df.groupby('uuid')['res'].tranform(lambda x:  1 if x == x.shift(-1) else 0)

*ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().*

---

IIUC，您可以尝试使用

系列。每组重复的
：

f = lambda x: (x.eq(x.shift()) | x.eq(x.shift(-1))).astype(int)
df['flag'] = df.groupby('uuid')['res'].transform(f)


组内重复的或连续的。例如，一个组内的[AA，bb，aaa ]是否考虑了AA标记？“ALollz怎么说，假设你在一个组中重复了2个值，在这种情况下你想返回什么？”安基你是对的，没有代码>复制（）/<代码>函数的解决方案，你可以在一个组中添加更多的重复，并显示你想要的吗？对于组111，如果您有两个重复的
aa，aa，xx，xx，yy
，是否要
1,2,1,2,1
？或者
0,0,1,1,2
？我想标记和比较最近的行，我已经更新了我的数据框以向您说明清楚。@阿迪尔布兰科没有意识到您更新了这个问题，您能用
f=lambda x:（x.eq（x.shift（））| x.eq（x.shift（-1））.astype（int）
替换
f
吗？编辑了我的答案