英文:
conditionally append value to list of lists in pandas
问题
以下是您提供的代码的中文翻译：
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]dfA B0 1 [[1], [1], [1]]1 2 [[1], [1], [1]]2 3 [[1], [1], [1]]# 尝试将B列中的第一个列表添加2df['B'] = df['B'].mask(df.A == 2, df['B'].apply(lambda x: x[0].append(2)))dfA B0 1 [[1, 2, 2, 2], [1], [1]]1 2 None2 3 [[1, 2, 2, 2], [1], [1]]# 期望的结果是:df['B'] = [[[1],[1],[1]],[[1,2],[1],[1]],[[1],[1],[1]]]dfA B0 1 [[1], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1], [1], [1]]
请注意，我已经按照您的要求，只返回代码部分的中文翻译。如果您有任何其他问题或需要进一步的帮助，请随时提问。
英文:
I'm trying to conditionally append a list of lists in pandas:
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]dfA B0 1 [[1], [1], [1]]1 2 [[1], [1], [1]]2 3 [[1], [1], [1]]# attempting to append 1st list of lists in B column with 2df['B'] = df['B'].mask(df.A == 2, df['B'].apply(lambda x: x[0].append(2)))dfA B0 1 [[1, 2, 2, 2], [1], [1]]1 2 None2 3 [[1, 2, 2, 2], [1], [1]]#expected result I'm hoping for is:df['B'] = [[[1],[1],[1]],[[1,2],[1],[1]],[[1],[1],[1]]]dfA B0 1 [[1], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1], [1], [1]]
答案1
得分: 2
尝试使用lambda函数条件性地修改列表的列表，如下所示：
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]df['B'] = df.apply(lambda row: row['B'] if row['A'] != 2 else [row['B'][0] + [2]] + row['B'][1:], axis=1)print(df)
在这里，我们创建了一个新的列表，其中包括修改后的列表，然后返回它。使用apply方法，以axis=1对整个DataFrame进行逐行应用lambda函数。
英文:
try using a lambda function to conditionally modify the list of lists like this:
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]df['B'] = df.apply(lambda row: row['B'] if row['A'] != 2 else [row['B'][0] + [2]] + row['B'][1:], axis=1)print(df)
here we create a new list of lists that includes the modified list and then return that. apply method is called on the entire DataFrame with axis=1 to apply the lambda function row-wise.
答案2
得分: 1
list.append 是原地操作，所以实际上返回的是 None 而不是列表。这就是为什么你的新 df 在第二行有 None。
以下是一种向列表添加 2 的方法。我们取第二行中的第一个列表，添加 [2]，然后展开其余的列表以形成预期输出：
df['B'].mask(df['A'].eq(2), lambda x: x.map(lambda x: [x[0] + [2], *x[1:]]))
输出：
0 [[1], [1], [1]]1 [[1, 2], [1], [1]]2 [[1], [1], [1]]
英文:
list.append works in place, so it actually returns None instead of a list. This is why your new df has None on the second row.
Below is a way to add 2 to the list. We take the first list in the second row and add [2], then unpack the rest of the lists to form the expected output:
df['B'].mask(df['A'].eq(2),lambda x: x.map(lambda x: [x[0] + [2],*x[1:]]))
Output:
0 [[1], [1], [1]]1 [[1, 2], [1], [1]]2 [[1], [1], [1]]
答案3
得分: 0
问题出在你生成数据框的方式上：
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]df
[[[1],[1],[1]]] * df.shape[0] 在Python中有点棘手。因为df 中不同行的所有[[1],[1],[1]]都指向相同的对象，即列表[[1], [1], [1]]。
尝试这样做：
df.at[1, 'B'][0] = [1, 2]
.at[]允许你同时使用索引号和列名进行访问，[0]选择访问后返回的列表的第一个元素。所以这里不需要使用lambda。
有人可能认为只有第二行的'B'列列表中的第一个元素[[1], [1], [1]]被更改为[[1, 2], [1], [1]]。但如果你查看整个数据框：
df
它返回：
A B0 1 [[1, 2], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1, 2], [1], [1]]
因为'B'中的元素都指向相同的对象，它们一起被改变。
这就是为什么在生成这种结构时应避免使用[ ] * <number>。而应该使用列表推导式：
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]] for _ in range(df.shape[0])]df# 然后通过以下方式进行更改：df.at[1, 'B'][0] = [1, 2]df
避免在lambda中使用append：
你可以尝试：
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]] for _ in range(df.shape[0])]dfimport copydef myfun(x):l = copy.deepcopy(x)l[0] = l[0] + [2]return ldf.at[1, 'B'] = myfun(df.at[1, 'B'])df
你也可以使用np.where()来获取行索引：
df.at[np.where(df['A'] == 2)[0][0], 'B'] = myfun(df.at[np.where(df['A'] == 2)[0][0], 'B'])
英文:
The problem is the way how you generate your data frame:
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]df
The [[[1],[1],[1]]] * df.shape[0] is a tricky thing in Python.
Because all [[1],[1],[1]] in the different rows of you df point to the
same object, a list [[1], [1], [1]].
Try this:
df.at[1, 'B'][0] = [1, 2]
.at[ ] allows you to access using index number and column name at the same time. the [0] chooses the first element of the list returned after access.
So no lambda needed here.
One thinks that only the first element in the second row's 'B' columns' list
[[1], [1], [1]] is changed to [[1, 2], [1], [1]].
but if you look at the entire data frame:
df
it returns:
A B0 1 [[1, 2], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1, 2], [1], [1]]
Because the elements in B point all to the identical object, they all get mutated at once.
That is why you should avoid [ ] * <number> when generating such constructs.
Instead, use e.g. a list comprehension:
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]] for _ in range(df.shape[0])]df# and mutate by:df.at[1, 'B'][0] = [1, 2]dfA B0 1 [[1], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1], [1], [1]]
Avoid append in your lambda
How about:
import pandas as pddf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]] for _ in range(df.shape[0])]dfimport copydef myfun(x):l = copy.deepcopy(x)l[0] = l[0] + [2]return ldf.at[1, 'B'] = myfun(df.at[1, 'B'])dfA B0 1 [[1], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1], [1], [1]]
One could use np.where() to get row index:
df.at[np.where(df['A'] == 2)[0][0], 'B'] = myfun(df.at[np.where(df['A'] == 2)[0][0], 'B'])
答案4
得分: 0
以下是您提供的代码的翻译部分：
import pandas as pdimport copydf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]i, j = 1, 1l = copy.deepcopy(df.iat[i, j]) # 使用深拷贝以避免指针问题l[0] = [1,2]df.iat[i, j] = lprint(df)
A B0 1 [[1], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1], [1], [1]]
英文:
import pandas as pdimport copydf = pd.DataFrame(data={'A': [1, 2, 3]})df['B'] = [[[1],[1],[1]]] * df.shape[0]i,j = 1,1l = copy.deepcopy(df.iat[i, j]) # Deepcopy to avoid pointer probleml[0] = [1,2]df.iat[i, j] = lprint(df)
A B0 1 [[1], [1], [1]]1 2 [[1, 2], [1], [1]]2 3 [[1], [1], [1]]
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