相当于Python中的R group_by()+ rleid() [英] R group_by() + rleid() equivalent in Python
问题描述
我在Python中有以下数据框:
I've got a following data frame in Python:
df = pd.DataFrame.from_dict({'measurement_id': np.repeat([1, 2], [6, 6]),
'min': np.concatenate([np.repeat([1, 2, 3], [2, 2, 2]),
np.repeat([1, 2, 3], [2, 2, 2])]),
'obj': list('AB' * 6),
'var': [1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1]})
首先，在由object定义的每个组中，我想将ID分配给measurement_id和var列的唯一运行.如果这些列的任何值发生更改，它将启动应指定新ID的新运行.
First, within each group defined by object, I'd like to assign id to unique run of measurement_id and var columns. If any value of those columns changes, it starts new run that should be assigned with new id. So the
df['rleid_output'] = [1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 3]
然后，对于由rleid_output定义的每个组，我想检查运行持续了多少分钟(min列)，给了我expected_output列:
Then, for each group defined by rleid_output I'd like to check how many minutes (min column) the run lasted giving me expected_output column:
df['expected_output'] = [2, 2, 2, 2, 1, 1, 2, 3, 2, 3, 1, 3]
如果是R，我将按照以下步骤操作:
If it was R, I'd proceed as follows:
df <- data.frame(measurement_id = rep(1:2, each = 6),
min = rep(rep(1:3, each = 2), 2),
object = rep(LETTERS[1:2], 6),
var = c(1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1))
df %>%
group_by(object) %>%
mutate(rleid = data.table::rleid(measurement_id, var)) %>%
group_by(object, rleid) %>%
mutate(expected_output = last(min) - first(min) + 1)
所以我需要的主要是与Python pd.DataFrame.groupby子句一起工作的R data.table::rleid等效项.有什么想法可以解决这个问题吗?
So the main thing I need is R data.table::rleid equivalent that would work with Python pd.DataFrame.groupby clause. Any ideas how to solve this?
@Edit:数据框的新的更新示例:
@ new, updated example of data frame:
df = pd.DataFrame.from_dict({'measurement_id': np.repeat([1, 2], [6, 6]),
'min': np.concatenate([np.repeat([1, 2, 3], [2, 2, 2]),
np.repeat([1, 2, 3], [2, 2, 2])]),
'obj': list('AB' * 6),
'var': [1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1]})
df['rleid_output'] = [1, 1, 2, 1, 3, 2, 4, 3, 4, 3, 5, 3]
df['expected_output'] = [1, 2, 1, 2, 1, 1, 2, 3, 2, 3, 1, 3]
推荐答案
更新后的答案
问题在于，每个measurement_id, obj, var组中的min列都应保持顺序.我们可以在measurement_id, obj, var上按组进行检查，然后检查min列中的差异是否大于1.如果是这样，我们会在expected_output中将其标记为唯一的持续时间:
The problem is that the min column in each group of measurement_id, obj, var should be maintained order. We can check this by group by on measurement_id, obj, var and then checking if the difference in min column is greater than 1. If so, we mark it as a unique duration in expected_output:
df['grouper'] = (df.groupby(['measurement_id', 'obj', 'var'])['min']
.apply(lambda x: x.diff().fillna(1).eq(1))
)
df['expected_output'] = (
df.groupby(['measurement_id', 'obj', 'var'])['grouper'].transform('sum').astype(int)
)
df = df.drop(columns='grouper')
measurement_id min obj var expected_output
0 1 1 A 1 1
1 1 1 B 2 2
2 1 2 A 2 1
3 1 2 B 2 2
4 1 3 A 1 1
5 1 3 B 1 1
6 2 1 A 2 2
7 2 1 B 1 3
8 2 2 A 2 2
9 2 2 B 1 3
10 2 3 A 1 1
11 2 3 B 1 3
遵循OP的逻辑的旧答案
我们可以通过使用GroupBy.diff来获取您的rleid_output，基本上是每次measurement_id每次更改var时唯一的标识符. obj
We can achieve this by using GroupBy.diff to get your rleid_output, basically a unique identifier each time var changes for each measurement_id& obj
之后，使用GroupBy.nunique测量minutes的量:
rleid_output = df.groupby(['measurement_id', 'obj'])['var'].diff().abs().bfill()
df['expected_output'] = (df.groupby(['measurement_id', 'obj', rleid_output])['min']
.transform('nunique'))
measurement_id min obj var expected_output
0 1 1 A 1 2
1 1 1 B 2 2
2 1 2 A 1 2
3 1 2 B 2 2
4 1 3 A 2 1
5 1 3 B 1 1
6 2 1 A 2 2
7 2 1 B 1 3
8 2 2 A 2 2
9 2 2 B 1 3
10 2 3 A 1 1
11 2 3 B 1 3
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