使用Python中的奇数创建行计数
这里必须有一个简单的解决方案。我知道如何使用cumcount,但我希望它数数奇数。例如,我有以下DF。搜索非常毫无用处,因为它们都在计算“奇数”数字。
letters = [ "A", "A", "A", "B", "B", "B"]
df = pd.DataFrame(letters, columns=["letter"])
df['cumcount'] = df.groupby('letter').cumcount() + 1
df =
letter cumcount
A 1
A 2
A 3
B 1
B 2
B 3
我想做的就是输出这一点。
df =
letter cumcount odd_count
A 1 1
A 2 3
A 3 5
B 1 1
B 2 3
B 3 5
There has to be a simple solution here. I know how to use cumcount but I want it to count in odd numbers. For example I have the following DF. Searches have been pretty useless as they all pull up counting the "odd" numbers.
letters = [ "A", "A", "A", "B", "B", "B"]
df = pd.DataFrame(letters, columns=["letter"])
df['cumcount'] = df.groupby('letter').cumcount() + 1
df =
letter cumcount
A 1
A 2
A 3
B 1
B 2
B 3
What I'm looking to do is have the output be this.
df =
letter cumcount odd_count
A 1 1
A 2 3
A 3 5
B 1 1
B 2 3
B 3 5
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让我们做一些数学:
Lets do some math: