问题描述
我有一个 DF(从会计软件收集)，看起来像这样.
串行||日期 ||详情 ||价格-------------------------------1 ||第0308章安德鲁||1002 ||南||手套 ||NaN3 ||第0408章约翰逊||504 ||南||检票口 ||NaN我想合并连续的 2 行并创建一个新列产品"，第二行详细信息"值.预期的输出应该类似于 ---
串行||日期 ||详情 ||价格 ||产品-------------------------------------------------1 ||第0308章安德鲁||100 ||手套3 ||第0408章约翰逊||50 ||便门如何使用 Pandas 实现这一点?
这些答案基于数据帧的格式，总是呈现遵循 OP 呈现的相同模式的行对.第一行显示一个人，第二行显示一个产品和日期，价格列是 NaN.
使用shift然后dropna
df.assign(Product=df.Particulars.shift(-1)).dropna()系列 日期 详情 价格 产品0 1 308.0 安德鲁 100.0 手套2 3 408.0 约翰逊 50.0 检票口加入
完全相同但不同的东西
df.join(df.Particulars.shift(-1).rename('Product')).dropna()详情
每个请求
df.Particulars.shift(-1)将 Particulars 列的所有成员返回一行0 手套1 约翰逊2 检票口3 南名称:详细信息，dtype:对象当我将其分配给现有数据框
df.assign(Product=df.Particulars.shift(-1))时，它会添加一个具有新名称'Product 的列'此处的值是移动的详细信息.系列 日期 详情 价格 产品0 1 308.0 安德鲁 100.0 手套1 2 NaN 手套 NaN Johnson2 3 408.0 约翰逊 50.0 检票口3 4 NaN 检票口 NaN NaN剩下的就是删除带有
NaN值的行，我们就有了上面显示的内容.
灵感来自 @QuangHoang 的回答
如果我每隔一行切片，我就不需要依赖 dropna
df.assign(Product=df.Particulars.shift(-1))[::2]或者更简洁
df[::2].assign(Product=[*df.Particulars[1::2]])一种方法
这是我想到的第一种方式，而且很恶心
i = np.flatnonzero(df.Price.notna())j = i + 1df.iloc[i].assign(Product=df.iloc[j].Particulars.values)系列 日期 详情 价格 产品0 1 308.0 安德鲁 100.0 手套2 3 408.0 约翰逊 50.0 检票口I have a DF(collected from an accounting software) which looks like this.
Serial || Date || Particulars || Price
--------------------------------------
1 || 0308 || Andrew || 100
2 || NaN || Gloves || NaN
3 || 0408 || Johnson || 50
4 || NaN || Wicket || NaN
I want to merge the 2 consecutive rows and make a new column 'Product' with 2nd rows 'Particulars' value.The expected output should look like ---
Serial || Date || Particulars || Price || Product
-------------------------------------------------
1 || 0308 || Andrew || 100 || Gloves
3 || 0408 || Johnson || 50 || Wicket
How do I achieve this with pandas?
These answers are predicated on the format of the dataframe always presenting pairs of rows that follow the same pattern presented by OP. First row shows a person, second row shows a product and date, price columns are NaN.
Use shift then dropna
df.assign(Product=df.Particulars.shift(-1)).dropna()
Serial Date Particulars Price Product
0 1 308.0 Andrew 100.0 Gloves
2 3 408.0 Johnson 50.0 Wicket
join
Same exact thing but different
df.join(df.Particulars.shift(-1).rename('Product')).dropna()
Details
Per Request
df.Particulars.shift(-1)brings all members of the Particulars column back one row0 Gloves 1 Johnson 2 Wicket 3 NaN Name: Particulars, dtype: objectWhen I assign this to the existing dataframe
df.assign(Product=df.Particulars.shift(-1))it adds a column with a new name'Product'where the values are the shifted Particulars.Serial Date Particulars Price Product 0 1 308.0 Andrew 100.0 Gloves 1 2 NaN Gloves NaN Johnson 2 3 408.0 Johnson 50.0 Wicket 3 4 NaN Wicket NaN NaNAll that's left is to drop the rows withe the
NaNvalues and we have what is presented above.
Inspired by @QuangHoang's answer
I don't need to depend on dropna if I slice every other row
df.assign(Product=df.Particulars.shift(-1))[::2]
Or even more terse
df[::2].assign(Product=[*df.Particulars[1::2]])
One way to do it
This was the first way I thought of and it's gross
i = np.flatnonzero(df.Price.notna())
j = i + 1
df.iloc[i].assign(Product=df.iloc[j].Particulars.values)
Serial Date Particulars Price Product
0 1 308.0 Andrew 100.0 Gloves
2 3 408.0 Johnson 50.0 Wicket
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