Python 所选行和列的 Pandas min()
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Pandas min() of selected row and columns
提问by yash.trojan.25
I am trying to create a column which contains only the minimum of the one row and a few columns, for example:
我正在尝试创建一个仅包含一行和几列中的最小值的列，例如：
A0 A1 A2 B0 B1 B2 C0 C1
0 0.84 0.47 0.55 0.46 0.76 0.42 0.24 0.75
1 0.43 0.47 0.93 0.39 0.58 0.83 0.35 0.39
2 0.12 0.17 0.35 0.00 0.19 0.22 0.93 0.73
3 0.95 0.56 0.84 0.74 0.52 0.51 0.28 0.03
4 0.73 0.19 0.88 0.51 0.73 0.69 0.74 0.61
5 0.18 0.46 0.62 0.84 0.68 0.17 0.02 0.53
6 0.38 0.55 0.80 0.87 0.01 0.88 0.56 0.72
Here I am trying to create a column which contains the minimum for each row of columns B0, B1, B2.
在这里，我试图创建一个列，其中包含 B0、B1、B2 列的每一行的最小值。
The output would look like this:
输出将如下所示：
A0 A1 A2 B0 B1 B2 C0 C1 Minimum
0 0.84 0.47 0.55 0.46 0.76 0.42 0.24 0.75 0.42
1 0.43 0.47 0.93 0.39 0.58 0.83 0.35 0.39 0.39
2 0.12 0.17 0.35 0.00 0.19 0.22 0.93 0.73 0.00
3 0.95 0.56 0.84 0.74 0.52 0.51 0.28 0.03 0.51
4 0.73 0.19 0.88 0.51 0.73 0.69 0.74 0.61 0.51
5 0.18 0.46 0.62 0.84 0.68 0.17 0.02 0.53 0.17
6 0.38 0.55 0.80 0.87 0.01 0.88 0.56 0.72 0.01
Here is part of the code, but it is not doing what I want it to do:
这是代码的一部分，但它没有做我想要它做的事情：
for i in range(0,2):
df['Minimum'] = df.loc[0,'B'+str(i)].min()
回答by Marius
This is a one-liner, you just need to use the axisargument for minto tell it to work across the columns rather than down:
这是一个单行，您只需要使用axisfor 参数min告诉它跨列而不是向下工作：
df['Minimum'] = df.loc[:, ['B0', 'B1', 'B2']].min(axis=1)
If you need to use this solution for different numbers of columns, you can use a for loop or list comprehension to construct the list of columns:
如果您需要对不同数量的列使用此解决方案，您可以使用 for 循环或列表推导来构建列列表：
n_columns = 2
cols_to_use = ['B' + str(i) for i in range(n_columns)]
df['Minimum'] = df.loc[:, cols_to_use].min(axis=1)
回答by DenPO
For my tasks a universal and flexible approach is the following example:
对于我的任务，通用且灵活的方法是以下示例：
df['Minimum'] = df[['B0', 'B1', 'B2']].apply(lambda x: min(x[0],x[1],x[2]), axis=1)
The target column 'Minimum' is assigned the result of the lambda function based on the selected DF columns['B0', 'B1', 'B2']. Access elements in a function through the function alias and his new Index(if count of elements is more then one). Be sure to specify axis=1, which indicates line-by-line calculations. This is very convenient when you need to make complex calculations. However, I assume that such a solution may be inferior in speed.
根据选定的 DF 列 ['B0', 'B1', 'B2'] 为目标列 'Minimum' 分配 lambda 函数的结果。通过函数别名和他的新索引访问函数中的元素（如果元素数大于一）。一定要指定axis=1，表示逐行计算。当您需要进行复杂的计算时，这非常方便。但是，我认为这样的解决方案在速度上可能较差。
As for the selection of columns, in addition to the 'for' method, I can suggest using a filter like this:
至于列的选择，除了 'for' 方法，我可以建议使用这样的过滤器：
calls_to_use = list(filter(lambda f:'B' in f, df.columns))
literally, a filter is applied to the list of DF columns through a lambda function that checks for the occurrence of the letter 'B'.
从字面上看，过滤器通过检查字母“B”出现的 lambda 函数应用于 DF 列的列表。
after that the first example can be written as follows:
之后第一个例子可以写成如下：
calls_to_use = list(filter(lambda f:'B' in f, df.columns))
df['Minimum'] = df[calls_to_use].apply(lambda x: min(x), axis=1)
although after pre-selecting the columns, it would be preferable:
尽管在预先选择了列之后，最好是：
df['Minimum'] = df[calls_to_use].min(axis=1)
