# 获取一列的[0, x]元素的最小值[英] Get the min of [0, x] element wise for a column

### 问题描述

I need to compute a column where the value is the result of a vectorized operation over other columns:

```df["new_col"] = df["col1"] - min(0,df["col2"])
```

It turned out, however, that I cannot use min as in the above syntax. So, what is the right way to get the min between zero and a given value of pandas column?

## 推荐答案

you can use numpy.minimum to find the element-wise minimum of an array

```import numpy as np
df["new_col"] = df["col1"] - np.minimum(0,df["col2"])
```

## 其他推荐答案

You could use some masking and a temporary column. Totally ignoring the 'min' function.

```magicnumber = 0
tempcol = df['col2']
df['col1'] - tempcol
```

Or you can use a lambda function:

```magicnumber = 0
df['col1'] - df['col2'].apply(lambda x: np.min(magicnumber, x))
```

OR you can apply over two columns:

```df['magicnumber'] = 0
df['col1'] - df[['col2', 'magicnumber']].apply(np.min, axis=1)
```

## 其他推荐答案

I think that the other answers aren't what you meant. They take the minimum value in df['col2'] and compare it to 0 (and thus always return the same value), while you wanted the minimum between each value in col2 and 0:

```df = pd.DataFrame(data={'a': [2, 3], 'b': [-1, 1]})

df['new_col'] = map(lambda a, b: a - min(0, b), df['a'], df['b'])

print df

>>    a  b  new_col
0  2 -1        3
1  3  1        3
```