pandas 将我的列转换为 2 个小数位
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Converting my column to 2 decimal places
提问by Omer Qureshi
I have a dataset:
我有一个数据集：
df = pd.read_excel('/Users/Adeel/Desktop/ECON628-01-omerqureshi84/datasets/main-data.xlsx')
It has columns with names such as "lerate" which is the log of the exchange rates for countries. It's in 5 decimal places and and I'm trying to convert it to 2 decimal places.
它具有名称为“lerate”的列，它是各国汇率的日志。它有 5 个小数位，我正在尝试将其转换为 2 个小数位。
I'm using the following code:
我正在使用以下代码：
for i in range(len(df.lerate)):
print ("%.2f" % df.lerate[i])
It's giving me an error.
它给了我一个错误。
Can anyone help? I don't know what's wrong here.
任何人都可以帮忙吗？我不知道这里出了什么问题。
回答by mechanical_meat
You can use round:
您可以使用round：
df.lerate = df.lerate.round(2)
Example:
例子：
>>> df = pd.DataFrame(np.random.random([3, 3]),
columns=['A', 'B', 'C'], index=['first', 'second', 'third'])
>>> df.A = df.A.round(2)
>>> df
A B C
first 0.82 0.581855 0.548373
second 0.21 0.536690 0.986906
third 0.78 0.100343 0.576521
