擅长:python、mysql、java
<p>我首先将<code>Knownvalue</code>系列放入一个整数列表中,这个整数等于它的截断值除以10(例如27.87//10=2)。这些桶表示所需列位置的整数。因为<code>Knownvalue</code>在第一列中,所以我在这些值中添加了一个。在</p>
<p>接下来,我通过这些bin值进行枚举,这有效地为我提供了行和列整数索引的元组对。我使用<code>iat</code>将这些位置的值设置为1。在</p>
<pre><code>import pandas as pd
import numpy as np
# Create some sample data.
df_vals = pd.DataFrame({'Knownvalue': np.random.random(5) * 50})
df = pd.concat([df_vals, pd.DataFrame(np.zeros((5, 5)), columns=list('ABCDE'))], axis=1)
# Create desired column locations based on the `Knownvalue`.
bins = (df.Knownvalue // 10).astype('int').tolist()
>>> bins
[4, 3, 0, 1, 0]
# Set these locations equal to 1.
for idx, col in enumerate(bins):
df.iat[idx, col + 1] = 1 # The first column is the `Knownvalue`, hence col + 1
>>> df
Knownvalue A B C D E
0 47.353937 0 0 0 0 1
1 37.460338 0 0 0 1 0
2 3.797964 1 0 0 0 0
3 18.323131 0 1 0 0 0
4 7.927030 1 0 0 0 0
</code></pre>