我有 df_surf_dist=
index surface_relle_bati
0 (8.718, 27.733] 442
1 (27.733, 46.467] 1485
2 (46.467, 65.2] 1296
3 (65.2, 83.933] 927
4 (83.933, 102.667] 288
5 (102.667, 121.4] 81
6 (121.4, 140.133] 22
7 (140.133, 158.867] 8
8 (158.867, 177.6] 10
9 (177.6, 196.333] 1
10 (196.333, 215.067] 1
11 (215.067, 233.8] 1
12 (233.8, 252.533] 0
13 (252.533, 271.267] 0
14 (271.267, 290.0] 1
我想对范围进行舍入，以便绘制它并使比例看起来不错。
我尝试了df_surf_dist.index.round(decimals=-1)
但是我得到了AttributeError:'RangeIndex'对象没有属性'round'
我还尝试了df_surf_dist.index = df_surf_dist.index.to_series().apply(lambda x: x.round(1))
但是我得到了AttributeError:'int'对象没有属性'round'
最佳答案
您可以使用自定义函数重新创建间隔:
def round_interval(i, ndigits=2):
return pd.Interval(round(i.left, ndigits), round(i.right, ndigits), i.closed)
df['range'].apply(round_interval, ndigits=1)
输出:
0 (8.7, 27.7]
1 (27.7, 46.5]
2 (46.5, 65.2]
Name: range, dtype: interval
使用的输入:
from pandas import Interval
d = {'range': {0: Interval(8.718, 27.733, closed='right'),
1: Interval(27.733, 46.467, closed='right'),
2: Interval(46.467, 65.2, closed='right')},
'surface_relle_bati': {0: 442, 1: 1485, 2: 1296}}
df = pd.DataFrame(d)
关于python - 围绕 pandas 的范围，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/72290949/