我的熊猫数据框为:
df3 = pd.DataFrame({
'T': [11.0,22.0,11.23,20.03],
'v2': [11.0,13.0,55.1,33.0],
'v3' : [112.1,2.0,2.1,366.0],
'v4': [np.nan, "blue", 1.0, 2.0]
})
T v2 v3 v4
0 11.00 11.0 112.1 NaN
1 22.00 13.0 2.0 blue
2 11.23 55.1 2.1 1.0
3 20.03 33.0 366.0 2.0
我必须有:
T v2 v3 v4
0 11 11.0 112.1 NaN
1 22 13.0 2.0 blue
2 11.23 55.1 2.1 1.0
3 20.03 33.0 366.0 2.0
所以我只能在'T'上将float转换为整数。
答案 0 :(得分:5)
有可能,但有点hack,因为有必要转换为object
:
df3['T'] = np.array([int(x) if int(x) == x else x for x in df3['T']], dtype=object)
print (df3)
T v2 v3 v4
0 11 11.0 112.1 NaN
1 22 13.0 2.0 blue
2 11.23 55.1 2.1 1
3 20.03 33.0 366.0 2
print (df3['T'].tolist())
[11, 22, 11.23, 20.03]
如果可能缺少值:
df3 = pd.DataFrame({
'T': [11.0,22.0,11.23,np.nan],
'v2': [11.0,13.0,55.1,33.0],
'v3' : [112.1,2.0,2.1,366.0],
'v4': [np.nan, "blue", 1.0, 2.0]
})
df3['T'] = np.array([int(x) if x % 1 == 0 else x for x in df3['T']], dtype=object)
print (df3)
T v2 v3 v4
0 11 11.0 112.1 NaN
1 22 13.0 2.0 blue
2 11.23 55.1 2.1 1
3 NaN 33.0 366.0 2
print (df3['T'].tolist())
[11, 22, 11.23, nan]
答案 1 :(得分:0)
使用与@jezrael相同的想法,但使用is_integer:
import numpy as np
import pandas as pd
df3 = pd.DataFrame({
'T': [11.0, 22.0, 11.23, 20.03],
'v2': [11.0, 13.0, 55.1, 33.0],
'v3': [112.1, 2.0, 2.1, 366.0],
'v4': [np.nan, "blue", 1.0, 2.0]
})
df3['T'] = np.array([int(x) if float(x).is_integer() else x for x in df3['T']], dtype=object)
print(df3)
输出
T v2 v3 v4
0 11 11.0 112.1 NaN
1 22 13.0 2.0 blue
2 11.23 55.1 2.1 1
3 20.03 33.0 366.0 2
或将numpy.where与numpy.fmod一起使用:
mask = np.fmod(df3['T'].values, 1) == 0
df3['T'] = np.where(mask, df3['T'].values.astype(np.int), df3['T']).astype(dtype=object)
print(df3)
答案 2 :(得分:0)
或者为什么不呢?
df3=df3.apply(lambda x: int(x) if int(x)==x and x==x and isinstance(x,float) else x)
现在:
print(df3)
将被期望输出:
T v2 v3 v4
0 11 11.0 112.1 NaN
1 22 13.0 2.0 blue
2 11.23 55.1 2.1 1.0
3 20.03 33.0 366.0 2.0