我有一个熊猫数据框,其中有16,777,216行。这是介于0和255之间的三列(红色,绿色和蓝色)的所有可能组合。
我想在此数据框中添加一列,这是该行三个值的十六进制代码。我认为以下内容将是最佳解决方案:
df["Hex"] = "#{0:02x}{1:02x}{2:02x}".format(df["Red"],df["Green"],df["Blue"])
但是,您似乎无法将一系列传递给字符串格式方法。
是否有解决此问题的方法?此外,鉴于数据帧相当大,那将是最有效的方法吗?
答案 0 :(得分:2)
您可以使用.apply
,例如:
df = pd.DataFrame(np.random.randint(256, size=(10, 3)), columns=['Red', 'Green', 'Blue'])
例如:
Red Green Blue
0 125 100 174
1 107 247 235
2 230 254 33
3 91 107 33
4 209 220 232
5 175 10 47
6 120 66 44
7 21 136 254
8 226 237 32
9 89 57 71
然后:
df.apply('#{Red:02X}{Green:02X}{Blue:02X}'.format_map, axis=1)
给你:
0 #7D64AE
1 #6BF7EB
2 #E6FE21
3 #5B6B21
4 #D1DCE8
5 #AF0A2F
6 #78422C
7 #1588FE
8 #E2ED20
9 #593947
dtype: object
答案 1 :(得分:1)
对于python 3.6+
,可以使用非常快的f-string
s:
z = zip(df['Red'], df['Blue'], df['Green'])
df["Hex"] = [f'#{R:02X}{B:02X}{G:02X}' for R,B,G in z]
对于较低版本:
df["Hex"] = ['#{0:02X}{1:02X}{2:02X}'.format(R,B,G) for R,B,G in z]
感谢@Jon改进解决方案:
df["Hex"] = ['#{0:02X}{1:02X}{2:02X}'.format(*el) for el in z]
性能:
#10000 rows
df = pd.DataFrame(np.random.randint(256, size=(10000, 3)), columns=['Red', 'Green', 'Blue'])
In [244]: %%timeit
...: z = zip(df['Red'], df['Green'], df['Blue'])
...: df["Hex"] = [f'#{R:02X}{B:02X}{G:02X}' for R,B,G in z]
...:
12.9 ms ± 45.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [245]: %%timeit
...: z = zip(df['Red'], df['Green'], df['Blue'])
...: df["Hex"] = ['#{0:02X}{1:02X}{2:02X}'.format(R,B,G) for R,B,G in z]
...:
12.4 ms ± 1.14 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [246]: %%timeit
...: z = zip(df['Red'], df['Green'], df['Blue'])
...: df["Hex"] = ['#{0:02X}{1:02X}{2:02X}'.format(*el) for el in z]
...:
11.3 ms ± 55 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [246]: %%timeit
...: df["Hex"] = df.apply('#{Red:02X}{Green:02X}{Blue:02X}'.format_map, axis=1)
...:
346 ms ± 42.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)