我的数据框中有一个列，它的值在2100到8000之间。我想把这一列分成几列，每列间隔500。所以让我举个例子：
column
2100
2105
2119
.
8000
我想这样分开：
column1 column2 column3 . . column n
2100 0 0 . . 0
. 0 . . . 0
2600 0 0
2601 0 . . .
. .
3101 0
3102 0
.
3602
8000
请提出解决办法。
最佳答案
以下是一种使用pd.cut和DataFrame.pivot的方法：
df = pd.DataFrame(list(range(2100, 8000+1)), columns=['column'])
# create the bins to be used in pd.cut
bins = list(range(df.column.min(), df.column.max()+50, 50))
# array([2100, 2150, 2200, 2250, 2300 ...
# Create the labels for pd.cut, which will be used as column names
labels = [f'column{i}' for i in range(len(bins)-1)]
# ['column0', 'column1', 'column2', 'column3', 'column4', ...
df['bins'] = pd.cut(df.column, bins, labels=labels, include_lowest=True)
它会给你：
column bins
0 2100 column0
1 2101 column0
2 2102 column0
3 2103 column0
4 2104 column0
5 2105 column0
6 2106 column0
7 2107 column0
8 2108 column0
现在使用
pivot获得最终结果：ix = df.groupby('bins').column.cumcount()
df.pivot(columns = 'bins', index=ix).fillna(0)
bins column0 column1 column2 column3 column4 column5 column6 column7 column8 ...
0 2100.0 2151.0 2201.0 2251.0 2301.0 2351.0 2401.0 2451.0 2501.0
1 2101.0 2152.0 2202.0 2252.0 2302.0 2352.0 2402.0 2452.0 2502.0
2 2102.0 2153.0 2203.0 2253.0 2303.0 2353.0 2403.0 2453.0 2503.0
3 2103.0 2154.0 2204.0 2254.0 2304.0 2354.0 2404.0 2454.0 2504.0
4 2104.0 2155.0 2205.0 2255.0 2305.0 2355.0 2405.0 2455.0 2505.0
5 2105.0 2156.0 2206.0 2256.0 2306.0 2356.0 2406.0 2456.0 2506.0
6 2106.0 2157.0 2207.0 2257.0 2307.0 2357.0 2407.0 2457.0 2507.0
7 2107.0 2158.0 2208.0 2258.0 2308.0 2358.0 2408.0 2458.0 2508.0
8 2108.0 2159.0 2209.0 2259.0 2309.0 2359.0 2409.0 2459.0 2509.0
9 2109.0 2160.0 2210.0 2260.0 2310.0 2360.0 2410.0 2460.0 2510.0
10 2110.0 2161.0 2211.0 2261.0 2311.0 2361.0 2411.0 2461.0 2511.0
...
让我们将其全部封装在一个函数中，并尝试使用一个更简单的示例来更好地了解其工作原理：
def binning_and_pivot(df, bin_size):
bins = list(range(df.column.min(), df.column.max()+bin_size, bin_size))
labels = [f'column{i}' for i in range(len(bins)-1)]
df['bins'] = pd.cut(df.column, bins, labels=labels, include_lowest=True)
ix = df.groupby('bins').column.cumcount()
return df.pivot(columns = 'bins', index=ix).fillna(0)
df = pd.DataFrame(list(range(100+1)), columns=['column'])
df = df.sample(frac=0.7).reset_index(drop=True)
binning_and_pivot(df, bin_size=10)
bins column0 column1 column2 column3 column4 column5 column6 column7 column8
0 2.0 16.0 32.0 39.0 45.0 55.0 69.0 81.0 87.0
1 6.0 21.0 29.0 42.0 46.0 59.0 72.0 76.0 92.0
2 3.0 13.0 31.0 36.0 49.0 61.0 68.0 74.0 91.0
3 12.0 20.0 25.0 41.0 52.0 56.0 70.0 78.0 86.0
4 8.0 17.0 30.0 37.0 43.0 62.0 64.0 73.0 89.0
5 7.0 19.0 27.0 38.0 50.0 53.0 71.0 77.0 83.0
6 0.0 22.0 28.0 0.0 0.0 54.0 65.0 82.0 90.0
7 0.0 18.0 24.0 0.0 0.0 60.0 63.0 80.0 0.0
8 0.0 14.0 26.0 0.0 0.0 0.0 0.0 75.0 0.0
bins column9
0 95.0
1 100.0
2 96.0
3 0.0
4 0.0
5 0.0
6 0.0
7 0.0
8 0.0
关于python - Python根据范围将一列分为多列，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/56020778/