如何计算连续减少（增加）的数量

Question

如何计算连续减少（增加）的数量

我有一个数据帧，它有500K行和12列，每个月包括开始和结束月份。每列代表一个月。我需要比较范围（startMonth，endmonth）中的每一行，I-th月和（I+1）-th个月。（Ps：范围不是恒定的。每个行有不同的范围大小。）

条件：如果start month>；end month，我应该看到“Neg99=-999”

以下是我的示例数据：

import pandas as pd
import numpy as np

idx = [1001,1002,1003,1004,1005,1006,1007,1008,1009,1010,1011,1012,1013,1014,1015,1016,1017,1018]
data = {'M_1': [3, 1, 0, 0, 1, 0, 1, 1, 1, 0, 6, 6, 6,0,0,2,0,2],
        'M_2': [2, 2, 3, 1, 1, 0, 1, 2, 0, 1, 5, 5, 5,1,1,1,1,2],
        'M_3': [1, 3, 2, 2, 1, 0, 1, 2, 1, 0, 4, 4, 4,1,1,0,2,2],
        'M_4': [0, 4, 1, 3, 1, 0, 1, 2, 0, 1, 3, 0, 3,1,1,0,0,0],
        'M_5': [1, 0, 0, 4, 2, 0, 1, 3, 1, 0, 2, 1, 2,1,1,0,0,0],
        'M_6': [2, 0, 0, 0, 3, 0, 1, 3, 0, 1, 1, 2, 1,1,1,0,0,0],
        'M_7': [3, 0, 0, 0, 2, 0, 1, 2, 1, 0, 0, 3, 0,0,1,0,0,0],
        'M_8': [0, 1, 0, 0, 2, 0, 1, 2, 0, 1, 1, 1, 1,0,0,0,0,0],
        'M_9': [0, 2, 0, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0,0,0,0,0,0],
        'M_10': [0, 3, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0,0,0,0,0,0],
        'M_11': [0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0,0,0,0,0,0],
        'M_12': [0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0,0,0,0,0,0]}



startMonth = pd.DataFrame([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 5,1,1,1,1,1],
                          columns=['start'],index=idx)
endMonth = pd.DataFrame([12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 2,12,12,2,2,2],
                        columns=['end'], index=idx)

df1 = pd.DataFrame(data, index=idx)
Neg99 = -999

我为日期范围写了bool数组

^{pr2}$

我需要找出每一行的连续减少和增加。在

• 这里是减少代码

# Consecutive Decreases
decr = (np.diff(np.hstack((masked.values, np.zeros((masked.values.shape[0], 1)))), axis=1) > 0).argmin(axis=1)

final_decr = pd.DataFrame(decr,
                      index=idx, columns=['decr'])


final_decr.decr= np.select( condlist = [startMonth.start > endMonth.end],
                           choicelist = [Neg99],
                           default = final_decr.decr)

• 这里是增量代码

incr = (np.diff(np.hstack((masked.values, np.zeros((masked.values.shape[0], 1)))), axis=1) < 0).argmin(axis=1)

final_incr = pd.DataFrame(incr,
                      index=idx, columns=['incr'])
final_incr.incr= np.select( condlist = [startMonth.start > endMonth.end],
                           choicelist = [Neg99],
                           default = final_incr.incr)

最后，我的预期产出是：

Final increase table (.csv);
idx,my_results,expected_result
1001,0,0
1002,3,3
1003,1,1
1004,4,4
1005,0,0
1006,0,0
1007,0,0
1008,1,1
1009,0,0
1010,1,1
1011,0,3
1012,0,0
1013,-999,-999
1014,5,1
1015,6,1
1016,0,0
1017,0,2
1018,0,0

Final decrease table (.csv);
idx,my_result,expected_result
1001,3,3
1002,0,0
1003,0,0
1004,0,0
1005,0,0
1006,0,0
1007,0,0
1008,0,0
1009,1,1
1010,0,0
1011,0,0
1012,0,3
1013,-999,-999
1014,0,0
1015,0,0
1016,0,2
1017,0,0
1018,0,0

Final NoChange table (.csv);
idx,my_result,expected_result
1001,0,0
1002,0,0
1003,0,0
1004,0,0
1005,3,3
1006,11,11
1007,11,11
1008,0,0
1009,0,0
1010,0,0
1011,0,0
1012,0,0
1013,-999,-999
1014,0,0
1015,0,0
1016,0,0
1017,0,0
1018,2,0

Thanks for your advice!

Answer 1

因此，我认为不使用argmin，您可以在重命名列之后使用^{}来获得位置的整数值。然后删除startMonth的值，例如：

incr = (df1.rename(columns={col:int(col.split('_')[1]) for col in masked.columns})
           .diff(-1, axis=1) < 0).mask(~arr_bool).idxmin(axis=1) - startMonth.start
decr = (df1.rename(columns={col:int(col.split('_')[1]) for col in masked.columns})
           .diff(-1, axis=1) > 0).mask(~arr_bool).idxmin(axis=1) - startMonth.start

然后你可以像你一样做np.select，或者也许只要一个.illna(-999)就足够了，因为我认为这个解决方案在任何地方都有nan，结果就是满足你的条件startMonth.start > endMonth.end