只有当当前单元格中的条件匹配时，我才需要添加具有前几行的序列。这是数据帧：
import pandas as pd
data = {'col1': [1, 2, 1, 0, 0, 0, 0, 3, 2, 2, 0, 0]}
df = pd.DataFrame(data, columns=['col1'])
df['continuous'] = df.col1
print(df)
如果是value > 0，则我需要+1具有先前和的单元格，否则为-1。所以，我期待的结果是；
col1 continuous
0 1 1//+1 as its non-zero
1 2 2//+1 as its non-zero
2 1 3//+1 as its non-zero
3 0 2//-1 as its zero
4 0 1
5 0 0
6 0 0// not to go less than 0
7 3 1
8 2 2
9 2 3
10 0 2
11 0 1
情况2：在我想要>0的地方，我需要<-0.1
data = {'col1': [-0.097112634,-0.092674324,-0.089176841,-0.087302284,-0.087351866,-0.089226185,-0.092242213,-0.096446987,-0.101620036,-0.105940337,-0.109484752,-0.113515648,-0.117848816,-0.121133266,-0.123824577,-0.126030136,-0.126630895,-0.126015218,-0.124235003,-0.122715224,-0.121746573,-0.120794916,-0.120291174,-0.120323152,-0.12053229,-0.121491186,-0.122625851,-0.123819704,-0.125751858,-0.127676591,-0.129339428,-0.132342431,-0.137119556,-0.142040092,-0.14837848,-0.15439201,-0.159282645,-0.161271982,-0.162377701,-0.162838307,-0.163204393,-0.164095634,-0.165496071,-0.167224488,-0.167057078,-0.165706164,-0.163301617,-0.161423938,-0.158669389,-0.156508912,-0.15508329,-0.15365104,-0.151958972,-0.150317528,-0.149234892,-0.148259354,-0.14737422,-0.145958527,-0.144633388,-0.143120273,-0.14145652,-0.139930163,-0.138774126,-0.136710524,-0.134692221,-0.132534879,-0.129921444,-0.127974949,-0.128294058,-0.129241763,-0.132263506,-0.137828981,-0.145549768,-0.154244588,-0.163125109,-0.171814857,-0.179911465,-0.186223859,-0.190653162,-0.194761064,-0.197988536,-0.200500606,-0.20260121,-0.204797089,-0.208281065,-0.211846904,-0.215312626,-0.218696339,-0.221489975,-0.221375209,-0.220996031,-0.218558429,-0.215936558,-0.213933531,-0.21242896,-0.209682125,-0.208196607,-0.206243585,-0.202190476,-0.19913106,-0.19703291,-0.194244664,-0.189609518,-0.186600526,-0.18160171,-0.175875689,-0.170767095,-0.167453329,-0.163516985,-0.161168703,-0.158197984,-0.156378046,-0.154794499,-0.153236804,-0.15187487,-0.151623385,-0.150628282,-0.149039072,-0.14826268,-0.147535739,-0.145557646,-0.142223729,-0.139343068,-0.135355686,-0.13047743,-0.125999173,-0.12218752,-0.117021996,-0.111542982,-0.106409901,-0.101904095,-0.097910825,-0.094683375,-0.092079967,-0.088953862,-0.086268097,-0.082907394,-0.080723466,-0.078117426,-0.075431993,-0.072079536,-0.068962411,-0.064831759,-0.061257701,-0.05830671,-0.053889968,-0.048972414,-0.044763431,-0.042162829,-0.039328369,-0.038968862,-0.040450835,-0.041974942,-0.042161609,-0.04280523,-0.042702428,-0.042593856,-0.043166561,-0.043691795,-0.044093492,-0.043965231,-0.04263305,-0.040836102,-0.039605133,-0.037204273,-0.034368645,-0.032293737,-0.029037983,-0.025509509,-0.022704668,-0.021346266,-0.019881524,-0.018675734,-0.017509566,-0.017148129,-0.016671088,-0.016015011,-0.016241862,-0.016416445,-0.016548878,-0.016475455,-0.016405742,-0.015567737,-0.014190101,-0.012373151,-0.010370329,-0.008131459,-0.006729419,-0.005667607,-0.004883919,-0.004841328,-0.005403019,-0.005343759,-0.005377974,-0.00548823,-0.004889709,-0.003884973,-0.003149113,-0.002975268,-0.00283163,-0.00322658,-0.003546589,-0.004233582,-0.004448617,-0.004706967,-0.007400356,-0.010104064,-0.01230257,-0.014430498,-0.016499501,-0.015348355,-0.013974229,-0.012845464,-0.012688459,-0.012552231,-0.013719074,-0.014404172,-0.014611632,-0.013401283,-0.011807386,-0.007417753,-0.003321279,0.000363954,0.004908491,0.010151584,0.013223831,0.016746553,0.02106351,0.024571507,0.027588073,0.031313637,0.034419301,0.037016545,0.038172954,0.038237253,0.038094387,0.037783779,0.036482515,0.036080763,0.035476154,0.034107081,0.03237083,0.030934259,0.029317076,0.028236195,0.027850758,0.024612491,0.01964433,0.015153308,0.009684456,0.003336172]}
df = pd.DataFrame(data, columns=['col1'])
lim = float(-0.1)
s = df['col1'].lt(lim)
out = s.where(s, -1).cumsum()
df['sol'] = out - out.where((out < 0) & (~s)).ffill().fillna(0)
print(df)
对我来说，这里的关键问题是控制输出不低于零。考虑到这一点，我们可以屏蔽负输出，并进行相应调整：
# a little longer data for corner case
df = pd.DataFrame({'col1': [1, 2, 1, 0, 0, 0, 0, 3, 2, 2, 0, 0,0,0,0,2,3,4]})
s = df.col1.gt(0)
out = s.where(s,-1).cumsum()
df['continuous'] = out - out.where((out<0)&(~s)).ffill().fillna(0)
输出：
col1 continuous
0 1 1
1 2 2
2 1 3
3 0 2
4 0 1
5 0 0
6 0 0
7 3 1
8 2 2
9 2 3
10 0 2
11 0 1
12 0 0
13 0 0
14 0 0
15 2 1
16 3 2
17 4 3
您可以在布尔值上使用cumsum函数：
每当col1不为零时给我一个+1：
(df.col1 != 0 ).cumsum()
每当col1为零时给我一个-1：
- (df.col1 == 0 ).cumsum()
那就把它们加在一起吧！
df['continuous'] = (df.col1 != 0 ).cumsum() - (df.col1 == 0 ).cumsum()
然而，这并不能解决你提到的低于零的标准