René
Reputation: 79
Select n rows after specific number
I work with a data.frame like this:
Country Date balance_of_payment business_confidence_indicator consumer_confidence_indicator CPI Crisis_IMF
1 Australia 1980-01-01 -0.87 100.215 99.780 25.4 0
2 Australia 1980-04-01 -1.62 100.061 99.746 26.2 0
3 Australia 1980-07-01 -3.70 100.599 100.049 26.6 0
4 Australia 1980-10-01 -3.13 100.597 100.735 27.2 0
5 Australia 1981-01-01 -2.73 101.149 101.016 27.8 0
6 Australia 1981-04-01 -4.11 100.936 100.150 28.4 0
I want to create a summary statistic with describe(dataset)from the HmiscPackage.
I need to differentiate between the timespans n-quarters before Crisis_IMF is 1, the time in which Crisis_IMF is 1 and the state n-quaters after Crisis_IMF is 1.
To select the time in which Crisis_IMF is 1, I did describe(dataset[dataset$Crisis_IMF==1,"balance_of_payment"]).
But I do not know how to make the command over the timespan of n-quarters (e.g. 8) after the event.
Edit:
dataset$Crisis_IMF
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[60] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[119] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[178] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[237] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[296] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[355] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[591] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[650] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[768] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[827] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[886] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[1535] 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
[1594] 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1653] 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[1771] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0
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[2420] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[2656] 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[3423] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3482] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3541] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3600] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3659] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3718] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3777] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3836] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3895] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
[3954] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1
[4013] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4072] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4131] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4190] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4249] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4308] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Edit2; further information on the dataset:
Country Date balance_of_payment Crisis_IMF
1 Australia 1980-01-01 -0.87 0
2 Australia 1980-04-01 -1.62 0
3 Australia 1980-07-01 -3.70 0
4 Australia 1980-10-01 -3.13 0
5 Australia 1981-01-01 -2.73 0
6 Australia 1981-04-01 -4.11 0
7 Australia 1981-07-01 -3.98 0
8 Australia 1981-10-01 -5.27 0
9 Australia 1982-01-01 -5.31 0
10 Australia 1982-04-01 -4.67 0
11 Australia 1982-07-01 -3.30 0
12 Australia 1982-10-01 -3.24 0
13 Australia 1983-01-01 -3.45 0
14 Australia 1983-04-01 -2.86 0
15 Australia 1983-07-01 -3.58 0
...
137 Australia 2014-01-01 -2.18 0
138 Australia 2014-04-01 -3.44 0
139 Australia 2014-07-01 -3.04 0
140 Australia 2014-10-01 -2.39 0
141 Austria 1980-01-01 -3.97 0
142 Austria 1980-04-01 -3.89 0
143 Austria 1980-07-01 -1.84 0
144 Austria 1980-10-01 -1.60 0
145 Austria 1981-01-01 -2.74 0
146 Austria 1981-04-01 -2.88 0
147 Austria 1981-07-01 -2.83 0
148 Austria 1981-10-01 -2.06 0
149 Austria 1982-01-01 -0.63 0
150 Austria 1982-04-01 0.61 0
151 Austria 1982-07-01 2.42 0
152 Austria 1982-10-01 2.70 0
There can be more then one crisis period for one country. That e.g. in Australia are Crisis from 1990-01-01 to 1991-04-01 and 2002-01-01 to 2005-01-01. I want to create 3 different describe commands, which show the behaviour of the variable in the above mentioned states.
Upvotes: 3
Views: 271
Answers (1)
bgoldst
Reputation: 35314
You haven't provided your full data, so I have to guess that your Crisis_IMF column has an unbroken sequence of zeroes (before the crisis), followed by an unbroken sequence of ones (during which the IMF crisis was considered to be in effect), finally followed by an unbroken sequence of zeroes (after the crisis).
Below I've synthesized my own data for testing. I only synthesized columns Crisis_IMF and balance_of_payment, because those appear to be the only columns relevant to your problem. I used 30 rows, with the first 10 before, the next 10 during, and the last 10 after the crisis. I used sort of a random parabolic arc for the balance_of_payment, but that was entirely random.
library('Hmisc');
set.seed(1);
N <- 30;
df <- data.frame(balance_of_payment=-5+2*seq(-1.5,1.5,len=N)^2+rnorm(N,0,0.2), Crisis_IMF=c(rep(0,N/3),rep(1,N/3),rep(0,N/3)) );
df;
## balance_of_payment Crisis_IMF
## 1 -0.6252908 0
## 2 -1.0625579 0
## 3 -1.8228927 0
## 4 -1.8503850 0
## 5 -2.5744076 0
## 6 -3.2324647 0
## 7 -3.3561408 0
## 8 -3.6484112 0
## 9 -3.9805631 0
## 10 -4.4136342 0
## 11 -4.2642312 1
## 12 -4.6598435 1
## 13 -4.9904788 1
## 14 -5.3947830 1
## 15 -4.7696630 1
## 16 -5.0036359 1
## 17 -4.9550811 1
## 18 -4.6774634 1
## 19 -4.5735679 1
## 20 -4.4478071 1
## 21 -4.1687610 0
## 22 -3.9392921 0
## 23 -3.7811631 0
## 24 -3.8514970 0
## 25 -2.9444058 0
## 26 -2.6515349 0
## 27 -2.2006002 0
## 28 -1.9499174 0
## 29 -1.1949166 0
## 30 -0.4164117 0
crisisRange <- range(which(df$Crisis_IMF==1));
crisisRange;
## [1] 11 20
df$Off_Crisis <- c((1-crisisRange[1]):-1,rep(0,diff(crisisRange)+1),1:(nrow(df)-crisisRange[2]));
df;
## balance_of_payment Crisis_IMF Off_Crisis
## 1 -0.6252908 0 -10
## 2 -1.0625579 0 -9
## 3 -1.8228927 0 -8
## 4 -1.8503850 0 -7
## 5 -2.5744076 0 -6
## 6 -3.2324647 0 -5
## 7 -3.3561408 0 -4
## 8 -3.6484112 0 -3
## 9 -3.9805631 0 -2
## 10 -4.4136342 0 -1
## 11 -4.2642312 1 0
## 12 -4.6598435 1 0
## 13 -4.9904788 1 0
## 14 -5.3947830 1 0
## 15 -4.7696630 1 0
## 16 -5.0036359 1 0
## 17 -4.9550811 1 0
## 18 -4.6774634 1 0
## 19 -4.5735679 1 0
## 20 -4.4478071 1 0
## 21 -4.1687610 0 1
## 22 -3.9392921 0 2
## 23 -3.7811631 0 3
## 24 -3.8514970 0 4
## 25 -2.9444058 0 5
## 26 -2.6515349 0 6
## 27 -2.2006002 0 7
## 28 -1.9499174 0 8
## 29 -1.1949166 0 9
## 30 -0.4164117 0 10
n <- 8;
describe(df[df$Off_Crisis>=-n&df$Off_Crisis<=-1,'balance_of_payment']);
## df[df$Off_Crisis >= -n & df$Off_Crisis <= -1, "balance_of_payment"]
## n missing unique Info Mean
## 8 0 8 1 -3.11
##
## -4.41363415781177 (1, 12%), -3.98056311135777 (1, 12%), -3.64841115885525 (1, 12%), -3.35614082447269 (1, 12%), -3.23246466374394 (1, 12%), -2.57440760140387 (1, 12%), -1.85038498107066 (1, 12%), -1.82289266659616 (1, 12%)
describe(df[df$Off_Crisis==0,'balance_of_payment']);
## df[df$Off_Crisis == 0, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 10 0 10 1 -4.774 -5.219 -5.043 -4.982 -4.724 -4.595 -4.429 -4.347
##
## -5.39478302143074 (1, 10%), -5.00363594891363 (1, 10%), -4.99047879387293 (1, 10%), -4.95508109661503 (1, 10%), -4.76966304348196 (1, 10%), -4.67746343562751 (1, 10%), -4.65984348113626 (1, 10%), -4.57356788939893 (1, 10%), -4.44780713171369 (1, 10%), -4.26423116226702 (1, 10%)
describe(df[df$Off_Crisis>=1&df$Off_Crisis<=n,'balance_of_payment']);
## df[df$Off_Crisis >= 1 & df$Off_Crisis <= n, "balance_of_payment"]
## n missing unique Info Mean
## 8 0 8 1 -3.186
##
## -4.16876100605885 (1, 12%), -3.93929212154225 (1, 12%), -3.85149697413106 (1, 12%), -3.78116310320806 (1, 12%), -2.94440583734139 (1, 12%), -2.65153490367274 (1, 12%), -2.20060024283928 (1, 12%), -1.949917420894 (1, 12%)
The solution works by first computing the range of indexes during which the crisis was in effect as crisisRange. Then it appends to the data.frame a new column Off_Crisis which captures how many quarters offset from the crisis the row is, using negative numbers for before and positive numbers for after, and assuming each row represents exactly one quarter.
The describe() calls can then be made by subsetting on the Off_Crisis column, getting just the quarters offset from the crisis that you want for each call.
Edit: Whew! That was tough. Pretty sure I got it though:
library('Hmisc');
set.seed(1);
N <- 60;
df <- data.frame(balance_of_payment=rep(-5+2*seq(-1.5,1.5,len=N/2)^2,2)+rnorm(N,0,0.2), Crisis_IMF=c(rep(0,N/6),rep(1,N/6),rep(0,N/3),rep(1,N/6),rep(0,N/6)) );
df;
## balance_of_payment Crisis_IMF
## 1 -0.6252908 0
## 2 -1.0625579 0
## 3 -1.8228927 0
## 4 -1.8503850 0
## 5 -2.5744076 0
## 6 -3.2324647 0
## 7 -3.3561408 0
## 8 -3.6484112 0
## 9 -3.9805631 0
## 10 -4.4136342 0
## 11 -4.2642312 1
## 12 -4.6598435 1
## 13 -4.9904788 1
## 14 -5.3947830 1
## 15 -4.7696630 1
## 16 -5.0036359 1
## 17 -4.9550811 1
## 18 -4.6774634 1
## 19 -4.5735679 1
## 20 -4.4478071 1
## 21 -4.1687610 0
## 22 -3.9392921 0
## 23 -3.7811631 0
## 24 -3.8514970 0
## 25 -2.9444058 0
## 26 -2.6515349 0
## 27 -2.2006002 0
## 28 -1.9499174 0
## 29 -1.1949166 0
## 30 -0.4164117 0
## 31 -0.2282641 0
## 32 -1.1198441 0
## 33 -1.5782326 0
## 34 -2.1802021 0
## 35 -2.9157211 0
## 36 -3.1513699 0
## 37 -3.5324846 0
## 38 -3.8079388 0
## 39 -3.8757143 0
## 40 -4.1999213 0
## 41 -4.5994921 1
## 42 -4.7884845 1
## 43 -4.7268380 1
## 44 -4.8405104 1
## 45 -5.1324004 1
## 46 -5.1361483 1
## 47 -4.8789267 1
## 48 -4.7125241 1
## 49 -4.7602814 1
## 50 -4.3903659 1
## 51 -4.2729353 0
## 52 -4.2181247 0
## 53 -3.7278522 0
## 54 -3.6794993 0
## 55 -2.7817662 0
## 56 -2.2442292 0
## 57 -2.2428854 0
## 58 -1.8645939 0
## 59 -0.9853426 0
## 60 -0.5270109 0
df$Off_Crisis <- ifelse(df$Crisis_IMF==1,0,with(rle(df$Crisis_IMF),{ mids <- lengths[-c(1,length(lengths))]; c(-lengths[1]:-1,sequence(mids)-rep(rbind(0,mids+1),rbind(ceiling(mids/2),floor(mids/2))),1:lengths[length(lengths)]); }));
df;
## balance_of_payment Crisis_IMF Off_Crisis
## 1 -0.6252908 0 -10
## 2 -1.0625579 0 -9
## 3 -1.8228927 0 -8
## 4 -1.8503850 0 -7
## 5 -2.5744076 0 -6
## 6 -3.2324647 0 -5
## 7 -3.3561408 0 -4
## 8 -3.6484112 0 -3
## 9 -3.9805631 0 -2
## 10 -4.4136342 0 -1
## 11 -4.2642312 1 0
## 12 -4.6598435 1 0
## 13 -4.9904788 1 0
## 14 -5.3947830 1 0
## 15 -4.7696630 1 0
## 16 -5.0036359 1 0
## 17 -4.9550811 1 0
## 18 -4.6774634 1 0
## 19 -4.5735679 1 0
## 20 -4.4478071 1 0
## 21 -4.1687610 0 1
## 22 -3.9392921 0 2
## 23 -3.7811631 0 3
## 24 -3.8514970 0 4
## 25 -2.9444058 0 5
## 26 -2.6515349 0 6
## 27 -2.2006002 0 7
## 28 -1.9499174 0 8
## 29 -1.1949166 0 9
## 30 -0.4164117 0 10
## 31 -0.2282641 0 -10
## 32 -1.1198441 0 -9
## 33 -1.5782326 0 -8
## 34 -2.1802021 0 -7
## 35 -2.9157211 0 -6
## 36 -3.1513699 0 -5
## 37 -3.5324846 0 -4
## 38 -3.8079388 0 -3
## 39 -3.8757143 0 -2
## 40 -4.1999213 0 -1
## 41 -4.5994921 1 0
## 42 -4.7884845 1 0
## 43 -4.7268380 1 0
## 44 -4.8405104 1 0
## 45 -5.1324004 1 0
## 46 -5.1361483 1 0
## 47 -4.8789267 1 0
## 48 -4.7125241 1 0
## 49 -4.7602814 1 0
## 50 -4.3903659 1 0
## 51 -4.2729353 0 1
## 52 -4.2181247 0 2
## 53 -3.7278522 0 3
## 54 -3.6794993 0 4
## 55 -2.7817662 0 5
## 56 -2.2442292 0 6
## 57 -2.2428854 0 7
## 58 -1.8645939 0 8
## 59 -0.9853426 0 9
## 60 -0.5270109 0 10
n <- 8;
describe(df[df$Off_Crisis>=-n&df$Off_Crisis<=-1,'balance_of_payment']);
## df[df$Off_Crisis >= -n & df$Off_Crisis <= -1, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 16 0 16 1 -3.133 -4.253 -4.090 -3.825 -3.294 -2.476 -1.837 -1.762
##
## -4.41363415781177 (1, 6%), -4.19992133068899 (1, 6%), -3.98056311135777 (1, 6%), -3.87571430729169 (1, 6%), -3.80793877922333 (1, 6%), -3.64841115885525 (1, 6%)
## -3.53248462570045 (1, 6%), -3.35614082447269 (1, 6%), -3.23246466374394 (1, 6%), -3.15136989958027 (1, 6%), -2.91572106713267 (1, 6%), -2.57440760140387 (1, 6%)
## -2.1802021496148 (1, 6%), -1.85038498107066 (1, 6%), -1.82289266659616 (1, 6%), -1.57823262180228 (1, 6%)
describe(df[df$Off_Crisis==0,'balance_of_payment']);
## df[df$Off_Crisis == 0, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 20 0 20 1 -4.785 -5.149 -5.133 -4.964 -4.765 -4.645 -4.442 -4.384
##
## lowest : -5.395 -5.136 -5.132 -5.004 -4.990, highest: -4.599 -4.574 -4.448 -4.390 -4.264
describe(df[df$Off_Crisis>=1&df$Off_Crisis<=n,'balance_of_payment']);
## df[df$Off_Crisis >= 1 & df$Off_Crisis <= n, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 16 0 16 1 -3.157 -4.232 -4.193 -3.873 -3.312 -2.244 -2.075 -1.929
##
## -4.27293530430708 (1, 6%), -4.21812466033862 (1, 6%), -4.16876100605885 (1, 6%), -3.93929212154225 (1, 6%), -3.85149697413106 (1, 6%), -3.78116310320806 (1, 6%)
## -3.72785216159621 (1, 6%), -3.67949925417454 (1, 6%), -2.94440583734139 (1, 6%), -2.78176624658013 (1, 6%), -2.65153490367274 (1, 6%), -2.24422917606577 (1, 6%)
## -2.24288543679152 (1, 6%), -2.20060024283928 (1, 6%), -1.949917420894 (1, 6%), -1.86459386937746 (1, 6%)
For this demo I synthesized five periods: 10 rows of non-crisis, 10 rows of crisis (the first), 20 rows of non-crisis, 10 rows of crisis (the second), and 10 rows of non-crisis. The algorithm is the same, namely to compute an Off_Crisis column (which was much more difficult this time!) and then use it to subset the data.frame for each describe() call. Only now, data points from different crises will be combined in the subsets.
Upvotes: 3
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