 Matlab: Comparing two vectors with different length and different values?
By : Mr.xiong
Date : March 29 2020, 07:55 AM
help you fix your problem It sounds like what you are trying to do is have an ismember function for use on real valued data. That is, check for each value B(i) in your vector B whether B(i) is within the tolerance threshold T of at least one value in your vector A code :
tf = false(1, length(b)); %//the result vector, true if that element of b is in a
t = 0.01; %// the tolerance threshold
for i = 1:length(b)
%// is the absolute difference between the
%//element of a and b less that the threshold?
matches = abs(a - b(i)) < t;
%// if b(i) matches any of the elements of a
tf(i) = any(matches);
end
t = 0.01;
tf = arrayfun(@(bi) any(abs(a - bi) < t), b);
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 Comparing node values and writing the value of another node based on that comparison with XSLT
By : harish
Date : March 29 2020, 07:55 AM
I wish this help you The transform I have written so far is just not right and apparently oversimplified. code :
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="part" match="strow" use="stentry[@props='part-number']" />
<!-- identity transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="stentry[@props='part-name']">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:variable name="matching-part" select="key('part', ../stentry[@props='part-number'], document('otherfile.xml'))" />
<xsl:choose>
<xsl:when test="$matching-part">
<xsl:value-of select="$matching-part/stentry[@props='part-name']"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="."/>
</xsl:otherwise>
</xsl:choose>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
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 Comparing the values in the same position of two equal length data.frames in R
By : Cluchi97
Date : March 29 2020, 07:55 AM
seems to work fine This is what tapply() is designed for. Here's a demo using randomized data (so the lat/lon positions and state names won't correspond to reality): code :
states <- c('Bayern','Saarland','BadenW','SAnhalt','Sachsen','MVorpommern','NRWestfalen','Berlin','Hamburg','Bremen','SHolstein','Niedersachsen','Hessen','Thueringen','RPfalz','Brandenburg');
lats <- seq(51.30,by=-0.05,len=281);
lons <- seq(8.70,by=0.05,len=321);
set.seed(1);
Laender <- as.data.frame(matrix(sample(states,length(lats)*length(lons),replace=T),length(lats),dimnames=list(sprintf('%.2f',lats),sprintf('%.2f',lons))));
df <- as.data.frame(matrix(pmax(0,round(rnorm(length(lats)*length(lons),5,20))),length(lats),dimnames=list(sprintf('%.2f',lats),sprintf('%.2f',lons))));
Laender[1:6,1:6];
## 8.70 8.75 8.80 8.85 8.90 8.95
## 51.30 Sachsen Hamburg MVorpommern Berlin Hessen Hessen
## 51.25 MVorpommern RPfalz Niedersachsen Berlin RPfalz Berlin
## 51.20 Bremen MVorpommern RPfalz NRWestfalen Sachsen Bayern
## 51.15 RPfalz Bayern Niedersachsen Bayern Berlin BadenW
## 51.10 SAnhalt BadenW SAnhalt SHolstein Sachsen BadenW
## 51.05 RPfalz MVorpommern SHolstein Niedersachsen MVorpommern Saarland
df[1:6,1:6];
## 8.70 8.75 8.80 8.85 8.90 8.95
## 51.30 16 1 0 14 0 5
## 51.25 24 0 11 0 27 0
## 51.20 15 0 0 13 0 25
## 51.15 0 21 0 21 2 0
## 51.10 30 0 0 15 0 0
## 51.05 0 0 0 31 0 0
tapply(as.matrix(df),as.matrix(Laender),mean);
## BadenW Bayern Berlin Brandenburg Bremen
## 10.35327 10.30455 10.80498 11.09401 10.57423
## Hamburg Hessen MVorpommern Niedersachsen NRWestfalen
## 11.05088 10.55788 10.66969 10.90239 11.09304
## RPfalz Saarland Sachsen SAnhalt SHolstein
## 10.54924 10.48975 10.87170 10.49251 10.51719
## Thueringen
## 10.52608
Laender <- as.matrix(read.csv('path/file.csv',row.names=1,check.names=F));
df <- as.matrix(read.csv('path/file1.csv',row.names=1,check.names=F));
mean(df[Laender=='BadenW']);
## [1] 10.35327
sapply(unique(c(as.matrix(Laender))),function(s) mean(df[Laender==s]));
## Sachsen MVorpommern Bremen RPfalz SAnhalt
## 10.87170 10.66969 10.57423 10.54924 10.49251
## Brandenburg SHolstein Bayern BadenW NRWestfalen
## 11.09401 10.51719 10.30455 10.35327 11.09304
## Hessen Berlin Niedersachsen Thueringen Saarland
## 10.55788 10.80498 10.90239 10.52608 10.48975
## Hamburg
## 11.05088
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 copy values from one dataframe to another dataframe(different length) by comparing row values in python
By : Hải Nguyên Trần
Date : March 29 2020, 07:55 AM
hop of those help? As @Merlin has already mentioned joining (using pd.merge() method) should be pretty straightforward: code :
In [126]: pd.merge(daily.drop('Val', 1), monthly.drop('Date', 1), on=['Year','Month'])
Out[126]:
Date Year Month val Val
0 2016-01-01 2016 1 0 0.00
1 2016-01-02 2016 1 0 0.00
2 2016-01-03 2016 1 0 0.00
3 2016-01-04 2016 1 0 0.00
4 2016-01-05 2016 1 0 0.00
5 2016-01-06 2016 1 0 0.00
6 2016-01-07 2016 1 0 0.00
7 2016-01-08 2016 1 0 0.00
8 2016-01-09 2016 1 0 0.00
9 2016-01-10 2016 1 0 0.00
10 2016-01-11 2016 1 0 0.00
11 2016-01-12 2016 1 0 0.00
12 2016-01-13 2016 1 0 0.00
13 2016-01-14 2016 1 0 0.00
14 2016-01-15 2016 1 0 0.00
.. ... ... ... ... ...
137 2016-05-17 2016 5 0 0.28
138 2016-05-18 2016 5 0 0.28
139 2016-05-19 2016 5 0 0.28
140 2016-05-20 2016 5 0 0.28
141 2016-05-21 2016 5 0 0.28
142 2016-05-22 2016 5 0 0.28
143 2016-05-23 2016 5 0 0.28
144 2016-05-24 2016 5 0 0.28
145 2016-05-25 2016 5 0 0.28
146 2016-05-26 2016 5 0 0.28
147 2016-05-27 2016 5 0 0.28
148 2016-05-28 2016 5 0 0.28
149 2016-05-29 2016 5 0 0.28
150 2016-05-30 2016 5 0 0.28
151 2016-05-31 2016 5 0 0.28
[152 rows x 5 columns]
In [108]: df
Out[108]:
Val Date Year Month
0 0.00 2016-01-31 2016 1
1 0.10 2016-02-28 2016 2
2 0.07 2016-03-31 2016 3
3 0.01 2016-04-30 2016 4
4 0.28 2016-05-31 2016 5
In [117]: df.set_index('Date').resample('MS').mean().append(x.iloc[[-1]]).resample('D').pad().reset_index()
Out[117]:
Date Val Year Month
0 2016-01-01 0.00 2016 1
1 2016-01-02 0.00 2016 1
2 2016-01-03 0.00 2016 1
3 2016-01-04 0.00 2016 1
4 2016-01-05 0.00 2016 1
5 2016-01-06 0.00 2016 1
6 2016-01-07 0.00 2016 1
7 2016-01-08 0.00 2016 1
8 2016-01-09 0.00 2016 1
9 2016-01-10 0.00 2016 1
10 2016-01-11 0.00 2016 1
11 2016-01-12 0.00 2016 1
12 2016-01-13 0.00 2016 1
13 2016-01-14 0.00 2016 1
14 2016-01-15 0.00 2016 1
.. ... ... ... ...
137 2016-05-17 0.28 2016 5
138 2016-05-18 0.28 2016 5
139 2016-05-19 0.28 2016 5
140 2016-05-20 0.28 2016 5
141 2016-05-21 0.28 2016 5
142 2016-05-22 0.28 2016 5
143 2016-05-23 0.28 2016 5
144 2016-05-24 0.28 2016 5
145 2016-05-25 0.28 2016 5
146 2016-05-26 0.28 2016 5
147 2016-05-27 0.28 2016 5
148 2016-05-28 0.28 2016 5
149 2016-05-29 0.28 2016 5
150 2016-05-30 0.28 2016 5
151 2016-05-31 0.28 2016 5
[152 rows x 4 columns]
In [112]: df.set_index('Date').resample('MS').mean()
Out[112]:
Val Year Month
Date
2016-01-01 0.00 2016 1
2016-02-01 0.10 2016 2
2016-03-01 0.07 2016 3
2016-04-01 0.01 2016 4
2016-05-01 0.28 2016 5
In [113]: df.set_index('Date').resample('MS').mean().append(x.iloc[[-1]])
Out[113]:
Val Year Month
Date
2016-01-01 0.00 2016 1
2016-02-01 0.10 2016 2
2016-03-01 0.07 2016 3
2016-04-01 0.01 2016 4
2016-05-01 0.28 2016 5
2016-05-31 0.28 2016 5
|
 XSLT Filter comparing two values in same sub node
By : Joe Delaney
Date : March 29 2020, 07:55 AM
help you fix your problem To remove Film elements where FilmPostDate and FilmActiveDate you can actually nest the conditions in the match attribute code :
<xsl:template
match="m:Film[m:FilmPostings/m:FilmPosting/m:FilmPostingDates[m:FilmPostDate = m:FilmActiveDate]]" />
<xsl:template
match="m:Film[not(m:FilmPostings/m:FilmPosting/m:FilmPostingDates[m:FilmPostDate != m:FilmActiveDate])]" />
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