Here’s one way to get the min and max values: $min = min(array_column($array, ‘Price’)); $max = max(array_column($array, ‘Price’)); To return the nested array for the min and max: $prices = array_column($array, ‘Price’); $min_array = $array[array_search(min($prices), $prices)]; $max_array = $array[array_search(max($prices), $prices)]; You could do each in one line since that looked like what you were trying … Read more
Using C++11/C++0x compile flags, you can auto it = max_element(std::begin(cloud), std::end(cloud)); // C++11 Otherwise, write your own: template <typename T, size_t N> const T* mybegin(const T (&a)[N]) { return a; } template <typename T, size_t N> const T* myend (const T (&a)[N]) { return a+N; } See it live at http://ideone.com/aDkhW: #include <iostream> #include <algorithm> … Read more
In [2]: min(data, key = lambda t: t[1]) Out[2]: (5, 0.01) or: In [3]: import operator In [4]: min(data, key=operator.itemgetter(1)) Out[4]: (5, 0.01)
You can use apply to go through each row apply(df, 1, FUN = min) Where 1 means to apply FUN to each row of df, 2 would mean to apply FUN to columns. To remove missing values, use: apply(df, 1, FUN = min, na.rm = TRUE)
Pairwise maximum, pmax(a, b), will give c(3,4,6). a <- c(3,3,5,NA,1) b <- c(2,4,6,0,NA) pmax(a, b) # [1] 3 4 6 NA NA pmax(a, b, na.rm = TRUE) # [1] 3 4 6 0 1 There is also a pairwise minimum pmin(a, b) # [1] 2 3 5 NA NA pmin(a, b, na.rm = TRUE) # … Read more
Store the items in a list of lists,[word,num] pairs and then apply min on that list of list. Use key parameter of min to specify the which item must be used for comparison of item.: with open(‘abc’) as f: lis = [[line.strip(),next(f).strip()] for line in f] minn = min(lis, key = lambda x: float(x[1])) print … Read more
Here’s a possible solution: public class NumInput { public static void main(String [] args) { int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; Scanner s = new Scanner(System.in); while (true) { System.out.print(“Enter a Value: “); int val = s.nextInt(); if (val == 0) { break; } if (val < min) { min = val; } … Read more
from __future__ import division somelist = [1,12,2,53,23,6,17] max_value = max(somelist) min_value = min(somelist) avg_value = 0 if len(somelist) == 0 else sum(somelist)/len(somelist) If you want to manually find the minimum as a function: somelist = [1,12,2,53,23,6,17] def my_min_function(somelist): min_value = None for value in somelist: if not min_value: min_value = value elif value < min_value: … Read more
You can include your which.max call as the first argument to your subsetting call: df[which.max(df$Temp),]
def clamp(n, minn, maxn): return max(min(maxn, n), minn) or functionally equivalent: clamp = lambda n, minn, maxn: max(min(maxn, n), minn) now, you use: n = clamp(n, 7, 42) or make it perfectly clear: n = minn if n < minn else maxn if n > maxn else n even clearer: def clamp(n, minn, maxn): if … Read more