We can use loc to reorder by passing a list: In [27]: # get a list of columns cols = list(df) # move the column to head of list using index, pop and insert cols.insert(0, cols.pop(cols.index(‘Mid’))) cols Out[27]: [‘Mid’, ‘Net’, ‘Upper’, ‘Lower’, ‘Zsore’] In [28]: # use ix to reorder df = df.loc[:, cols] df … Read more
$numbers = array(1,2,3,4); array_push($numbers, array_shift($numbers)); print_r($numbers); Output Array ( [0] => 2 [1] => 3 [2] => 4 [3] => 1 )
try this $(“textarea”).keydown(function(e){ // Enter was pressed without shift key if (e.keyCode == 13 && !e.shiftKey) { // prevent default behavior e.preventDefault(); } }); update your fiddle to $(“.Post_Description_Text”).keydown(function(e){ if (e.keyCode == 13 && !e.shiftKey) { // prevent default behavior e.preventDefault(); //alert(“ok”); return false; } });
int x = (number >> (8*n)) & 0xff; where n is 0 for the first byte, 1 for the second byte, etc.
You need eq with shift: df[‘match’] = df.col1.eq(df.col1.shift()) print (df) col1 match 0 1 False 1 3 False 2 3 True 3 1 False 4 2 False 5 3 False 6 2 False 7 2 True Or instead eq use ==, but it is a bit slowier in large DataFrame: df[‘match’] = df.col1 == df.col1.shift() … Read more
Logically it does not work and you should reverse your loop: for (int i = position-1; i >= 0; i–) { array[i+1] = array[i]; } Alternatively you can use System.arraycopy(array, 0, array, 1, position);
The test combines the 2 bits in codeword at offset 2 * i and 2 * i + 1 and evaluates the body if they are not both 0. The expression is parsed as: int i; _int8 codeword[64800]; //loop running through codeword if ((codeword[i << 1] << 1) | codeword[(i << 1) | 1]) { … Read more