问题
I am trying to take the following R statement and convert it to Python using NumPy:
1 + apply(tmp,1,function(x) length(which(x[1:k] < x[k+1])))
Is there a Python equivalent to which()? Here, x is row in matrix tmp, and k corresponds to the number of columns in another matrix.
Previously, I tried the following Python code, and received a Value Error (operands could not be broadcast together with shapes):
for row in tmp:
print np.where(tmp[tmp[:,range(k)] < tmp[:,k]])
回答1:
The Python code below answers my question:
np.array([1 + np.sum(row[range(k)] < row[k]) for row in tmp])
Here tmp is a 2d array, and k is a variable which was set for column comparison.
Thanks to https://stackoverflow.com/users/601095/doboy for inspiring me with the answer!
回答2:
>>> which = lambda lst:list(np.where(lst)[0]) Example: >>> lst = map(lambda x:x<5, range(10)) >>> lst [True, True, True, True, True, False, False, False, False, False] >>> which(lst) [0, 1, 2, 3, 4]
回答3:
From http://effbot.org/zone/python-list.htm:
To get the index for all matching items, you can use a loop, and pass in a start index:
i = -1
try:
while 1:
i = L.index(value, i+1)
print "match at", i
except ValueError:
pass
来源:https://stackoverflow.com/questions/12207014/python-equivalent-of-which-in-r