CODEWITHSUNDEEP
pythonpandas
ubuntu_noob
ubuntu_noob

Reputation: 2365

Find the nearest non zero number in a column of a specific index

I have a pandas dataframe which has a column like 

df = pd.DataFrame({'A':[0,0,15,0,0,0,0,0,0,5]})

    A
0   0
1   0
2   15
3   0
4   0
5   0
6   0
7   0
8   0
9   5

Now based on the index(lets say 5) I want to determine the nearest non zero number in the column(here index 2 and value 15) and return the hops it takes to get there from given index. In the given example it would be +3 as it comes from index 2 to index 5 and if the given index is 7 then the answer would be -2 as index is 9 value 5

Upvotes: 1

Views: 771

Answers (4)

piRSquared
piRSquared

Reputation: 294508

Use numpy.flatnonzero to return back the positions of where the array has non zero values.
Subtract the index you are referencing from to get the direction and distance from those positions.

d = np.flatnonzero(df.A.values) - 5
i = d[np.abs(d).argmin()] + 5

df.iloc[i]

A    15
Name: 2, dtype: int64

Upvotes: 1

andrew_reece
andrew_reece

Reputation: 21274

With a boolean vector of zero/non-zero, you can calculate the distance from a given index to the non-zero values using subtract(). That will give you the "hops", and idxmin() can locate the actual value.

def get_closest(df, target):
    df["not_zero"] = df.ne(0)
    not_zero = pd.Series(df.index[df.not_zero])
    dist = not_zero.subtract(target).abs()
    minidx = not_zero.loc[dist.idxmin()]
    steps = dist.min() if minidx < target else -dist.min()
    print("Closest steps to non-zero:", steps)
    print("Closest non-zero value:", df.A[minidx])

get_closest(df, 5)
# Closest steps to non-zero: 3
# Closest non-zero value: 15

get_closest(df, 7)
# Closest steps to non-zero: -2
# Closest non-zero value: 5

Upvotes: 1

DYZ
DYZ

Reputation: 57105

Start by finding the indexes of all non-zero elements:

nonzeros = df[df.A != 0]).index

Calculate the distances from your row to all of them:

anchor = 5
dists = anchor - nonzeros

Find the smallest distance (by the absolute value):

nhops = min(dists, key=abs)

Altogether, in one line:

hnops = min((anchor - df[df.A != 0].index), key=abs)
#3

The index of the closest non-zero value can be calculated by recombining nhops and anchor:

min_index = anchor - nhops
#2

Upvotes: 2

BENY
BENY

Reputation: 323356

IIUC nonzero with argmin

a=df.A.nonzero()[0]
a[abs(np.argmin(a-5))]
Out[950]: 2

Upvotes: 1

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