# pandas行數 - python dataframe取值

## 熊貓dataframe:如何計算一個二進制列中的1行的數量? (2)

``````import pandas as pd
import numpy as np

df = pd.DataFrame({"first_column": [0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0]})

>>> df
first_column
0              0
1              0
2              0
3              1
4              1
5              1
6              0
7              0
8              1
9              1
10             0
11             0
12             0
13             0
14             1
15             1
16             1
17             1
18             1
19             0
20             0``````

`first_column`是一個0和1的二進制列。 有連續的“集群”，它們總是以至少兩對的形式出現。

``````>>> df
first_column    counts
0              0        0
1              0        0
2              0        0
3              1        3
4              1        3
5              1        3
6              0        0
7              0        0
8              1        2
9              1        2
10             0        0
11             0        0
12             0        0
13             0        0
14             1        5
15             1        5
16             1        5
17             1        5
18             1        5
19             0        0
20             0        0``````

``````def cumsum_bincount(a):
# Append 0 & look for a [0,1] pattern. Form a binned array based off 1s groups
ids = a*(np.diff(np.r_[0,a])==1).cumsum()

# Get the bincount, index into the count with ids and finally mask out 0s
return a*np.bincount(ids)[ids]``````

``````In [88]: df['counts'] = cumsum_bincount(df.first_column.values)

In [89]: df
Out[89]:
first_column  counts
0              0       0
1              0       0
2              0       0
3              1       3
4              1       3
5              1       3
6              0       0
7              0       0
8              1       2
9              1       2
10             0       0
11             0       0
12             0       0
13             0       0
14             1       5
15             1       5
16             1       5
17             1       5
18             1       5
19             0       0
20             0       0``````

``````In [101]: df.first_column.values[:5] = 1

In [102]: df['counts'] = cumsum_bincount(df.first_column.values)

In [103]: df
Out[103]:
first_column  counts
0              1       6
1              1       6
2              1       6
3              1       6
4              1       6
5              1       6
6              0       0
7              0       0
8              1       2
9              1       2
10             0       0
11             0       0
12             0       0
13             0       0
14             1       5
15             1       5
16             1       5
17             1       5
18             1       5
19             0       0
20             0       0``````

``````# Relevant column -> grouper needs to be 1-Dimensional
col_vals = df['first_column']

# Group by sequence of consecutive values and value in the sequence.
grouped = df.groupby(((col_vals!=col_vals.shift(1)).cumsum(), col_vals))

# Get the length of consecutive values if they are different from zero, else zero
df['counts'] = grouped['first_column'].transform(lambda group: len(group))\
.where(col_vals!=0, 0)``````

``````for key, group in grouped:
print key, group

(1, 0)    first_column
0             0
1             0
2             0
(2, 1)    first_column
3             1
4             1
5             1
(3, 0)    first_column
6             0
7             0
(4, 1)    first_column
8             1
9             1
(5, 0)     first_column
10             0
11             0
12             0
13             0
(6, 1)     first_column
14             1
15             1
16             1
17             1
18             1
(7, 0)     first_column
19             0
20             0``````