Question

我有一个类似的数据框：

对于每个ID，我需要连续出现0次的最大次数。在这种情况下，由于0连续三次显示ID 111，连续两次显示222，因此所需的输出应为：

ID   count_max_0
111    3
222    2

value_counts没有做我想要的事情，因为它计算了列中的所有值。

我该怎么做？

Answer 1

您可以使用

iszero = (df['value']==0)
df['group'] = (iszero.diff()==1).cumsum()

分配每行的组号：

In [115]: df
Out[115]: 
     ID  value  group
0   111      1      0
1   111      0      1
2   111      1      2
3   111      0      3
4   111      0      3
5   111      0      3
6   111      1      4
7   222      1      4
8   222      0      5
9   222      0      5
10  222      1      6

现在，您可以按ID和group分组，以获得所需的值计数：

import pandas as pd

df = pd.DataFrame({'ID': [111, 111, 111, 111, 111, 111, 111, 222, 222, 222, 222],
 'value': [1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1]})
iszero = (df['value']==0)
df['group'] = (iszero.diff()==1).cumsum()

counts = (df.loc[iszero]             # restrict to rows which have 0 value
          .groupby('ID')['group']    # group by ID, inspect the group column
          .value_counts()            # count the number of 0s for each (ID, group)
          .groupby(level='ID')       # group by ID only
          .first())                  # select the first (and highest) value count

print(counts)

产量

ID
111    3
222    2
Name: group, dtype: int64

Answer 2

这应该有效：

import numpy as np

# load data etc
...

def get_count_max_0(df):
    """
    Computes the max length of a sequence of zeroes
    broken by ones.
    """
    values = np.array(df['value'].tolist())
    # compute change points where 0 -> 1
    cps_1 = np.where(
        (values[1:] != values[:-1]) &
        (values[1:] == 1)
    )[0]
    # compute change points where 1 -> 0
    cps_0 = np.where(
        (values[1:] != values[:-1]) &
        (values[1:] == 0)
    )[0]

    # find lengths of zero chains
    deltas = cps_1 - cps_0
    # get index of max length
    idx = np.where(deltas == deltas.max())[0][0]
    # return max length
    return deltas[idx]

# group by ID, apply get_count_max_0 to each group and 
# convert resulting series back to data frame to match your expected output.
max_counts = df.groupby("ID").apply(get_count_max_0).to_frame("count_max_0")

print(max_counts)

输出结果为：

     count_max_0
ID              
111            3
222            2

Answer 3

aggregations = {
    'value': {
        'total': 'sum'
    }
}
dftwo = df.groupby('ID').agg(aggregations)

dataframe

熊猫：在列中计算一些值

3 个答案: