按容差变化按列比较不同的熊猫数据帧

时间：2020-01-14 12:32:30

标签： python pandas dataframe

我有3个这样的熊猫数据框：

#0
                 A     C     G     T          uA          uC          uG          uT     cmA     cmC     cmG     cmT
    seq_1_0   47.0  47.0  54.0  52.0  100.978723  100.957447  100.370370  99.788462   5147.0  5144.0  5055.0  4968.0
    seq_1_50  47.0  47.0  54.0  52.0  101.829787  101.680851  99.092593   99.692308   5279.0  5256.0  4864.0  4953.0
    seq_2_0   47.0  47.0  54.0  52.0  100.978723  100.957447  100.370370  99.788462   5147.0  5144.0  5055.0  4968.0
    seq_2_50  47.0  47.0  54.0  52.0  101.468085  101.425532  99.000000   100.346154  5223.0  5216.0  4850.0  5052.0
    seq_3_0   47.0  47.0  54.0  52.0  100.212766  99.680851   100.870370  101.115385  5030.0  4952.0  5131.0  5169.0
    seq_3_50  46.0  47.0  53.0  54.0  100.173913  100.978723  100.924528  99.944444   5026.0  5148.0  5139.0  4990.0
    seq_4_0   45.0  47.0  54.0  54.0  99.044444   99.000000   101.407407  102.111111  4856.0  4851.0  5214.0  5323.0
    seq_4_50  47.0  47.0  53.0  53.0  101.872340  104.382979  97.849057   98.490566   5285.0  5686.0  4684.0  4776.0
    seq_5_0   54.0  34.0  37.0  75.0  90.462963   91.647059   90.756757   116.546667  3700.0  3848.0  3737.0  7915.0
    seq_5_50  48.0  33.0  37.0  82.0  94.937500   113.636364  113.162162  92.756098   4277.0  7337.0  7245.0  3990.0
    seq_6_0   60.0  50.0  48.0  42.0  98.500000   93.900000   106.125000  104.785714  4777.0  4139.0  5976.0  5752.0
    seq_6_50  59.0  46.0  52.0  43.0  98.338983   98.826087   102.615385  102.697674  4754.0  4825.0  5402.0  5415.0
#1
                 A     C     G     T          uA          uC          uG          uT     cmA     cmC     cmG     cmT
    seq_1_0   47.0  47.0  54.0  52.0  100.978723  100.957447  100.370370  99.788462   5147.0  5144.0  5055.0  4968.0
    seq_1_50  47.0  47.0  54.0  52.0  101.829787  101.680851  99.092593   99.692308   5279.0  5256.0  4864.0  4953.0
    seq_2_0   47.0  47.0  54.0  52.0  100.978723  100.957447  100.370370  99.788462   5147.0  5144.0  5055.0  4968.0
    seq_2_50  47.0  47.0  54.0  52.0  101.468085  101.425532  99.000000   100.346154  5223.0  5216.0  4850.0  5052.0
    seq_3_0   47.0  47.0  54.0  52.0  100.212766  99.680851   100.870370  101.115385  5030.0  4952.0  5131.0  5169.0
    seq_3_50  46.0  47.0  53.0  54.0  100.173913  100.978723  100.924528  99.944444   5026.0  5148.0  5139.0  4990.0
    seq_4_0   45.0  47.0  54.0  54.0  99.044444   99.000000   101.407407  102.111111  4856.0  4851.0  5214.0  5323.0
    seq_4_50  47.0  47.0  53.0  53.0  101.872340  104.382979  97.849057   98.490566   5285.0  5686.0  4684.0  4776.0
    seq_5_0   54.0  34.0  37.0  75.0  90.462963   91.647059   90.756757   116.546667  3700.0  3848.0  3737.0  7915.0
    seq_5_50  48.0  33.0  37.0  82.0  94.937500   113.636364  113.162162  92.756098   4277.0  7337.0  7245.0  3990.0
#2
                 A     C     G     T          uA          uC          uG          uT     cmA     cmC     cmG     cmT
    seq_1_0   48.0  48.0  53.0  51.0  100.291667  99.208333   101.943396  100.411765  5042.0  4882.0  5297.0  5062.0
    seq_1_50  48.0  47.0  54.0  51.0  100.083333  101.680851  99.092593   101.294118  5012.0  5256.0  4864.0  5196.0
    seq_2_0   47.0  47.0  54.0  52.0  100.978723  100.957447  100.370370  99.788462   5147.0  5144.0  5055.0  4968.0
    seq_2_50  47.0  47.0  54.0  52.0  101.468085  101.425532  99.000000   100.346154  5223.0  5216.0  4850.0  5052.0
    seq_3_0   50.0  47.0  53.0  50.0  98.980000   99.680851   101.490566  101.740000  4847.0  4952.0  5226.0  5265.0
    seq_3_50  49.0  47.0  52.0  52.0  95.857143   100.978723  102.519231  102.423077  4403.0  5148.0  5387.0  5371.0

我想将第一个数据帧（＃0）的所有列与其他两个数据帧（＃1和＃2）进行比较，以识别哪个索引具有不同的列值（例如，索引seq_6_0和seq_6_50存在于＃0数据帧中，而其他两个数据帧中不存在。

但是我也想对每列进行公差变化，以将不同数据帧的列视为相等，例如：

数据帧＃0的索引seq_1_0具有以下值：

A     C     G     T          uA          uC          uG          uT     cmA     cmC     cmG     cmT
47.0  47.0  54.0  52.0  100.978723  100.957447  100.370370  99.788462   5147.0  5144.0  5055.0  4968.0

daframe＃2的索引seq_1_0具有：

A     C     G     T          uA          uC          uG          uT     cmA     cmC     cmG     cmT
48.0  48.0  53.0  51.0  100.291667  99.208333   101.943396  100.411765  5042.0  4882.0  5297.0  5062.0

所以我想为每列输入差异容差值，例如对于列["A","C","T","G"]，我需要比较值之间的公差值为90％，但对于其他列，我需要比较值之间的百分比不同。

我有可用于此功能的任何熊猫功能吗？

最好

1 个答案:

答案 0 :(得分：2)

使用np.isclose，可让您精确控制比较的绝对和相对公差。

我假设您只想比较行和两个数据框中都存在的标签。存在于一个而不存在于另一行中的行将被忽略。另外，由于您对A，C，G，T使用相对标准，所以services: traefik: container_name: traefik image: traefik ports: - 40080:40080 - 48080:8080 - 40443:40443 restart: unless-stopped volumes: - /etc/docker/container-data/traefik:/etc/traefik - /var/run/docker.sock:/var/run/docker.sock - /etc/localtime:/etc/localtime:ro environment: - OVH_ENDPOINT=ovh-eu - OVH_APPLICATION_KEY=G... - OVH_APPLICATION_SECRET=Z... - OVH_CONSUMER_KEY=B... version: "3" 与compare(df0,df1)不同。假定第二个参数是参考值。这与compare(df1,df0)的工作方式是一致的。

np.isclose

为了轻松直观地显示结果：

def compare(dfa, dfb):
    s = pd.Series(['A','C','G','T'])
    tmp = dfa.join(dfb, how='inner', lsuffix='_a', rsuffix='_b')

    # The A, C, G, T columns: within 90% of dfb
    lhs = tmp[s + '_a'].values
    rhs = tmp[s + '_b'].values
    compare1 = np.isclose(lhs, rhs, atol=0, rtol=0.9)

    # The uA, uC, uG, uT columns: within 1e-5
    lhs = tmp['u' + s + '_a'].values
    rhs = tmp['u' + s + '_b'].values
    compare2 = np.isclose(lhs, rhs, atol=1e-5, rtol=0)

    # The cmA, cmC, cmG, cmT columns: within 1e-3
    lhs = tmp['cm' + s + '_a'].values
    rhs = tmp['cm' + s + '_b'].values
    compare3 = np.isclose(lhs, rhs, atol=1e-3, rtol=0)

    # Assemble the result
    data = np.concatenate([compare1, compare2, compare3], axis=1)
    cols = pd.concat([s, 'u'+s, 'cm'+s])    
    result = pd.DataFrame(data, columns=cols, index=tmp.index)

    return result

compare(df0, df2)
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