Python 如何有效地将数据帧转换为多维numpy数组？

我正在尝试将大熊猫数据帧转换为具有特定结构的numpy数组。熊猫数据帧的形状是（56930255），这里是df

    trial   channel digit   Column7 Column8 Column9 Column10    Column11    Column12    Column13    ... Column249   Column250   Column251   Column252   Column253   Column254   Column255   Column256   Column257   Column258
0   0.0 AF3 0.0 4259.487179 4237.948717 4247.179487 4242.051282 4233.333333 4251.282051 4232.820512 ... 4275.897435 4288.717948 4287.692307 4273.333333 4287.179487 4260.000000 4271.282051 4286.666666 4255.384615 4257.948717
1   0.0 AF4 0.0 4103.076923 4100.512820 4102.564102 4087.692307 4074.358974 4095.897435 4093.846153 ... 4210.256410 4234.358974 4252.820512 4238.461538 4253.333333 4244.615384 4241.538461 4229.743589 4214.871794 4237.948717
2   0.0 T7  0.0 4245.128205 4218.461538 4242.051282 4245.128205 4233.333333 4257.435897 4241.025641 ... 4270.256410 4259.487179 4259.487179 4277.435897 4292.307692 4250.256410 4263.076923 4260.000000 4232.307692 4277.435897
3   0.0 T8  0.0 4208.717948 4188.717948 4204.102564 4198.461538 4179.487179 4203.589743 4194.871794 ... 4215.897435 4236.410256 4222.051282 4191.282051 4231.794871 4226.153846 4218.461538 4216.410256 4202.051282 4220.000000
4   0.0 PZ  0.0 4189.230769 4203.589743 4188.717948 4186.666666 4198.461538 4177.435897 4192.820512 ... 4220.512820 4224.102564 4217.435897 4237.948717 4205.641025 4214.358974 4212.307692 4185.128205 4199.487179 4196.923076
5   1.0 AF3 6.0 4273.846153 4265.641025 4270.256410 4284.615384 4265.128205 4279.487179 4268.717948 ... 4283.076923 4289.743589 4308.717948 4277.948717 4296.410256 4303.076923 4291.794871 4312.820512 4306.153846 4319.487179
6   1.0 AF4 6.0 4233.846153 4252.307692 4262.564102 4244.102564 4217.435897 4262.564102 4239.487179 ... 4315.897435 4318.461538 4320.000000 4302.564102 4338.461538 4339.487179 4330.256410 4358.461538 4331.794871 4331.794871
7   1.0 T7  6.0 4301.025641 4301.025641 4293.333333 4289.230769 4285.128205 4320.000000 4302.051282 ... 4287.692307 4294.871794 4291.794871 4255.384615 4278.461538 4276.923076 4262.564102 4290.256410 4267.179487 4262.564102
8   1.0 T8  6.0 4209.743589 4210.769230 4198.974358 4215.897435 4218.974358 4238.461538 4218.974358 ... 4214.358974 4215.384615 4215.897435 4194.871794 4231.794871 4210.256410 4174.358974 4221.025641 4223.076923 4217.948717
9   1.0 PZ  6.0 4208.717948 4216.410256 4233.846153 4215.384615 4238.461538 4235.897435 4236.410256 ... 4204.102564 4228.205128 4205.641025 4218.461538 4204.615384 4204.615384 4226.153846 4210.256410 4236.410256 4225.128205

我一直在尝试做的是将它转换成一个维度为[numberTrials+1,5252]的numpy数组。这就形成了一个数组，其中每个试验都有自己的5个数组，代表每个“通道”，每个通道包含252个值

这就是我所尝试的

numberTrials = int(df.max(axis=0)[0])
x_train = np.zeros([numberTrials+1,5,252])

i = 0
j = 0
k = 0
l = 0
i_limit = x_train[0][0].size

while k <= numberTrials: 
    while j < 5: #There are 5 channels per trial
        while i < i_limit: #There are 252 values per channel

            x_df = df.iloc[j+l,3:]
            x_array = x_df.values #convert pandas df to array

            x_train[k][j][i] = x_array[i]

            i += 1
        i = 0
        j += 1

    i = 0
    j = 0
    l += 5
    k += 1

numberTrials=int（df.max（axis=0）[0]）
x_序列=np.零（[numberTrials+1,5252]）
i=0
j=0
k=0
l=0
i_limit=x_列[0][0]。大小
而k是部分重写，消除了
i
级迭代：

x_train = np.zeros([numberTrials+1,5,252])

# i_limit = x_train.shape(2)    # instead of x_train[0][0].size

j = 0
k = 0
l = 0

while k <= numberTrials: 
    while j < 5: #There are 5 channels per trial
        x_df = df.iloc[j+l,3:]
        x_train[k,j,:] = x_df.values
        j += 1
    j = 0
    l += 5
    k += 1

对于
j
：

for k in range(numberTrials): 
    for j in range(5):
        x_df = df.iloc[j+l,3:]
        x_train[k,j,:] = x_df.values
    l += 5

并进一步推断（无需测试）：

但由于试验行是5个连续的组，我认为我们可以简单地重塑整个2d
值
数组：

x_df = df.iloc[:, 3:].values
x_train = x_df.reshape(-1,5, x_df.shape[2])

你可以这样做：
df.set_index（['trial'，'channel']）。值
mazing，我真的很惊讶你把所有这些都减少到了两行代码。它工作得很好，我只需将x_df.shape[2]更改为x_df.shape[1]。感谢您展示您的流程，因为它帮助我更好地理解它。
for k in range(numberTrials): 
    x_df = df.iloc[l:l+5, 3:]
    x_train[k,:,:] = x_df.values
    l += 5

x_df = df.iloc[:, 3:].values
x_train = x_df.reshape(-1,5, x_df.shape[2])